Find first missing positive
Given an integer array nums (which may or may not be sorted), find the smallest missing positive integer.
Find first missing positive is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
- Task
- Example
- nums = [1,2,0], [3,4,-1,1], [7,8,9,11,12]
- output = 3, 2, 1
11l
V nums = [[1, 2, 0], [3, 4, -1, 1], [7, 8, 9, 11, 12]]
L(l) nums
L(n) 1..
I n !C l
print(l‘ -> ’n)
L.break- Output:
[1, 2, 0] -> 3 [3, 4, -1, 1] -> 2 [7, 8, 9, 11, 12] -> 1
Action!
DEFINE PTR="CARD"
BYTE FUNC Contains(INT ARRAY a INT len,x)
INT i
FOR i=0 TO len-1
DO
IF a(i)=x THEN
RETURN (1)
FI
OD
RETURN (0)
BYTE FUNC FindFirstPositive(INT ARRAY a INT len)
INT res
res=1
WHILE Contains(a,len,res)
DO
res==+1
OD
RETURN (res)
PROC PrintArray(INT ARRAY a INT len)
INT i
Put('[)
FOR i=0 TO len-1
DO
IF i>0 THEN Put(' ) FI
PrintI(a(i))
OD
Put('])
RETURN
PROC Test(PTR ARRAY arr INT ARRAY lengths INT count)
INT ARRAY a
INT i,len,first
FOR i=0 TO count-1
DO
a=arr(i) len=lengths(i)
PrintArray(a,len)
Print(" -> ")
first=FindFirstPositive(a,len)
PrintIE(first)
OD
RETURN
PROC Main()
DEFINE COUNT="5"
PTR ARRAY arr(COUNT)
INT ARRAY
lengths=[3 4 5 3 0],
a1=[1 2 0],
a2=[3 4 65535 1],
a3=[7 8 9 11 12],
a4=[65534 65530 65520]
arr(0)=a1 arr(1)=a2 arr(2)=a3
arr(3)=a4 arr(4)=a4
Test(arr,lengths,COUNT)
RETURN- Output:
Screenshot from Atari 8-bit computer
[1 2 0] -> 3 [3 4 -1 1] -> 2 [7 8 9 11 12] -> 1 [-2 -6 -16] -> 1 [] -> 1
ALGOL 68
Uses the observation in the J sample that the maximum possible minimum missing positive integer is one more than the length of the list.
BEGIN # find the lowest positive integer not present in various arrays #
# returns the lowest positive integer not present in r #
PROC min missing positive = ( []INT r )INT:
BEGIN
[]INT a = r[ AT 1 ]; # a is r wih lower bound 1 #
# as noted in the J sample, the maximum possible minimum #
# missing positive integer is one more than the length of the array #
# note the values between 1 and UPB a present in a #
[ 1 : UPB a ]BOOL present;
FOR i TO UPB a DO present[ i ] := FALSE OD;
FOR i TO UPB a DO
INT ai = a[ i ];
IF ai >= 1 AND ai <= UPB a THEN
present[ ai ] := TRUE
FI
OD;
# find the lowest value not in present #
INT result := UPB a + 1;
BOOL found := FALSE;
FOR i TO UPB a WHILE NOT found DO
IF NOT present[ i ] THEN
found := TRUE;
result := i
FI
OD;
result
END # min missing positive # ;
print( ( " ", whole( min missing positive( ( 1, 2, 0 ) ), 0 ) ) );
print( ( " ", whole( min missing positive( ( 3, 4, -1, 1 ) ), 0 ) ) );
print( ( " ", whole( min missing positive( ( 7, 8, 9, 11, 12 ) ), 0 ) ) )
END- Output:
3 2 1
APL
fmp ← ⊃(⍳1+⌈/)~⊢
- Output:
fmp¨ (1 2 0) (3 4 ¯1 1) (7 8 9 11 12) 3 2 1
AppleScript
Procedural
local output, aList, n
set output to {}
repeat with aList in {{1, 2, 0}, {3, 4, -1, 1}, {7, 8, 9, 11, 12}}
set n to 1
repeat while (aList contains n)
set n to n + 1
end repeat
set end of output to n
end repeat
return output
- Output:
{3, 2, 1}
Functional
Defining the value required in terms of pre-existing generic primitives:
--------------- FIRST MISSING NATURAL NUMBER -------------
-- firstGap :: [Int] -> Int
on firstGap(xs)
script p
on |λ|(x)
xs does not contain x
end |λ|
end script
find(p, enumFrom(1))
end firstGap
--------------------------- TEST -------------------------
on run
script test
on |λ|(xs)
showList(xs) & " -> " & (firstGap(xs) as string)
end |λ|
end script
unlines(map(test, ¬
{{1, 2, 0}, {3, 4, -1, 1}, {7, 8, 9, 11, 12}}))
--> {1, 2, 0} -> 3
--> {3, 4, -1, 1} -> 2
--> {7, 8, 9, 11, 12} -> 1
end run
------------------------- GENERIC ------------------------
-- enumFrom :: Enum a => a -> [a]
on enumFrom(x)
script
property v : missing value
on |λ|()
if missing value is not v then
set v to 1 + v
else
set v to x
end if
return v
end |λ|
end script
end enumFrom
-- find :: (a -> Bool) -> Gen [a] -> Maybe a
on find(p, gen)
-- The first match for the predicate p
-- in the generator stream gen, or missing value
-- if no match is found.
set mp to mReturn(p)
set v to gen's |λ|()
repeat until missing value is v or (|λ|(v) of mp)
set v to (|λ|() of gen)
end repeat
v
end find
-- intercalate :: String -> [String] -> String
on intercalate(delim, xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, delim}
set s to xs as text
set my text item delimiters to dlm
s
end intercalate
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
-- The list obtained by applying f
-- to each element of xs.
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- showList :: [a] -> String
on showList(xs)
script go
on |λ|(x)
x as string
end |λ|
end script
"{" & intercalate(", ", map(go, xs)) & "}"
end showList
-- unlines :: [String] -> String
on unlines(xs)
-- A single string formed by the intercalation
-- of a list of strings with the newline character.
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set s to xs as text
set my text item delimiters to dlm
s
end unlines
- Output:
{1, 2, 0} -> 3
{3, 4, -1, 1} -> 2
{7, 8, 9, 11, 12} -> 1
AutoHotkey
First_Missing_Positive(obj){
Arr := [], i := 0
for k, v in obj
Arr[v] := true
while (++i<Max(obj*))
if !Arr[i]{
m := i
break
}
m := m ? m : Max(obj*) + 1
return m>0 ? m : 1
}
Examples:
nums := [[1,2,0], [3,4,-1,1], [7,8,9,11,12], [-4,-2,-3], []]
for i, obj in nums