Longest increasing subsequence
You are encouraged to solve this task according to the task description, using any language you may know.
Calculate and show here a longest increasing subsequence of the list:
And of the list:
Note that a list may have more than one subsequence that is of the maximum length.
Other tasks related to string operations:
- Metrics
- Counting
- Word frequency
- Letter frequency
- Jewels and stones
- I before E except after C
- Bioinformatics/base count
- Count occurrences of a substring
- Count how many vowels and consonants occur in a string
- Remove/replace
- XXXX redacted
- Conjugate a Latin verb
- Remove vowels from a string
- String interpolation (included)
- Strip block comments
- Strip comments from a string
- Strip a set of characters from a string
- Strip whitespace from a string -- top and tail
- Strip control codes and extended characters from a string
- Anagrams/Derangements/shuffling
- Word wheel
- ABC problem
- Sattolo cycle
- Knuth shuffle
- Ordered words
- Superpermutation minimisation
- Textonyms (using a phone text pad)
- Anagrams
- Anagrams/Deranged anagrams
- Permutations/Derangements
- Find/Search/Determine
- ABC words
- Odd words
- Word ladder
- Semordnilap
- Word search
- Wordiff (game)
- String matching
- Tea cup rim text
- Alternade words
- Changeable words
- State name puzzle
- String comparison
- Unique characters
- Unique characters in each string
- Extract file extension
- Levenshtein distance
- Palindrome detection
- Common list elements
- Longest common suffix
- Longest common prefix
- Compare a list of strings
- Longest common substring
- Find common directory path
- Words from neighbour ones
- Change e letters to i in words
- Non-continuous subsequences
- Longest common subsequence
- Longest palindromic substrings
- Longest increasing subsequence
- Words containing "the" substring
- Sum of the digits of n is substring of n
- Determine if a string is numeric
- Determine if a string is collapsible
- Determine if a string is squeezable
- Determine if a string has all unique characters
- Determine if a string has all the same characters
- Longest substrings without repeating characters
- Find words which contains all the vowels
- Find words which contain the most consonants
- Find words which contains more than 3 vowels
- Find words whose first and last three letters are equal
- Find words with alternating vowels and consonants
- Formatting
- Substring
- Rep-string
- Word wrap
- String case
- Align columns
- Literals/String
- Repeat a string
- Brace expansion
- Brace expansion using ranges
- Reverse a string
- Phrase reversals
- Comma quibbling
- Special characters
- String concatenation
- Substring/Top and tail
- Commatizing numbers
- Reverse words in a string
- Suffixation of decimal numbers
- Long literals, with continuations
- Numerical and alphabetical suffixes
- Abbreviations, easy
- Abbreviations, simple
- Abbreviations, automatic
- Song lyrics/poems/Mad Libs/phrases
- Mad Libs
- Magic 8-ball
- 99 bottles of beer
- The Name Game (a song)
- The Old lady swallowed a fly
- The Twelve Days of Christmas
- Tokenize
- Text between
- Tokenize a string
- Word break problem
- Tokenize a string with escaping
- Split a character string based on change of character
- Sequences
- Ref
- Dynamic Programming #1: Longest Increasing Subsequence on YouTube
- An efficient solution can be based on Patience sorting.
11l
F longest_increasing_subsequence(x)
V n = x.len
V P = [0] * n
V M = [0] * (n + 1)
V l = 0
L(i) 0 .< n
V lo = 1
V hi = l
L lo <= hi
V mid = (lo + hi) I/ 2
I (x[M[mid]] < x[i])
lo = mid + 1
E
hi = mid - 1
V newl = lo
P[i] = M[newl - 1]
M[newl] = i
I (newl > l)
l = newl
[Int] s
V k = M[l]
L(i) (l - 1 .. 0).step(-1)
s.append(x[k])
k = P[k]
R reversed(s)
L(d) [[3, 2, 6, 4, 5, 1], [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]]
print(‘a L.I.S. of #. is #.’.format(d, longest_increasing_subsequence(d)))- Output:
a L.I.S. of [3, 2, 6, 4, 5, 1] is [2, 4, 5] a L.I.S. of [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] is [0, 2, 6, 9, 11, 15]
360 Assembly
* Longest increasing subsequence 04/03/2017
LNGINSQ CSECT
USING LNGINSQ,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) save previous context
ST R13,4(R15) link backward
ST R15,8(R13) link forward
LR R13,R15 set addressability
LA R6,1 i=1
DO WHILE=(CH,R6,LE,=H'2') do i=1 to 2
IF CH,R6,EQ,=H'1' THEN if i=1 then
MVC N,=AL2((A2-A1)/2) n=hbound(a1)
MVC AA(64),A1 a=a1
ELSE , else
MVC N,=AL2((AA-A2)/2) n=hbound(a2)
MVC AA(64),A2 a=a2
ENDIF , endif
MVC PG,=CL80': ' init buffer
LA R2,AA-2 @a
LH R3,N n
BAL R14,PRINT print a
MVC LL,=H'0' l=0
SR R7,R7 j=0
DO WHILE=(CH,R7,LE,N) do j=0 to n
MVC LO,=H'1' lo=1
MVC HI,LL hi=l
LH R4,LO lo
DO WHILE=(CH,R4,LE,HI) do while lo<=hi
LH R1,LO lo
AH R1,HI lo+hi
SRA R1,1 /2
STH R1,MIDDLE middle=(lo+hi)/2
SLA R1,1 *2
LH R1,MM(R1) m(middle+1)
SLA R1,1 *2
LH R3,AA(R1) r3=a(m(middle+1)+1)
LR R1,R7 j
SLA R1,1 *2
LH R4,AA(R1) r4=a(j+1)
LH R2,MIDDLE middle
IF CR,R3,LT,R4 THEN if a(m(middle+1)+1)<a(j+1) then
LA R2,1(R2) middle+1
STH R2,LO lo=middle+1
ELSE , else
BCTR R2,0 middle-1
STH R2,HI hi=middle-1
ENDIF , endif
LH R4,LO lo
ENDDO , end /*while*/
LH R10,LO newl=lo
LR R1,R10 newl
SLA R1,1 *2
LH R3,MM-2(R1) m(newl)
LR R1,R7 j
SLA R1,1 *2
STH R3,PP(R1) p(j+1)=m(newl)
LR R1,R10 newl
SLA R1,1 *2
STH R7,MM(R1) m(newl+1)=j
IF CH,R10,GT,LL if newl>l then
STH R10,LL l=newl
ENDIF , endif
LA R7,1(R7) j++
ENDDO , enddo j
LH R1,LL l
SLA R1,1 *2
LH R10,MM(R1) k=m(l+1)
LH R7,LL j=l
DO WHILE=(CH,R7,GE,=H'1') do j=l to 1 by -1
LR R1,R10 k
SLA R1,1 *2
LH R2,AA(R1) a(k+1)
LR R1,R7 j
SLA R1,1 *2
STH R2,SS-2(R1) s(j)=a(k+1)
LR R1,R10 k
SLA R1,1 *2
LH R10,PP(R1) k=p(k+1)
BCTR R7,0 j--
ENDDO , enddo j
MVC PG,=CL80'> ' init buffer
LA R2,SS-2 @s
LH R3,LL l
BAL R14,PRINT print a
LA R6,1(R6) i++
ENDDO , enddo i
L R13,4(0,R13) restore previous savearea pointer
LM R14,R12,12(R13) restore previous context
XR R15,R15 rc=0
BR R14 exit
PRINT LA R10,PG ---- print subroutine
LA R10,2(R10) pgi=2
LA R7,1 j=1
DO WHILE=(CR,R7,LE,R3) do j=1 to nx
LR R1,R7 j
SLA R1,1 *2
LH R1,0(R2,R1) x(j)
XDECO R1,XDEC edit x(j)
MVC 0(3,R10),XDEC+9 output x(j)
LA R10,3(R10) pgi+=3
LA R7,1(R7) j++
ENDDO , enddo j
XPRNT PG,L'PG print buffer
BR R14 ---- return
A1 DC H'3',H'2',H'6',H'4',H'5',H'1'
A2 DC H'0',H'8',H'4',H'12',H'2',H'10',H'6',H'14'
DC H'1',H'9',H'5',H'13',H'3',H'11',H'7',H'15'
AA DS 32H a(32)
PP DS 32H p(32)
MM DS 32H m(32)
SS DS 32H s(32)
N DS H n
LL DS H l
LO DS H lo
HI DS H hi
MIDDLE DS H middle
PG DS CL80 buffer
XDEC DS CL12 temp for xdeco
YREGS
END LNGINSQ- Output:
: 3 2 6 4 5 1 > 2 4 5 : 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 > 0 2 6 9 11 15
AppleScript
… modified to return multiple co-longest sequences where found. It's not clear how equal values should be treated. Here the behaviour happens to be as in the demo code at the end.
on longestIncreasingSubsequences(aList)
script o
property inputList : aList
property m : {} -- End indices of identified subsequences.
property p : {} -- 'Predecessor' indices for each point in each subsequence.
property subsequence : {} -- Reconstructed longest sequence.
end script
-- Set m and p to lists of the same length as the input. Their initial contents don't matter!
copy aList to o's m
copy aList to o's p
set bestLength to 0
repeat with i from 1 to (count o's inputList)
-- Comments adapted from those in the Wikipedia article — as far as they can be understood!
-- Binary search for the largest possible 'lo' ≤ bestLength such that inputList[m[lo]] ≤ inputList[i].
set lo to 1
set hi to bestLength
repeat until (lo > hi)
set mid to (lo + hi + 1) div 2
if (item (item mid of o's m) of o's inputList < item i of o's inputList) then
set lo to mid + 1
else
set hi to mid - 1
end if
end repeat
-- After searching, lo is 1 greater than the length of the longest prefix of inputList[i].
-- The predecessor of inputList[i] is the last index of the subsequence of length lo - 1.
if (lo > 1) then set item i of o's p to item (lo - 1) of o's m
set item lo of o's m to i
-- If we found a subsequence longer than or the same length as any we've found yet,
-- update bestLength and store the end index associated with it.
if (lo > bestLength) then
set bestLength to lo
set bestEndIndices to {item bestLength of o's m}
else if (lo = bestLength) then
set end of bestEndIndices to item bestLength of o's m
end if
end repeat
-- Reconstruct the longest increasing subsequence(s).
set output to {}
if (bestLength > 0) then
repeat with k in bestEndIndices
set o's subsequence to {}
repeat bestLength times
set beginning of o's subsequence to item k of o's inputList
set k to item k of o's p
end repeat
set end of output to o's subsequence
end repeat
end if
return output
end longestIncreasingSubsequences
-- Task code and other tests:
local tests, output, input
set tests to {{3, 2, 6, 4, 5, 1}, {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}, ¬
{9, 10, 11, 3, 8, 9, 6, 7, 4, 5}, {4, 5, 5, 6}, {5, 5}}
set output to {}
repeat with input in tests
set end of output to {finds:longestIncreasingSubsequences(input's contents)}
end repeat
return output
- Output:
{{finds:{{2, 4, 5}}}, {finds:{{0, 2, 6, 9, 11, 15}}}, {finds:{{9, 10, 11}, {3, 8, 9}, {3, 6, 7}, {3, 4, 5}}}, {finds:{{4, 5, 6}}}, {finds:{{5}, {5}}}}
Arturo
lis: function [d][
l: new [[]]
loop d 'num [
x: []
loop l 'seq [
if positive? size seq [
if and? num > last seq
(size seq) > size x ->
x: seq
]
]
'l ++ @[x ++ @[num]]
]
result: []
loop l 'x [
if (size x) > size result ->
result: x
]
return result
]
loop [
[3 2 6 4 5 1]
[0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15]
] 'seq [
print ["LIS of" seq "=>" lis seq]
]
- Output:
LIS of [3 2 6 4 5 1] => [3 4 5] LIS of [0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15] => [0 4 6 9 13 15]
AutoHotkey
Lists := [[3,2,6,4,5,1], [0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]]
for k, v in Lists {
D := LIS(v)
MsgBox, % D[D.I].seq
}
LIS(L) {
D := []
for i, v in L {
D[i, "Length"] := 1, D[i, "Seq"] := v, D[i, "Val"] := v
Loop, % i - 1 {
if(D[A_Index].Val < v && D[A_Index].Length + 1 > D[i].Length) {
D[i].Length := D[A_Index].Length + 1
D[i].Seq := D[A_Index].Seq ", " v
if (D[i].Length > MaxLength)
MaxLength := D[i].Length, D.I := i
}
}
}
return, D
}
Output:
3, 4, 5 0, 4, 6, 9, 13, 15
C
Using an array that doubles as linked list (more like reversed trees really). O(n) memory and O(n2) runtime.
#include <stdio.h>
#include <stdlib.h>
struct node {
int val, len;
struct node *next;
};
void lis(int *v, int len)
{
int i;
struct node *p, *n = calloc(len, sizeof *n);
for (i = 0; i < len; i++)
n[i].val = v[i];
for (i = len; i--; ) {
// find longest chain that can follow n[i]
for (p = n + i; p++ < n + len; ) {
if (p->val > n[i].val && p->len >= n[i].len) {
n[i].next = p;
n[i].len = p->len + 1;
}
}
}
// find longest chain
for (i = 0, p = n; i < len; i++)
if (n[i].len > p->len) p = n + i;
do printf(" %d", p->val); while ((p = p->next));
putchar('\n');
free(n);
}
int main(void)
{
int x[] = { 3, 2, 6, 4, 5, 1 };
int y[] = { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 };
lis(x, sizeof(x) / sizeof(int));
lis(y, sizeof(y) / sizeof(int));
return 0;
}
- Output:
3 4 5 0 4 6 9 13 15
C#
Recursive
using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
public static class LIS
{
public static IEnumerable<T> FindRec<T>(IList<T> values, IComparer<T> comparer = null) =>
values == null ? throw new ArgumentNullException() :
FindRecImpl(values, Sequence<T>.Empty, 0, comparer ?? Comparer<T>.Default).Reverse();
private static Sequence<T> FindRecImpl<T>(IList<T> values, Sequence<T> current, int index, IComparer<T> comparer) {
if (index == values.Count) return current;
if (current.Length > 0 && comparer.Compare(values[index], current.Value) <= 0)
return FindRecImpl(values, current, index + 1, comparer);
return Max(
FindRecImpl(values, current, index + 1, comparer),
FindRecImpl(values, current + values[index], index + 1, comparer)
);
}
private static Sequence<T> Max<T>(Sequence<T> a, Sequence<T> b) => a.Length < b.Length ? b : a;
class Sequence<T> : IEnumerable<T>
{
public static readonly Sequence<T> Empty = new Sequence<T>(default(T), null);
public Sequence(T value, Sequence<T> tail)
{
Value = value;
Tail = tail;
Length = tail == null ? 0 : tail.Length + 1;
}
public T Value { get; }
public Sequence<T> Tail { get; }
public int Length { get; }
public static Sequence<T> operator +(Sequence<T> s, T value) => new Sequence<T>(value, s);
public IEnumerator<T> GetEnumerator()
{
for (var s = this; s.Length > 0; s = s.Tail) yield return s.Value;
}
IEnumerator IEnumerable.GetEnumerator() => GetEnumerator();
}
}
Patience sorting
public static class LIS
{
public static T[] Find<T>(IList<T> values, IComparer<T> comparer = null) {
if (values == null) throw new ArgumentNullException();
if (comparer == null) comparer = Comparer<T>.Default;
var pileTops = new List<T>();
var pileAssignments = new int[values.Count];
for (int i = 0; i < values.Count; i++) {
T element = values[i];
int pile = pileTops.BinarySearch(element, comparer);
if (pile < 0) pile = ~pile;
if (pile == pileTops.Count) pileTops.Add(element);
else pileTops[pile] = element;
pileAssignments[i] = pile;
}
T[] result = new T[pileTops.Count];
for (int i = pileAssignments.Length - 1, p = pileTops.Count - 1; p >= 0; i--) {
if (pileAssignments[i] == p) result[p--] = values[i];
}
return result;
}
public static void Main() {
Console.WriteLine(string.Join(",", LIS.Find(new [] { 3, 2, 6, 4, 5, 1 })));
Console.WriteLine(string.Join(",", LIS.Find(new [] { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 })));
}
}
- Output:
2, 4, 5 0, 2, 6, 9, 11, 15
C++
Patience sorting
C++11
#include <vector>
#include <list>
#include <algorithm>
#include <iostream>
template <typename T>
struct Node {
T value;
Node* prev_node;
};
template <typename Container>
Container lis(const Container& values) {
using E = typename Container::value_type;
using NodePtr = Node<E>*;
using ConstNodePtr = const NodePtr;
std::vector<NodePtr> pileTops;
std::vector<Node<E>> nodes(values.size());
// sort into piles
auto cur_node = std::begin(nodes);
for (auto cur_value = std::begin(values); cur_value != std::end(values); ++cur_value, ++cur_node)
{
auto node = &*cur_node;
node->value = *cur_value;
// lower_bound returns the first element that is not less than 'node->value'
auto lb = std::lower_bound(pileTops.begin(), pileTops.end(), node,
[](ConstNodePtr& node1, ConstNodePtr& node2) -> bool { return node1->value < node2->value; });
if (lb != pileTops.begin())
node->prev_node = *std::prev(lb);
if (lb == pileTops.end())
pileTops.push_back(node);
else
*lb = node;
}
// extract LIS from piles
// note that LIS length is equal to the number of piles
Container result(pileTops.size());
auto r = std::rbegin(result);
for (NodePtr node = pileTops.back(); node != nullptr; node = node->prev_node, ++r)
*r = node->value;
return result;
}
template <typename Container>
void show_lis(const Container& values)
{
auto&& result = lis(values);
for (auto& r : result) {
std::cout << r << ' ';
}
std::cout << std::endl;
}
int main()
{
show_lis(std::list<int> { 3, 2, 6, 4, 5, 1 });
show_lis(std::vector<int> { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 });
}
C++98
#include <vector>
#include <list>
#include <algorithm>
#include <iostream>
template <typename T>
struct Node {
T value;
Node* prev_node;
};
template <typename T>
bool compare (const T& node1, const T& node2)
{
return node1->value < node2->value;
}
template <typename Container>
Container lis(const Container& values) {
typedef typename Container::value_type E;
typedef typename Container::const_iterator ValueConstIter;
typedef typename Container::iterator ValueIter;
typedef Node<E>* NodePtr;
typedef const NodePtr ConstNodePtr;
typedef std::vector<Node<E> > NodeVector;
typedef std::vector<NodePtr> NodePtrVector;
typedef typename NodeVector::iterator NodeVectorIter;
typedef typename NodePtrVector::iterator NodePtrVectorIter;
std::vector<NodePtr> pileTops;
std::vector<Node<E> > nodes(values.size());
// sort into piles
NodeVectorIter cur_node = nodes.begin();
for (ValueConstIter cur_value = values.begin(); cur_value != values.end(); ++cur_value, ++cur_node)
{
NodePtr node = &*cur_node;
node->value = *cur_value;
// lower_bound returns the first element that is not less than 'node->value'
NodePtrVectorIter lb = std::lower_bound(pileTops.begin(), pileTops.end(), node, compare<NodePtr>);
if (lb != pileTops.begin())
node->prev_node = *(lb - 1);
if (lb == pileTops.end())
pileTops.push_back(node);
else
*lb = node;
}
// extract LIS from piles
// note that LIS length is equal to the number of piles
Container result(pileTops.size());
std::reverse_iterator<ValueIter> r = std::reverse_iterator<ValueIter>(result.rbegin());
for (NodePtr node = pileTops.back(); node; node = node->prev_node, ++r)
*r = node->value;
return result;
}
template <typename Container>
void show_lis(const Container& values)
{
const Container& result = lis(values);
for (typename Container::const_iterator it = result.begin(); it != result.end(); ++it) {
std::cout << *it << ' ';
}
std::cout << std::endl;
}
int main()
{
const int arr1[] = { 3, 2, 6, 4, 5, 1 };
const int arr2[] = { 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 };
std::vector<int> vec1(arr1, arr1 + sizeof(arr1) / sizeof(arr1[0]));
std::vector<int> vec2(arr2, arr2 + sizeof(arr2) / sizeof(arr2[0]));
show_lis(vec1);
show_lis(vec2);
}
- Output:
2 4 5 0 2 6 9 11 15
Clojure
Implementation using the Patience Sort approach. The elements (newelem) put on a pile combine the "card" with a reference to the top of the previous stack, as per the algorithm. The combination is done using cons, so what gets put on a pile is a list -- a descending subsequence.
(defn place [piles card]
(let [[les gts] (->> piles (split-with #(<= (ffirst %) card)))
newelem (cons card (->> les last first))
modpile (cons newelem (first gts))]
(concat les (cons modpile (rest gts)))))
(defn a-longest [cards]
(let [piles (reduce place '() cards)]
(->> piles last first reverse)))
(println (a-longest [3 2 6 4 5 1]))
(println (a-longest [0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15]))
- Output:
(2 4 5)
(0 2 6 9 11 15)
Common Lisp
Common Lisp: Using the method in the video
Slower and more memory usage compared to the patience sort method.
(defun longest-increasing-subseq (list)
(let ((subseqs nil))
(dolist (item list)
(let ((longest-so-far (longest-list-in-lists (remove-if-not #'(lambda (l) (> item (car l))) subseqs))))
(push (cons item longest-so-far) subseqs)))
(reverse (longest-list-in-lists subseqs))))
(defun longest-list-in-lists (lists)
(let ((longest nil)
(longest-len 0))
(dolist (list lists)
(let ((len (length list)))
(when (> len longest-len)
(setf longest list
longest-len len))))
longest))
(dolist (l (list (list 3 2 6 4 5 1)
(list 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15)))
(format t "~A~%" (longest-increasing-subseq l))))
- Output:
(2 4 5) (0 2 6 9 11 15)
Common Lisp: Using the Patience Sort approach
This is 5 times faster and and uses a third of the memory compared to the approach in the video.
(defun lis-patience-sort (input-list)
(let ((piles nil))
(dolist (item input-list)
(setf piles (insert-item item piles)))
(reverse (caar (last piles)))))
(defun insert-item (item piles)
(let ((not-found t))
(loop
while not-found
for pile in piles
and prev = nil then pile
and i from 0
do (when (<= item (caar pile))
(setf (elt piles i) (push (cons item (car prev)) (elt piles i))
not-found nil)))
(if not-found
(append piles (list (list (cons item (caar (last piles))))))
piles)))
(dolist (l (list (list 3 2 6 4 5 1)
(list 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15)))
(format t "~A~%" (lis-patience-sort l)))
- Output:
(2 4 5) (0 2 6 9 11 15)
Common Lisp: Using the Patience Sort approach (alternative)
This is a different version of the code above.
(defun insert-item (item piles)
(multiple-value-bind
(i prev)
(do* ((prev nil (car x))
(x piles (cdr x))
(i 0 (1+ i)))
((or (null x) (<= item (caaar x))) (values i prev)))
(if (= i (length piles))
(append piles (list (list (cons item (caar (last piles))))))
(progn (push (cons item (car prev)) (elt piles i))
piles))))
(defun longest-inc-seq (input)
(do* ((piles nil (insert-item (car x) piles))
(x input (cdr x)))
((null x) (reverse (caar (last piles))))))
(dolist (l (list (list 3 2 6 4 5 1)
(list 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15)))
(format t "~A~%" (longest-inc-seq l)))
- Output:
(2 4 5) (0 2 6 9 11 15)
D
Simple Version
Uses the second powerSet function from the Power Set Task.
import std.stdio, std.algorithm, power_set2;
T[] lis(T)(T[] items) pure nothrow {
//return items.powerSet.filter!isSorted.max!q{ a.length };
return items
.powerSet
.filter!isSorted
.minPos!q{ a.length > b.length }
.front;
}
void main() {
[3, 2, 6, 4, 5, 1].lis.writeln;
[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15].lis.writeln;
}
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Patience sorting
From the second Python entry, using the Patience sorting method.
import std.stdio, std.algorithm, std.array;
/// Return one of the Longest Increasing Subsequence of
/// items using patience sorting.
T[] lis(T)(in T[] items) pure nothrow
if (__traits(compiles, T.init < T.init))
out(result) {
assert(result.length <= items.length);
assert(result.isSorted);
assert(result.all!(x => items.canFind(x)));
} body {
if (items.empty)
return null;
static struct Node { T val; Node* back; }
auto pile = [[new Node(items[0])]];
OUTER: foreach (immutable di; items[1 .. $]) {
foreach (immutable j, ref pj; pile)
if (pj[$ - 1].val > di) {
pj ~= new Node(di, j ? pile[j - 1][$ - 1] : null);
continue OUTER;
}
pile ~= [new Node(di, pile[$ - 1][$ - 1])];
}
T[] result;
for (auto ptr = pile[$ - 1][$ - 1]; ptr != null; ptr = ptr.back)
result ~= ptr.val;
result.reverse();
return result;
}
void main() {
foreach (d; [[3,2,6,4,5,1],
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]])
d.lis.writeln;
}
The output is the same.
Faster Version
With some more optimizations.
import std.stdio, std.algorithm, std.range, std.array;
T[] lis(T)(in T[] items) pure nothrow
if (__traits(compiles, T.init < T.init))
out(result) {
assert(result.length <= items.length);
assert(result.isSorted);
assert(result.all!(x => items.canFind(x)));
} body {
if (items.empty)
return null;
static struct Node {
T value;
Node* pointer;
}
Node*[] pileTops;
auto nodes = minimallyInitializedArray!(Node[])(items.length);
// Sort into piles.
foreach (idx, x; items) {
auto node = &nodes[idx];
node.value = x;
immutable i = pileTops.length -
pileTops.assumeSorted!q{a.value < b.value}
.upperBound(node)
.length;
if (i != 0)
node.pointer = pileTops[i - 1];
if (i != pileTops.length)
pileTops[i] = node;
else
pileTops ~= node;
}
// Extract LIS from nodes.
size_t count = 0;
for (auto n = pileTops[$ - 1]; n != null; n = n.pointer)
count++;
auto result = minimallyInitializedArray!(T[])(count);
for (auto n = pileTops[$ - 1]; n != null; n = n.pointer)
result[--count] = n.value;
return result;
}
void main() {
foreach (d; [[3,2,6,4,5,1],
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]])
d.writeln;
}
The output is the same.
Déjà Vu
in-pair:
if = :nil dup:
false drop
else:
@in-pair &> swap &< dup
get-last lst:
get-from lst -- len lst
lis-sub pile i di:
for j range 0 -- len pile:
local :pj get-from pile j
if > &< get-last pj di:
push-to pj & di if j get-last get-from pile -- j :nil
return
push-to pile [ & di get-last get-last pile ]
lis d:
local :pile [ [ & get-from d 0 :nil ] ]
for i range 1 -- len d:
lis-sub pile i get-from d i
[ for in-pair get-last get-last pile ]
!. lis [ 3 2 6 4 5 1 ]
!. lis [ 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 ]- Output:
[ 2 4 5 ] [ 0 2 6 9 11 15 ]
EasyLang
func[] lis x[] .
n = len x[]
len p[] n
len m[] n
for i to n
lo = 1
hi = lng
while lo <= hi
mid = (lo + hi) div 2
if x[m[mid]] < x[i]
lo = mid + 1
else
hi = mid - 1
.
.
if lo > 1
p[i] = m[lo - 1]
.
m[lo] = i
if lo > lng
lng = lo
.
.
len res[] lng
if lng > 0
k = m[lng]
for i = lng downto 1
res[i] = x[k]
k = p[k]
.
.
return res[]
.
tests[][] = [ [ 3 2 6 4 5 1 ] [ 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 ] ]
for x to len tests[][]
print lis tests[x][]
.- Output:
[ 2 4 5 ] [ 0 2 6 9 11 15 ]
Elixir
Naive version
very slow
defmodule Longest_increasing_subsequence do
# Naive implementation
def lis(l) do
(for ss <- combos(l), ss == Enum.sort(ss), do: ss)
|> Enum.max_by(fn ss -> length(ss) end)
end
defp combos(l) do
Enum.reduce(1..length(l), [[]], fn k, acc -> acc ++ (combos(k, l)) end)
end
defp combos(1, l), do: (for x <- l, do: [x])
defp combos(k, l) when k == length(l), do: [l]
defp combos(k, [h|t]) do
(for subcombos <- combos(k-1, t), do: [h | subcombos]) ++ combos(k, t)
end
end
IO.inspect Longest_increasing_subsequence.lis([3,2,6,4,5,1])
IO.inspect Longest_increasing_subsequence.lis([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15])
- Output:
[3, 4, 5] [0, 4, 6, 9, 13, 15]
Patience sort version
defmodule Longest_increasing_subsequence do
# Patience sort implementation
def patience_lis(l), do: patience_lis(l, [])
defp patience_lis([h | t], []), do: patience_lis(t, [[{h,[]}]])
defp patience_lis([h | t], stacks), do: patience_lis(t, place_in_stack(h, stacks, []))
defp patience_lis([], []), do: []
defp patience_lis([], stacks), do: get_previous(stacks) |> recover_lis |> Enum.reverse
defp place_in_stack(e, [stack = [{h,_} | _] | tstacks], prevstacks) when h > e do
prevstacks ++ [[{e, get_previous(prevstacks)} | stack] | tstacks]
end
defp place_in_stack(e, [stack | tstacks], prevstacks) do
place_in_stack(e, tstacks, prevstacks ++ [stack])
end
defp place_in_stack(e, [], prevstacks) do
prevstacks ++ [[{e, get_previous(prevstacks)}]]
end
defp get_previous(stack = [_|_]), do: hd(List.last(stack))
defp get_previous([]), do: []
defp recover_lis({e, prev}), do: [e | recover_lis(prev)]
defp recover_lis([]), do: []
end
IO.inspect Longest_increasing_subsequence.patience_lis([3,2,6,4,5,1])
IO.inspect Longest_increasing_subsequence.patience_lis([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15])
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Erlang
Four implementations:
- Naive version
- Memoization
- Patience sort version
- Patience sort version2
Function combos is copied from panduwana blog.
Function maxBy is copied from Hynek -Pichi- Vychodil's answer.
Function memo and patience2 by Roman Rabinovich.
-module(longest_increasing_subsequence).
-export([test_naive/0, test_memo/0, test_patience/0, test_patience2/0, test_compare/1]).
% **************************************************
% Interface to test the implementation
% **************************************************
test_compare(N) when N =< 20 ->
Funs = [
{"Naive", fun lis/1},
{"Memo", fun memo/1},
{"Patience", fun patience_lis/1},
{"Patience2", fun patience2/1}
],
do_compare(Funs, N);
test_compare(N) when N =< 500 ->
Funs = [
{"Memo", fun memo/1},
{"Patience", fun patience_lis/1},
{"Patience2", fun patience2/1}
],
do_compare(Funs, N);
test_compare(N) ->
Funs = [
{"Patience", fun patience_lis/1},
{"Patience2", fun patience2/1}
],
do_compare(Funs, N).
do_compare(Funs, N) ->
List = [rand:uniform(1000) || _ <- lists:seq(1,N)],
Results = [{Name, timer:tc(fun() -> F(List) end)} || {Name,F} <- Funs],
Times = [{Name, Time} || {Name, {Time, _Result}} <- Results],
io:format("Result Times: ~p~n", [Times]).
test_naive() ->
test_gen(fun lis/1).
test_memo() ->
test_gen(fun memo/1).
test_patience() ->
test_gen(fun patience_lis/1).
test_patience2() ->
test_gen(fun patience2/1).
test_gen(F) ->
show_result(F([3,2,6,4,5,1])),
show_result(F([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15])).
show_result(Res) ->
io:format("~w\n", [Res]).
% **************************************************
% **************************************************
% Naive implementation
% **************************************************
lis(L) ->
maxBy(
fun(SS) -> length(SS) end,
[ lists:usort(SS)
|| SS <- combos(L),
SS == lists:sort(SS)]
).
% **************************************************
% Copied from http://stackoverflow.com/a/4762387/4162959
% **************************************************
maxBy(F, L) ->
element(
2,
lists:max([ {F(X), X} || X <- L])
).
% **************************************************
% Copied from https://panduwana.wordpress.com/2010/04/21/combination-in-erlang/
% **************************************************
combos(L) ->
lists:foldl(
fun(K, Acc) -> Acc++(combos(K, L)) end,
[[]],
lists:seq(1, length(L))
).
combos(1, L) ->
[[X] || X <- L];
combos(K, L) when K == length(L) ->
[L];
combos(K, [H|T]) ->
[[H | Subcombos]
|| Subcombos <- combos(K-1, T)]
++ (combos(K, T)).
% **************************************************
% **************************************************
% Memoization implementation, Roman Rabinovich
% **************************************************
memo(S) ->
put(test, #{}),
memo(S, -1).
memo([], _) -> [];
memo([H | Tail] = S, Min) when H > Min ->
case maps:get({S,Min}, get(test), undefined) of
undefined ->
L1 = [H | memo(Tail, H)],
L2 = memo(Tail, Min),
case length(L1) >= length(L2) of
true ->
Map = get(test),
put(test, Map#{{S, Min} => L1}),
L1;
_ ->
Map = get(test),
put(test, Map#{{S, Min} => L2}),
L2
end;
X -> X
end;
memo([_|Tail], Min) ->
memo(Tail, Min).
% **************************************************
% **************************************************
% Patience sort implementation
% **************************************************
patience_lis(L) ->
patience_lis(L, []).
patience_lis([H | T], Stacks) ->
NStacks =
case Stacks of
[] ->
[[{H,[]}]];
_ ->
place_in_stack(H, Stacks, [])
end,
patience_lis(T, NStacks);
patience_lis([], Stacks) ->
case Stacks of
[] ->
[];
[_|_] ->
lists:reverse( recover_lis( get_previous(Stacks) ) )
end.
place_in_stack(E, [Stack = [{H,_} | _] | TStacks], PrevStacks) when H > E ->
PrevStacks ++ [[{E, get_previous(PrevStacks)} | Stack] | TStacks];
place_in_stack(E, [Stack = [{H,_} | _] | TStacks], PrevStacks) when H =< E ->
place_in_stack(E, TStacks, PrevStacks ++ [Stack]);
place_in_stack(E, [], PrevStacks)->
PrevStacks ++ [[{E, get_previous(PrevStacks)}]].
get_previous(Stack = [_|_]) ->
hd(lists:last(Stack));
get_previous([]) ->
[].
recover_lis({E,Prev}) ->
[E|recover_lis(Prev)];
recover_lis([]) ->
[].
% **************************************************
% **************************************************
% Patience2 by Roman Rabinovich, improved performance over above
% **************************************************
patience2([]) -> [];
patience2([H|L]) ->
Piles = [[{H, undefined}]],
patience2(L, Piles, []).
patience2([], Piles, _) ->
get_seq(lists:reverse(Piles));
patience2([H|T], [[{PE,_}|_Rest] = Pile| Piles], PrevPiles) when H =< PE ->
case PrevPiles of
[] -> patience2(T, [[{H, undefined}|Pile]|Piles], []);
[[{K,_}|_]|_] -> patience2(T, lists:reverse(PrevPiles) ++ [[{H, K}|Pile]|Piles], [])
end;
patience2([H|_T] = L, [[{PE,_}|_Rest] = Pile| Piles], PrevPiles) when H > PE ->
patience2(L, Piles, [Pile|PrevPiles]);
patience2([H|T], [], [[{K,_}|_]|_]=PrevPiles) ->
patience2(T, lists:reverse([[{H,K}]|PrevPiles]), []).
get_seq([]) -> [];
get_seq([[{K,P}|_]|Rest]) ->
get_seq(P, Rest, [K]).
get_seq(undefined, [], Seq) -> Seq;
get_seq(K, [Pile|Rest], Seq) ->
case lists:keyfind(K, 1, Pile) of
undefined -> get_seq(K, Rest, Seq);
{K, P} -> get_seq(P, Rest, [K|Seq])
end.
% **************************************************
Output naive:
[3,4,5] [0,4,6,9,13,15]
Output memoization:
[3,4,5] [0,4,6,9,13,15]
Output patience:
[2,4,5] [0,2,6,9,11,15]
Output patience2:
[2,4,5] [0,2,6,9,11,15]
FreeBASIC
Sub Lis(arr() As Integer)
Dim As Integer lb = Lbound(arr), ub = Ubound(arr)
Dim As Integer i, lo, hi, mitad, newl, l = 0
Dim As Integer p(ub), m(ub)
For i = lb To ub
lo = 1
hi = l
Do While lo <= hi
mitad = Int((lo+hi)/2)
If arr(m(mitad)) < arr(i) Then
lo = mitad + 1
Else
hi = mitad - 1
End If
Loop
newl = lo
p(i) = m(newl-1)
m(newl) = i
If newL > l Then l = newl
Next i
Dim As Integer res(l)
Dim As Integer k = m(l)
For i = l-1 To 0 Step - 1
res(i) = arr(k)
k = p(k)
Next i
For i = Lbound(res) To Ubound(res)-1
Print res(i); " ";
Next i
End Sub
Dim As Integer arrA(5) => {3,2,6,4,5,1}
Lis(arrA())
Print
Dim As Integer arrB(15) => {0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15}
Lis(arrB())
Sleep
- Output:
2 4 5 0 2 6 9 11 15
Go
Patience sorting
package main
import (
"fmt"
"sort"
)
type Node struct {
val int
back *Node
}
func lis (n []int) (result []int) {
var pileTops []*Node
// sort into piles
for _, x := range n {
j := sort.Search(len(pileTops), func (i int) bool { return pileTops[i].val >= x })
node := &Node{ x, nil }
if j != 0 { node.back = pileTops[j-1] }
if j != len(pileTops) {
pileTops[j] = node
} else {
pileTops = append(pileTops, node)
}
}
if len(pileTops) == 0 { return []int{} }
for node := pileTops[len(pileTops)-1]; node != nil; node = node.back {
result = append(result, node.val)
}
// reverse
for i := 0; i < len(result)/2; i++ {
result[i], result[len(result)-i-1] = result[len(result)-i-1], result[i]
}
return
}
func main() {
for _, d := range [][]int{{3, 2, 6, 4, 5, 1},
{0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}} {
fmt.Printf("an L.I.S. of %v is %v\n", d, lis(d))
}
}
- Output:
an L.I.S. of [3 2 6 4 5 1] is [2 4 5] an L.I.S. of [0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15] is [0 2 6 9 11 15]
Haskell
Naive implementation
import Data.Ord ( comparing )
import Data.List ( maximumBy, subsequences )
import Data.List.Ordered ( isSorted, nub )
lis :: Ord a => [a] -> [a]
lis = maximumBy (comparing length) . map nub . filter isSorted . subsequences
-- longest <-- unique <-- increasing <-- all
main = do
print $ lis [3,2,6,4,5,1]
print $ lis [0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]
print $ lis [1,1,1,1]
- Output:
[2,4,5] [0,2,6,9,11,15] [1]
Patience sorting
{-# LANGUAGE FlexibleContexts, UnicodeSyntax #-}
module Main (main, lis) where
import Control.Monad.ST ( ST, runST )
import Control.Monad ( (>>=), (=<<), foldM )
import Data.Array.ST ( Ix, STArray, readArray, writeArray, newArray )
import Data.Array.MArray ( MArray )
infix 4 ≡
(≡) :: Eq α ⇒ α → α → Bool
(≡) = (==)
(∘) = (.)
lis ∷ Ord α ⇒ [α] → [α]
lis xs = runST $ do
let lxs = length xs
pileTops ← newSTArray (min 1 lxs , lxs) []
i ← foldM (stack pileTops) 0 xs
readArray pileTops i >>= return ∘ reverse
stack ∷ (Integral ι, Ord ε, Ix ι, MArray α [ε] μ)
⇒ α ι [ε] → ι → ε → μ ι
stack piles i x = do
j ← bsearch piles x i
writeArray piles j ∘ (x:) =<< if j ≡ 1 then return []
else readArray piles (j-1)
return $ if j ≡ i+1 then i+1 else i
bsearch ∷ (Integral ι, Ord ε, Ix ι, MArray α [ε] μ)
⇒ α ι [ε] → ε → ι → μ ι
bsearch piles x = go 1
where go lo hi | lo > hi = return lo
| otherwise =
do (y:_) ← readArray piles mid
if y < x then go (succ mid) hi
else go lo (pred mid)
where mid = (lo + hi) `div` 2
newSTArray ∷ Ix ι ⇒ (ι,ι) → ε → ST σ (STArray σ ι ε)
newSTArray = newArray
main ∷ IO ()
main = do
print $ lis [3, 2, 6, 4, 5, 1]
print $ lis [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]
print $ lis [1, 1, 1, 1]
- Output:
[2,4,5] [0,2,6,9,11,15] [1]
Icon and Unicon
The following works in both languages:
procedure main(A)
every writes((!lis(A)||" ") | "\n")
end
procedure lis(A)
r := [A[1]] | fail
every (put(pt := [], [v := !A]), p := !pt) do
if put(p, p[-1] < v) then r := (*p > *r, p)
else p[-1] := (p[-2] < v)
return r
end
Sample runs:
->lis 3 2 6 4 5 1 3 4 5 ->lis 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 0 4 6 9 11 15 ->
J
These examples are simple enough for brute force to be reasonable:
increasing=: (-: /:~)@#~"1 #:@i.@^~&2@#
longestinc=: ] #~ [: (#~ ([: (= >./) +/"1)) #:@I.@increasing
In other words: consider all 2^n bitmasks of length n, and select those which strictly select increasing sequences. Find the length of the longest of these and use the masks of that length to select from the original sequence.
Example use:
longestinc 3,2,6,4,5,1
2 4 5
3 4 5
longestinc 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15
0 2 6 9 11 15
0 2 6 9 13 15
0 4 6 9 11 15
0 4 6 9 13 15
Java
A solution based on patience sorting, except that it is not necessary to keep the whole pile, only the top (in solitaire, bottom) of the pile, along with pointers from each "card" to the top of its "previous" pile.
import java.util.*;
public class LIS {
public static <E extends Comparable<? super E>> List<E> lis(List<E> n) {
List<Node<E>> pileTops = new ArrayList<Node<E>>();
// sort into piles
for (E x : n) {
Node<E> node = new Node<E>();
node.value = x;
int i = Collections.binarySearch(pileTops, node);
if (i < 0) i = ~i;
if (i != 0)
node.pointer = pileTops.get(i-1);
if (i != pileTops.size())
pileTops.set(i, node);
else
pileTops.add(node);
}
// extract LIS from nodes
List<E> result = new ArrayList<E>();
for (Node<E> node = pileTops.size() == 0 ? null : pileTops.get(pileTops.size()-1);
node != null; node = node.pointer)
result.add(node.value);
Collections.reverse(result);
return result;
}
private static class Node<E extends Comparable<? super E>> implements Comparable<Node<E>> {
public E value;
public Node<E> pointer;
public int compareTo(Node<E> y) { return value.compareTo(y.value); }
}
public static void main(String[] args) {
List<Integer> d = Arrays.asList(3,2,6,4,5,1);
System.out.printf("an L.I.S. of %s is %s\n", d, lis(d));
d = Arrays.asList(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15);
System.out.printf("an L.I.S. of %s is %s\n", d, lis(d));
}
}
- Output:
an L.I.S. of [3, 2, 6, 4, 5, 1] is [2, 4, 5] an L.I.S. of [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] is [0, 2, 6, 9, 11, 15]
JavaScript
function getLis(input) {
if (input.length === 0) {
return [];
}
var lisLenPerIndex = [];
let max = { index: 0, length: 1 };
for (var i = 0; i < input.length; i++) {
lisLenPerIndex[i] = 1;
for (var j = i - 1; j >= 0; j--) {
if (input[i] > input[j] && lisLenPerIndex[j] >= lisLenPerIndex[i]) {
var length = lisLenPerIndex[i] = lisLenPerIndex[j] + 1;
if (length > max.length) {
max = { index: i, length };
}
}
}
}
var lis = [input[max.index]];
for (var i = max.index; i >= 0 && max.length !== 0; i--) {
if (input[max.index] > input[i] && lisLenPerIndex[i] === max.length - 1) {
lis.unshift(input[i]);
max.length--;
}
}
return lis;
}
console.log(getLongestIncreasingSubsequence([0, 7, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]));
console.log(getLongestIncreasingSubsequence([3, 2, 6, 4, 5, 1]));
- Output:
[ 0, 2, 6, 9, 11, 15 ] [ 2, 4, 5 ]
Patience sorting
function getLIS(input) {
if (input.length === 0) {
return 0;
}
const piles = [input[0]];
for (let i = 1; i < input.length; i++) {
const leftPileIdx = binarySearch(piles, input[i]);
if (leftPileIdx !== -1) {
piles[leftPileIdx] = input[i];
} else {
piles.push(input[i]);
}
}
return piles.length;
}
function binarySearch(arr, target) {
let lo = 0;
let hi = arr.length - 1;
while (lo <= hi) {
const mid = lo + Math.floor((hi - lo) / 2);
if (arr[mid] >= target) {
hi = mid - 1;
} else {
lo = mid + 1;
}
}
return lo < arr.length ? lo : -1;
}
console.log(getLongestIncreasingSubsequence([0, 7, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]));
console.log(getLongestIncreasingSubsequence([3, 2, 6, 4, 5, 1]));
- Output:
[ 0, 2, 6, 9, 11, 15 ] [ 2, 4, 5 ]
jq
Use the patience sorting method to find a longest (strictly) increasing subsequence.
Generic functions:
Recent versions of jq have functions that obviate the need for the two generic functions defined in this subsection.
def until(cond; update):
def _until:
if cond then . else (update | _until) end;
try _until catch if .== "break" then empty else . end;
# binary search for insertion point
def bsearch(target):
. as $in
| [0, length-1] # [low, high]
| until(.[0] > .[1];
.[0] as $low | .[1] as $high
| ($low + ($high - $low) / 2 | floor) as $mid
| if $in[$mid] >= target
then .[1] = $mid - 1
else .[0] = $mid + 1
end )
| .[0];lis:
def lis:
# Helper function:
# given a stream, produce an array of the items in reverse order:
def reverse(stream): reduce stream as $i ([]; [$i] + .);
# put the items into increasing piles using the structure:
# NODE = {"val": value, "back": NODE}
reduce .[] as $x
( []; # array of NODE
# binary search for the appropriate pile
(map(.val) | bsearch($x)) as $i
| setpath([$i];
{"val": $x,
"back": (if $i > 0 then .[$i-1] else null end) })
)
| .[length - 1]
| reverse( recurse(.back) | .val ) ;Examples:
( [3,2,6,4,5,1],
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]
) | lis- Output:
$ jq -c -n -f lis.jq
[2,4,5]
[0,2,6,9,11,15]
Julia
function lis(arr::Vector)
if length(arr) == 0 return copy(arr) end
L = Vector{typeof(arr)}(length(arr))
L[1] = [arr[1]]
for i in 2:length(arr)
nextL = []
for j in 1:i
if arr[j] < arr[i] && length(L[j]) ≥ length(nextL)
nextL = L[j]
end
end
L[i] = vcat(nextL, arr[i])
end
return L[indmax(length.(L))]
end
@show lis([3, 2, 6, 4, 5, 1])
@show lis([0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15])
- Output:
lis([3, 2, 6, 4, 5, 1]) = [2, 4, 5] lis([0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]) = [0, 2, 6, 9, 11, 15]
Kotlin
Uses the algorithm in the Wikipedia L.I.S. article:
// version 1.1.0
fun longestIncreasingSubsequence(x: IntArray): IntArray =
when (x.size) {
0 -> IntArray(0)
1 -> x
else -> {
val n = x.size
val p = IntArray(n)
val m = IntArray(n + 1)
var len = 0
for (i in 0 until n) {
var lo = 1
var hi = len
while (lo <= hi) {
val mid = Math.ceil((lo + hi) / 2.0).toInt()
if (x[m[mid]] < x[i]) lo = mid + 1
else hi = mid - 1
}
val newLen = lo
p[i] = m[newLen - 1]
m[newLen] = i
if (newLen > len) len = newLen
}
val s = IntArray(len)
var k = m[len]
for (i in len - 1 downTo 0) {
s[i] = x[k]
k = p[k]
}
s
}
}
fun main(args: Array<String>) {
val lists = listOf(
intArrayOf(3, 2, 6, 4, 5, 1),
intArrayOf(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15)
)
lists.forEach { println(longestIncreasingSubsequence(it).asList()) }
}
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Lua
function buildLIS(seq)
local piles = { { {table.remove(seq, 1), nil} } }
while #seq>0 do
local x=table.remove(seq, 1)
for j=1,#piles do
if piles[j][#piles[j]][1]>x then
table.insert(piles[j], {x, (piles[j-1] and #piles[j-1])})
break
elseif j==#piles then
table.insert(piles, {{x, #piles[j]}})
end
end
end
local t={}
table.insert(t, piles[#piles][1][1])
local p=piles[#piles][1][2]
for i=#piles-1,1,-1 do
table.insert(t, piles[i][p][1])
p=piles[i][p][2]
end
table.sort(t)
print(unpack(t))
end
buildLIS({3,2,6,4,5,1})
buildLIS({0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15})
- Output:
2 4 5 0 2 6 9 11 15
M2000 Interpreter
Using Stack objects in an array
stack:=stackitem(L(i)), ! stack(L(j)) returns a refence to a new stack object, with the first item on L(i) (which is a reference to stack object) and merge using ! the copy of L(j) stack.
Module LIS_example {
Function LIS {
LD=Stack.Size-1
dim L(0 to LD)
For i=0 to LD : Read V: L(i):=Stack:=V:next
M=1
M1=LD
for i=LD-1 to 0
for j=LD to i+1
if stackitem(L(i))<stackitem(L(j)) then
if len(L(i))<=len(L(j)) then L(i) =stack:=stackitem(L(i)), ! stack(L(j))
end if
next
if len(L(i))>=M then M1=i:M=Len(L(i))
next
=L(M1)
}
Const seq$="sequence", subseq$="Longest increasing subsequence"
Document doc$
Disp(seq$, Stack:=3,2,6,4,5,1)
Disp(subseq$, Lis(3,2,6,4,5,1))
Disp(seq$, Stack:=0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15)
Disp(subseq$, LIS(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15))
Print #-2,Doc$
Clipboard Doc$
Sub Disp(title$, m)
local n=each(m), s$
while n
s$+=", "+str$(stackitem(n),"")
end while
s$=trim$(mid$(s$, 2))
Doc$=title$+": "+s$+{
}
End Sub
}
LIS_exampleUsing arrays in an array
Module LIS_example {
Function LIS {
LD=Stack.Size-1
dim L(0 to LD)
For i=0 to LD : Read V: L(i):=(V,):next
M=1
M1=LD
for i=LD-1 to 0
for j=LD to i+1
if Array(L(i))<Array(L(j)) then
' you can use either is the same
' if len(L(i))<=len(L(j)) then L(i)=Cons((Array(L(i)),), L(j))
if len(L(i))<=len(L(j)) then L(i)=(Array(L(i)),): Append L(i), L(j)
end if
next
if len(L(i))>=M then M1=i:M=Len(L(i))
next
=L(M1)
}
Const seq$="sequence", subseq$="Longest increasing subsequence"
Document doc$
Disp(seq$, (3,2,6,4,5,1))
Disp(subseq$, LIS(3,2,6,4,5,1))
Disp(seq$, (0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15))
Disp(subseq$, LIS(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15))
Print #-2,Doc$
Clipboard Doc$
Sub Disp(title$, m)
local n=each(m), s$
while n
s$+=", "+str$(Array(n),"")
end while
s$=trim$(mid$(s$, 2))
Doc$=title$+": "+s$+{
}
End Sub
}
LIS_example- Output:
sequence: 3, 2, 6, 4, 5, 1 Longest increasing subsequence: 3, 4, 5 sequence: 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15 Longest increasing subsequence: 0, 2, 6, 9, 11, 15
Maple
# dynamic programming:
LIS := proc(L)
local i, j;
local index := 1;
local output := Array(1..numelems(L), i -> Array(1..0));
for i from 1 to numelems(L) do
for j from 1 to i - 1 do
if (L[j] < L[i]) and (upperbound(output[j]) > upperbound(output[i])) then
output[i] := copy(output[j]);
end if;
end do;
# append current value
output[i] ,= L[i];
end do;
#output longest subsequence using loop
for i from 2 to numelems(L) do
if (upperbound(output[i]) > upperbound(output[index])) then
index := i;
end if;
end do;
return output[index];
end proc:Alternatively, output the longest subsequence using built-in command max:
i := max[index](map(numelems,output));
output[i];L := [3, 2, 6, 4, 5, 1];
M := [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15];
LIS(L);
LIS(M);- Output:
[3 4 5] [0 4 6 9 13 15]
Mathematica /Wolfram Language
Although undocumented, Mathematica has the function LongestAscendingSequence which exactly does what the Task asks for:
LongestAscendingSequence/@{{3,2,6,4,5,1},{0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15}}
- Output:
{{2,4,5},{0,2,6,9,11,15}}
Nim
proc longestIncreasingSubsequence[T](d: seq[T]): seq[T] =
var l: seq[seq[T]]
for i in 0 .. d.high:
var x: seq[T]
for j in 0 ..< i:
if l[j][l[j].high] < d[i] and l[j].len > x.len:
x = l[j]
l.add x & @[d[i]]
for x in l:
if x.len > result.len:
result = x
for d in [@[3,2,6,4,5,1], @[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]]:
echo "A L.I.S. of ", d, " is ", longestIncreasingSubsequence(d)
- Output:
A L.I.S. of @[3, 2, 6, 4, 5, 1] is @[3, 4, 5] A L.I.S. of @[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] is @[0, 4, 6, 9, 13, 15]
Objective-C
Patience sorting
#import <Foundation/Foundation.h>
@interface Node : NSObject {
@public
id val;
Node *back;
}
@end
@implementation Node
@end
@interface NSArray (LIS)
- (NSArray *)longestIncreasingSubsequenceWithComparator:(NSComparator)comparator;
@end
@implementation NSArray (LIS)
- (NSArray *)longestIncreasingSubsequenceWithComparator:(NSComparator)comparator {
NSMutableArray *pileTops = [[NSMutableArray alloc] init];
// sort into piles
for (id x in self) {
Node *node = [[Node alloc] init];
node->val = x;
int i = [pileTops indexOfObject:node
inSortedRange:NSMakeRange(0, [pileTops count])
options:NSBinarySearchingInsertionIndex|NSBinarySearchingFirstEqual
usingComparator:^NSComparisonResult(Node *node1, Node *node2) {
return comparator(node1->val, node2->val);
}];
if (i != 0)
node->back = pileTops[i-1];
pileTops[i] = node;
}
// follow pointers from last node
NSMutableArray *result = [[NSMutableArray alloc] init];
for (Node *node = [pileTops lastObject]; node; node = node->back)
[result addObject:node->val];
return [[result reverseObjectEnumerator] allObjects];
}
@end
int main(int argc, const char *argv[]) {
@autoreleasepool {
for (NSArray *d in @[@[@3, @2, @6, @4, @5, @1],
@[@0, @8, @4, @12, @2, @10, @6, @14, @1, @9, @5, @13, @3, @11, @7, @15]])
NSLog(@"an L.I.S. of %@ is %@", d,
[d longestIncreasingSubsequenceWithComparator:^NSComparisonResult(id obj1, id obj2) {
return [obj1 compare:obj2];
}]);
}
return 0;
}
- Output:
an L.I.S. of (
3,
2,
6,
4,
5,
1
) is (
2,
4,
5
)
an L.I.S. of (
0,
8,
4,
12,
2,
10,
6,
14,
1,
9,
5,
13,
3,
11,
7,
15
) is (
0,
2,
6,
9,
11,
15
)
OCaml
Naïve implementation
let longest l = List.fold_left (fun acc x -> if List.length acc < List.length x
then x
else acc) [] l
let subsequences d l =
let rec check_subsequences acc = function
| x::s -> check_subsequences (if (List.hd (List.rev x)) < d
then x::acc
else acc) s
| [] -> acc
in check_subsequences [] l
let lis d =
let rec lis' l = function
| x::s -> lis' ((longest (subsequences x l)@[x])::l) s
| [] -> longest l
in lis' [] d
let _ =
let sequences = [[3; 2; 6; 4; 5; 1]; [0; 8; 4; 12; 2; 10; 6; 14; 1; 9; 5; 13; 3; 11; 7; 15]]
in
List.map (fun x -> print_endline (String.concat " " (List.map string_of_int
(lis x)))) sequences
- Output:
3 4 5 0 4 6 9 13 15
Patience sorting
let lis cmp list =
let pile_tops = Array.make (List.length list) [] in
let bsearch_piles x len =
let rec aux lo hi =
if lo > hi then
lo
else
let mid = (lo + hi) / 2 in
if cmp (List.hd pile_tops.(mid)) x < 0 then
aux (mid+1) hi
else
aux lo (mid-1)
in
aux 0 (len-1)
in
let f len x =
let i = bsearch_piles x len in
pile_tops.(i) <- x :: if i = 0 then [] else pile_tops.(i-1);
if i = len then len+1 else len
in
let len = List.fold_left f 0 list in
List.rev pile_tops.(len-1)
Usage:
# lis compare [3; 2; 6; 4; 5; 1];; - : int list = [2; 4; 5] # lis compare [0; 8; 4; 12; 2; 10; 6; 14; 1; 9; 5; 13; 3; 11; 7; 15];; - : int list = [0; 2; 6; 9; 11; 15]
Pascal
O(NLogN) version.
program LisDemo;
{$mode objfpc}{$h+}
uses
SysUtils;
function Lis(const A: array of Integer): specialize TArray<Integer>;
var
TailIndex: array of Integer;
function CeilIndex(Value, R: Integer): Integer;
var
L, M: Integer;
begin
L := 0;
while L < R do begin
{$PUSH}{$Q-}{$R-}M := (L + R) shr 1;{$POP}
if A[TailIndex[M]] < Value then L := M + 1
else R := M;
end;
Result := R;
end;
var
I, J, Len: Integer;
Parents: array of Integer;
begin
Result := nil;
if Length(A) = 0 then exit;
SetLength(TailIndex, Length(A));
SetLength(Parents, Length(A));
Len := 1;
for I := 1 to High(A) do
if A[I] < A[TailIndex[0]] then
TailIndex[0] := I
else
if A[TailIndex[Len-1]] < A[I] then begin
Parents[I] := TailIndex[Len - 1];
TailIndex[Len] := I;
Inc(Len);
end else begin
J := CeilIndex(A[I], Len - 1);
Parents[I] := TailIndex[J - 1];
TailIndex[J] := I;
end;
if Len < 2 then exit([A[0]]);
SetLength(Result, Len);
J := TailIndex[Len - 1];
for I := Len - 1 downto 0 do begin
Result[I] := A[J];
J := Parents[J];
end;
end;
procedure PrintArray(const A: array of Integer);
var
I: SizeInt;
begin
Write('[');
for I := 0 to High(A) - 1 do
Write(A[I], ', ');
WriteLn(A[High(A)], ']');
end;
begin
PrintArray(Lis([3, 2, 6, 4, 5, 1]));
PrintArray(Lis([0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]));
PrintArray(Lis([1, 1, 1, 1, 1, 0]));
end.
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15] [1]
Perl
Dynamic programming
use strict;
sub lis {
my @l = map [], 1 .. @_;
push @{$l[0]}, +$_[0];
for my $i (1 .. @_-1) {
for my $j (0 .. $i - 1) {
if ($_[$j] < $_[$i] and @{$l[$i]} < @{$l[$j]} + 1) {
$l[$i] = [ @{$l[$j]} ];
}
}
push @{$l[$i]}, $_[$i];
}
my ($max, $l) = (0, []);
for (@l) {
($max, $l) = (scalar(@$_), $_) if @$_ > $max;
}
return @$l;
}
print join ' ', lis 3, 2, 6, 4, 5, 1;
print join ' ', lis 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15;
- Output:
2 4 5 0 2 6 9 11 15
Patience sorting
sub lis {
my @pileTops;
# sort into piles
foreach my $x (@_) {
# binary search
my $low = 0, $high = $#pileTops;
while ($low <= $high) {
my $mid = int(($low + $high) / 2);
if ($pileTops[$mid]{val} >= $x) {
$high = $mid - 1;
} else {
$low = $mid + 1;
}
}
my $i = $low;
my $node = {val => $x};
$node->{back} = $pileTops[$i-1] if $i != 0;
$pileTops[$i] = $node;
}
my @result;
for (my $node = $pileTops[-1]; $node; $node = $node->{back}) {
push @result, $node->{val};
}
return reverse @result;
}
foreach my $r ([3, 2, 6, 4, 5, 1],
[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]) {
my @d = @$r;
my @lis = lis(@d);
print "an L.I.S. of [@d] is [@lis]\n";
}
- Output:
an L.I.S. of [3 2 6 4 5 1] is [2 4 5] an L.I.S. of [0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15] is [0 2 6 9 11 15]
Phix
Using the Wikipedia algorithm (converted to 1-based indexing)
with javascript_semantics function lis(sequence x, integer n = length(x)) sequence p = repeat(0,n), m = repeat(0,n) integer len = 0 for i=1 to n do integer lo = 1 integer hi = len while lo<=hi do integer mid = ceil((lo+hi)/2) if x[m[mid]]<x[i] then lo = mid + 1 else hi = mid - 1 end if end while if lo>1 then p[i] = m[lo-1] end if m[lo] = i if lo>len then len = lo end if end for sequence res = repeat(0,len) if len>0 then integer k = m[len] for i=len to 1 by -1 do res[i] = x[k] k = p[k] end for end if return res end function constant tests = {{3, 2, 6, 4, 5, 1}, {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}} for i=1 to length(tests) do ?lis(tests[i]) end for
- Output:
{2,4,5}
{0,2,6,9,11,15}
PHP
Patience sorting
<?php
class Node {
public $val;
public $back = NULL;
}
function lis($n) {
$pileTops = array();
// sort into piles
foreach ($n as $x) {
// binary search
$low = 0; $high = count($pileTops)-1;
while ($low <= $high) {
$mid = (int)(($low + $high) / 2);
if ($pileTops[$mid]->val >= $x)
$high = $mid - 1;
else
$low = $mid + 1;
}
$i = $low;
$node = new Node();
$node->val = $x;
if ($i != 0)
$node->back = $pileTops[$i-1];
$pileTops[$i] = $node;
}
$result = array();
for ($node = count($pileTops) ? $pileTops[count($pileTops)-1] : NULL;
$node != NULL; $node = $node->back)
$result[] = $node->val;
return array_reverse($result);
}
print_r(lis(array(3, 2, 6, 4, 5, 1)));
print_r(lis(array(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15)));
?>
- Output:
Array
(
[0] => 2
[1] => 4
[2] => 5
)
Array
(
[0] => 0
[1] => 2
[2] => 6
[3] => 9
[4] => 11
[5] => 15
)
Picat
Mode-directed tabling
table(+,+,max)
lis_mode(In, Out,OutLen) =>
one_is(In, [], Is),
Out = reverse(Is),
OutLen = Out.length.
one_is([], Current, Current2) => Current = Current2.
one_is([H|T], Current, Final) =>
( Current = [], one_is(T, [H], Final));
( Current = [H1|_], H1 @< H, one_is(T, [H|Current], Final));
one_is(T, Current, Final).Constraint modelling approach
For larger instances, the sat solver is generally faster than the cp solver.
lis_cp(S, Res,Z) =>
Len = S.len,
X = new_list(Len),
X :: 0..1,
increasing_except_0($[X[I]*S[I] : I in 1..Len]),
Z #= sum(X),
solve($[max(Z)],X),
% Extract the found LIS
Res = [S[I] : I in 1..Len, X[I] == 1].
%
% Ensures that array A is (strictly) increasing if we disregard any 0's
%
increasing_except_0(A) =>
N = A.len,
foreach(I in 1..N, J in I+1..N)
(A[I] #!= 0 #/\ A[J] #!= 0) #=> (A[I] #< A[J])
end.Test
import sat. % for lis_cp
% import cp. % Slower than sat on larger instances.
go =>
nolog,
Tests = [
[3,2,6,4,5,1],
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15],
[1,1,1,1],
[4,65,2,-31,0,99,83,782,1]
],
Funs = [lis_mode, lis_cp],
foreach(Fun in Funs)
println(fun=Fun),
foreach(Test in Tests)
call(Fun,Test,Lis,Len),
printf("%w: LIS=%w (len=%d)\n",Test, Lis,Len)
end,
nl,
end,
nl.- Output:
[3,2,6,4,5,1]: LIS=[3,4,5] (len=3) [0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]: LIS=[0,4,6,9,13,15] (len=6) [1,1,1,1]: LIS=[1] (len=1) [4,65,2,-31,0,99,83,782,1]: LIS=[4,65,99,782] (len=4)
The mode directed tabling tends to be the fastest of the two methods.
PicoLisp
Adapted patience sorting approach:
(de longinc (Lst)
(let (D NIL R NIL)
(for I Lst
(cond
((< I (last D))
(for (Y . X) D
(T (> X I) (set (nth D Y) I)) ) )
((< I (car R))
(set R I)
(when D (set (cdr R) (last D))) )
(T (when R (queue 'D (car R)))
(push 'R I) ) ) )
(flip R) ) )Original recursive glutton:
(de glutton (L)
(let N (pop 'L)
(maxi length
(recur (N L)
(ifn L
(list (list N))
(mapcan
'((R)
(if (> (car R) N)
(list (cons N R) R)
(list (list N) R) ) )
(recurse (car L) (cdr L)) ) ) ) ) ) )
(test (2 4 5)
(glutton (3 2 6 4 5 1)))
(test (2 6 9 11 15)
(glutton (8 4 12 2 10 6 14 1 9 5 13 3 11 7 15)))
(test (-31 0 83 782)
(glutton (4 65 2 -31 0 99 83 782 1)) )PowerShell
function Get-LongestSubsequence ( [int[]]$A )
{
If ( $A.Count -lt 2 ) { return $A }
# Start with an "empty" pile
# (We will only store the top value in each "pile".)
$Pile = @( [int]::MaxValue )
$Last = 0
# Hashtable to hold the back pointers
$BP = @{}
# For each number in the orginal sequence...
ForEach ( $N in $A )
{
# Find the first pile with a value greater than N
$i = 0..$Last | Where { $N -lt $Pile[$_] } | Select -First 1
# Place N on the pile
$Pile[$i] = $N
# Set the back pointer for this value to the value of the previous pile
$BP["$N"] = $Pile[$i-1]
# If this is the previously empty pile, add a new empty pile
If ( $i -eq $Last )
{
$Pile += @( [int]::MaxValue )
$Last++
}
}
# Ignore the empty pile
$Last--
# Start with the value of the last pile
$N = $Pile[$Last]
$S = @( $N )
# Add the remainder of the values by walking through the back pointers
ForEach ( $i in $Last..1 )
{
$S += ( $N = $BP["$N"] )
}
# Return the series (reversed into the correct order)
return $S[$Last..0]
}
( Get-LongestSubsequence 3, 2, 6, 4, 5, 1 ) -join ', '
( Get-LongestSubsequence 0, 8, 4, 12, 2, 10, 6, 16, 14, 1, 9, 5, 13, 3, 11, 7, 15 ) -join ', '
- Output:
2, 4, 5 0, 2, 6, 9, 11, 15
Prolog
Works with SWI-Prolog version 6.4.1
Naïve implementation.
lis(In, Out) :-
% we ask Prolog to find the longest sequence
aggregate(max(N,Is), (one_is(In, [], Is), length(Is, N)), max(_, Res)),
reverse(Res, Out).
% we describe the way to find increasing subsequence
one_is([], Current, Current).
one_is([H | T], Current, Final) :-
( Current = [], one_is(T, [H], Final));
( Current = [H1 | _], H1 < H, one_is(T, [H | Current], Final));
one_is(T, Current, Final).
Prolog finds the first longest subsequence
?- lis([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15], Out). Out = [0,4,6,9,13,15]. ?- lis([3,2,6,4,5,1], Out). Out = [3,4,5].
Python
Python: O(nlogn) Method from Wikipedia's LIS Article[1]
def longest_increasing_subsequence(X):
"""Returns the Longest Increasing Subsequence in the Given List/Array"""
N = len(X)
P = [0] * N
M = [0] * (N+1)
L = 0
for i in range(N):
lo = 1
hi = L
while lo <= hi:
mid = (lo+hi)//2
if (X[M[mid]] < X[i]):
lo = mid+1
else:
hi = mid-1
newL = lo
P[i] = M[newL-1]
M[newL] = i
if (newL > L):
L = newL
S = []
k = M[L]
for i in range(L-1, -1, -1):
S.append(X[k])
k = P[k]
return S[::-1]
if __name__ == '__main__':
for d in [[3,2,6,4,5,1], [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]]:
print('a L.I.S. of %s is %s' % (d, longest_increasing_subsequence(d)))
- Output:
a L.I.S. of [3, 2, 6, 4, 5, 1] is [2, 4, 5] a L.I.S. of [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] is [0, 2, 6, 9, 11, 15]
Python: Method from video
def longest_increasing_subsequence(d):
'Return one of the L.I.S. of list d'
l = []
for i in range(len(d)):
l.append(max([l[j] for j in range(i) if l[j][-1] < d[i]] or [[]], key=len)
+ [d[i]])
return max(l, key=len)
if __name__ == '__main__':
for d in [[3,2,6,4,5,1], [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]]:
print('a L.I.S. of %s is %s' % (d, longest_increasing_subsequence(d)))
- Output:
a L.I.S. of [3, 2, 6, 4, 5, 1] is [3, 4, 5] a L.I.S. of [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] is [0, 4, 6, 9, 13, 15]
Python: Patience sorting method
from collections import namedtuple
from functools import total_ordering
from bisect import bisect_left
@total_ordering
class Node(namedtuple('Node_', 'val back')):
def __iter__(self):
while self is not None:
yield self.val
self = self.back
def __lt__(self, other):
return self.val < other.val
def __eq__(self, other):
return self.val == other.val
def lis(d):
"""Return one of the L.I.S. of list d using patience sorting."""
if not d:
return []
pileTops = []
for di in d:
j = bisect_left(pileTops, Node(di, None))
pileTops[j:j+1] = [Node(di, pileTops[j-1] if j > 0 else None)]
return list(pileTops[-1])[::-1]
if __name__ == '__main__':
for d in [[3,2,6,4,5,1],
[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]]:
print('a L.I.S. of %s is %s' % (d, lis(d)))
- Output:
a L.I.S. of [3, 2, 6, 4, 5, 1] is [2, 4, 5] a L.I.S. of [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] is [0, 2, 6, 9, 11, 15]
Racket
Patience sorting. The program saves only the top card of each pile, with a link (cons) to the top of the previous pile at the time it was inserted. It uses binary search to find the correct pile.
#lang racket/base
(require data/gvector)
(define (gvector-last gv)
(gvector-ref gv (sub1 (gvector-count gv))))
(define (lis-patience-sort input-list)
(let ([piles (gvector)])
(for ([item (in-list input-list)])
(insert-item! piles item))
(reverse (gvector-last piles))))
(define (insert-item! piles item)
(if (zero? (gvector-count piles))
(gvector-add! piles (cons item '()))
(cond
[(not (<= item (car (gvector-last piles))))
(gvector-add! piles (cons item (gvector-last piles)))]
[(<= item (car (gvector-ref piles 0)))
(gvector-set! piles 0 (cons item '()))]
[else (let loop ([first 1] [last (sub1 (gvector-count piles))])
(if (= first last)
(gvector-set! piles first (cons item (gvector-ref piles (sub1 first))))
(let ([middle (quotient (+ first last) 2)])
(if (<= item (car (gvector-ref piles middle)))
(loop first middle)
(loop (add1 middle) last)))))])))
- Output:
'(2 4 5) '(0 2 6 9 11 15)
Raku
(formerly Perl 6)
Dynamic programming
Straight-forward implementation of the algorithm described in the video.
sub lis(@d) {
my @l = [].item xx @d;
@l[0].push: @d[0];
for 1 ..^ @d -> $i {
for ^$i -> $j {
if @d[$j] < @d[$i] && @l[$i] < @l[$j] + 1 {
@l[$i] = [ @l[$j][] ]
}
}
@l[$i].push: @d[$i];
}
return max :by(*.elems), @l;
}
say lis([3,2,6,4,5,1]);
say lis([0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15]);
- Output:
[2 4 5] [0 2 6 9 11 15]
Patience sorting
sub lis(@deck is copy) {
my @S = [@deck.shift() => Nil].item;
for @deck -> $card {
with first { @S[$_][*-1].key > $card }, ^@S -> $i {
@S[$i].push: $card => @S[$i-1][*-1] // Nil
} else {
@S.push: [ $card => @S[*-1][*-1] // Nil ].item
}
}
reverse map *.key, (
@S[*-1][*-1], *.value ...^ !*.defined
)
}
say lis <3 2 6 4 5 1>;
say lis <0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15>;
- Output:
[2 4 5] [0 2 6 9 11 15]
REXX
/*REXX program finds & displays the longest increasing subsequence from a list of #'s.*/
$.=; $.1= 3 2 6 4 5 1 /*define the 1st list to be examined. */
$.2= 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 /* " " 2nd " " " " */
do j=1 while $.j\==''; say /* [↓] process all of the list for LIS*/
say ' input: ' $.j /*display the (original) input list. */
call LIS $.j /*invoke the LIS function. */
say 'output: ' result /*display the output (result from LIS)*/
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
LIS: procedure; parse arg x; n= words(x); if n==0 then return ''
p.=; m.= p.
do #=1 to n; _= # - 1; @._= word(x, #) /*build an array (@) from input.*/
end /*#*/
L= 0
do j=0 to n-1; lo= 1
HI= L
do while LO<=HI; middle= (LO+HI) % 2
_= m.middle /*create a temporary value for @ index.*/
if @._<@.j then LO= middle + 1
else HI= middle - 1
end /*while*/
newLO= LO
_= newLO - 1 /*create a temporary value for M index.*/
p.j= m._
m.newLO= j
if newLO>L then L= newLO
end /*i*/
k= m.L; $= /* [↓] build a list for the result. */
do L; $= @.k $; k= p.k /*perform this DO loop L times. */
end /*i*/
return strip($) /*the result has an extra leading blank*/
- output when using the internal default input:
input: 3 2 6 4 5 1 output: 2 4 5 input: 0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15 output: 0 2 6 9 11 15
Ring
# Project : Longest increasing subsequence
tests = [[3, 2, 6, 4, 5, 1], [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]]
res = []
for x=1 to len(tests)
lis(tests[x])
showarray(res)
end
func lis(X)
N = len(X)
P = list(N)
M = list(N)
for nr = 1 to len(P)
P[nr] = 0
next
for nr = 1 to len(M)
P[nr] = 0
next
len = 0
for i=1 to N
lo = 1
hi = len
while lo <= hi
mid = floor((lo+hi)/2)
if X[M[mid]]<X[i]
lo = mid + 1
else
hi = mid - 1
ok
end
if lo>1
P[i] = M[lo-1]
ok
M[lo] = i
if lo>len
len = lo
ok
next
res = list(len)
if len>0
k = M[len]
for i=len to 1 step -1
res[i] = X[k]
k = P[k]
next
ok
return res
func showarray(vect)
see "{"
svect = ""
for n = 1 to len(vect)
svect = svect + vect[n] + ", "
next
svect = left(svect, len(svect) - 2)
see svect
see "}" + nlOutput:
{2, 4, 5}
{0, 2, 6, 9, 11, 15}
Ruby
Patience sorting
Node = Struct.new(:val, :back)
def lis(n)
pileTops = []
# sort into piles
for x in n
# binary search
low, high = 0, pileTops.size-1
while low <= high
mid = low + (high - low) / 2
if pileTops[mid].val >= x
high = mid - 1
else
low = mid + 1
end
end
i = low
node = Node.new(x)
node.back = pileTops[i-1] if i > 0
pileTops[i] = node
end
result = []
node = pileTops.last
while node
result.unshift(node.val)
node = node.back
end
result
end
p lis([3, 2, 6, 4, 5, 1])
p lis([0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15])
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Rust
fn lis(x: &[i32])-> Vec<i32> {
let n = x.len();
let mut m = vec![0; n];
let mut p = vec![0; n];
let mut l = 0;
for i in 0..n {
let mut lo = 1;
let mut hi = l;
while lo <= hi {
let mid = (lo + hi) / 2;
if x[m[mid]] <= x[i] {
lo = mid + 1;
} else {
hi = mid - 1;
}
}
let new_l = lo;
p[i] = m[new_l - 1];
m[new_l] = i;
if new_l > l {
l = new_l;
}
}
let mut o = vec![0; l];
let mut k = m[l];
for i in (0..l).rev() {
o[i] = x[k];
k = p[k];
}
o
}
fn main() {
let list = vec![3, 2, 6, 4, 5, 1];
println!("{:?}", lis(&list));
let list = vec![0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15];
println!("{:?}", lis(&list));
}
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Scala
Patience sorting
- Output:
See it in running in your browser by ScalaFiddle (JavaScript) or by Scastie (JVM).
object LongestIncreasingSubsequence extends App {
val tests = Map(
"3,2,6,4,5,1" -> Seq("2,4,5", "3,4,5"),
"0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15" -> Seq("0,2,6,9,11,15", "0,2,6,9,13,15", "0,4,6,9,13,15", "0,4,6,9,11,15")
)
def lis(l: Array[Int]): Seq[Array[Int]] =
if (l.length < 2) Seq(l)
else {
def increasing(done: Array[Int], remaining: Array[Int]): Seq[Array[Int]] =
if (remaining.isEmpty) Seq(done)
else
(if (remaining.head > done.last)
increasing(done :+ remaining.head, remaining.tail)
else Nil) ++ increasing(done, remaining.tail) // all increasing combinations
val all = (1 to l.length)
.flatMap(i => increasing(l take i takeRight 1, l.drop(i + 1)))
.sortBy(-_.length)
all.takeWhile(_.length == all.head.length) // longest of all increasing combinations
}
def asInts(s: String): Array[Int] = s split "," map (_.toInt)
assert(tests forall {
case (given, expect) =>
val allLongests: Seq[Array[Int]] = lis(asInts(given))
println(
s"$given has ${allLongests.length} longest increasing subsequences, e.g. ${
allLongests.last.mkString(",")}")
allLongests.forall(lis => expect.contains(lis.mkString(",")))
})
}
- Output:
3,2,6,4,5,1 has 2 longest increasing subsequences, e.g. 2,4,5 0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15 has 4 longest increasing subsequences, e.g. 0,2,6,9,11,15
Brute force solution
def powerset[A](s: List[A]) = (0 to s.size).map(s.combinations(_)).reduce(_++_)
def isSorted(l:List[Int])(f: (Int, Int) => Boolean) = l.view.zip(l.tail).forall(x => f(x._1,x._2))
def sequence(set: List[Int])(f: (Int, Int) => Boolean) = powerset(set).filter(_.nonEmpty).filter(x => isSorted(x)(f)).toList.maxBy(_.length)
sequence(set)(_<_)
sequence(set)(_>_)
Scheme
Patience sorting
(define (lis less? lst)
(define pile-tops (make-vector (length lst)))
(define (bsearch-piles x len)
(let aux ((lo 0)
(hi (- len 1)))
(if (> lo hi)
lo
(let ((mid (quotient (+ lo hi) 2)))
(if (less? (car (vector-ref pile-tops mid)) x)
(aux (+ mid 1) hi)
(aux lo (- mid 1)))))))
(let aux ((len 0)
(lst lst))
(if (null? lst)
(reverse (vector-ref pile-tops (- len 1)))
(let* ((x (car lst))
(i (bsearch-piles x len)))
(vector-set! pile-tops i (cons x (if (= i 0)
'()
(vector-ref pile-tops (- i 1)))))
(aux (if (= i len) (+ len 1) len) (cdr lst))))))
(display (lis < '(3 2 6 4 5 1))) (newline)
(display (lis < '(0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15))) (newline)
- Output:
(2 4 5) (0 2 6 9 11 15)
Sidef
Dynamic programming:
func lis(a) {
var l = a.len.of { [] }
l[0] << a[0]
for i in (1..a.end) {
for j in ^i {
if ((a[j] < a[i]) && (l[i].len < l[j].len+1)) {
l[i] = [l[j]...]
}
}
l[i] << a[i]
}
l.max_by { .len }
}
say lis(%i<3 2 6 4 5 1>)
say lis(%i<0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15>)
Patience sorting:
func lis(deck) {
var pileTops = []
deck.each { |x|
var low = 0;
var high = pileTops.end
while (low <= high) {
var mid = ((low + high) // 2)
if (pileTops[mid]{:val} >= x) {
high = mid-1
} else {
low = mid+1
}
}
var i = low
var node = Hash(val => x)
node{:back} = pileTops[i-1] if (i != 0)
pileTops[i] = node
}
var result = []
for (var node = pileTops[-1]; node; node = node{:back}) {
result << node{:val}
}
result.reverse
}
say lis(%i<3 2 6 4 5 1>)
say lis(%i<0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15>)
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
Standard ML
Patience sorting
fun lis cmp n =
let
val pile_tops = DynamicArray.array (length n, [])
fun bsearch_piles x =
let
fun aux (lo, hi) =
if lo > hi then
lo
else
let
val mid = (lo + hi) div 2
in
if cmp (hd (DynamicArray.sub (pile_tops, mid)), x) = LESS then
aux (mid+1, hi)
else
aux (lo, mid-1)
end
in
aux (0, DynamicArray.bound pile_tops)
end
fun f x =
let
val i = bsearch_piles x
in
DynamicArray.update (pile_tops, i,
x :: (if i = 0 then [] else DynamicArray.sub (pile_tops, i-1)))
end
in
app f n;
rev (DynamicArray.sub (pile_tops, DynamicArray.bound pile_tops))
end
Usage:
- lis Int.compare [3, 2, 6, 4, 5, 1]; val it = [2,4,5] : int list - lis Int.compare [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]; val it = [0,2,6,9,11,15] : int list
Swift
import Foundation
extension Array where Element: Comparable {
@inlinable
public func longestIncreasingSubsequence() -> [Element] {
var startI = [Int](repeating: 0, count: count)
var endI = [Int](repeating: 0, count: count + 1)
var len = 0
for i in 0..<count {
var lo = 1
var hi = len
while lo <= hi {
let mid = Int(ceil((Double(lo + hi)) / 2))
if self[endI[mid]] <= self[i] {
lo = mid + 1
} else {
hi = mid - 1
}
}
startI[i] = endI[lo-1]
endI[lo] = i
if lo > len {
len = lo
}
}
var s = [Element]()
var k = endI[len]
for _ in 0..<len {
s.append(self[k])
k = startI[k]
}
return s.reversed()
}
}
let l1 = [3, 2, 6, 4, 5, 1]
let l2 = [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]
print("\(l1) = \(l1.longestIncreasingSubsequence())")
print("\(l2) = \(l2.longestIncreasingSubsequence())")
- Output:
[3, 2, 6, 4, 5, 1] = [2, 4, 5] [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15] = [0, 2, 6, 9, 11, 15]
Swym
Based on the Python video solution. Interpreter at [[2]]
Array.'lis'
{
'stems' = Number.Array.mutableArray[ [] ]
forEach(this) 'value'->
{
'bestStem' = stems.where{==[] || .last < value}.max{.length}
stems.push( bestStem + [value] )
}
return stems.max{.length}
}
[3,2,6,4,5,1].lis.trace
[0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15].lis.trace- Output:
[3,4,5] [0,4,6,9,13,15]
Tcl
package require Tcl 8.6
proc longestIncreasingSubsequence {sequence} {
# Get the increasing subsequences (and their lengths)
set subseq [list 1 [lindex $sequence 0]]
foreach value $sequence {
set max {}
foreach {len item} $subseq {
if {[lindex $item end] < $value} {
if {[llength [lappend item $value]] > [llength $max]} {
set max $item
}
} elseif {![llength $max]} {
set max [list $value]
}
}
lappend subseq [llength $max] $max
}
# Pick the longest subsequence; -stride requires Tcl 8.6
return [lindex [lsort -stride 2 -index 0 $subseq] end]
}
Demonstrating:
puts [longestIncreasingSubsequence {3 2 6 4 5 1}]
puts [longestIncreasingSubsequence {0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15}]
- Output:
3 4 5 0 4 6 9 13 15
VBScript
Function LIS(arr)
n = UBound(arr)
Dim p()
ReDim p(n)
Dim m()
ReDim m(n)
l = 0
For i = 0 To n
lo = 1
hi = l
Do While lo <= hi
middle = Int((lo+hi)/2)
If arr(m(middle)) < arr(i) Then
lo = middle + 1
Else
hi = middle - 1
End If
Loop
newl = lo
p(i) = m(newl-1)
m(newl) = i
If newL > l Then
l = newl
End If
Next
Dim s()
ReDim s(l)
k = m(l)
For i = l-1 To 0 Step - 1
s(i) = arr(k)
k = p(k)
Next
LIS = Join(s,",")
End Function
WScript.StdOut.WriteLine LIS(Array(3,2,6,4,5,1))
WScript.StdOut.WriteLine LIS(Array(0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15))
- Output:
2,4,5, 0,2,6,9,11,15,
Wren
var longestIncreasingSubsequence = Fn.new { |x|
var n = x.count
if (n == 0) return []
if (n == 1) return x
var p = List.filled(n, 0)
var m = List.filled(n+1, 0)
var len = 0
for (i in 0...n) {
var lo = 1
var hi = len
while (lo <= hi) {
var mid = ((lo + hi)/2).ceil
if (x[m[mid]] < x[i]) {
lo = mid + 1
} else {
hi = mid - 1
}
}
var newLen = lo
p[i] = m[newLen - 1]
m[newLen] = i
if (newLen > len) len = newLen
}
var s = List.filled(len, 0)
var k = m[len]
for (i in len-1..0) {
s[i] = x[k]
k = p[k]
}
return s
}
var lists = [
[3, 2, 6, 4, 5, 1],
[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]
]
lists.each { |l| System.print(longestIncreasingSubsequence.call(l)) }
- Output:
[2, 4, 5] [0, 2, 6, 9, 11, 15]
zkl
fcn longestSequence(ns){ // based on Patience sorting
piles:=L();
backPtr:='wrap(np){ return(np-1,if(np) piles[np-1].len()-1 else -1) }; // maybe (-1,-1)
foreach n in (ns){ newPile:=True; // create list of sorted lists
foreach e,p in (piles.enumerate()){
if(n<p[-1][0]){
p.del(1,-1) // only need the first and last elements
.append(T(n,backPtr(e))); newPile=False;
break;
}
}
if(newPile) piles.append(L(T(n,backPtr(piles.len()))));
}
reg r=L(),p=-1,n=0;
do{ n,p=piles[p][n]; r.write(n); p,n=p; }while(p!=-1);
r.reverse()
}foreach ns in (T(T(1),T(3,2,6,4,5,1),T(4,65,2,-31,0,99,83,782,1),
T(0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15),"foobar")){
s:=longestSequence(ns);
println(s.len(),": ",s," from ",ns);
}- Output:
1: L(1) from L(1)
3: L(2,4,5) from L(3,2,6,4,5,1)
4: L(-31,0,83,782) from L(4,65,2,-31,0,99,83,782,1)
6: L(0,1,3,9,11,15) from L(0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15)
4: L("f","o","o","r") from foobar
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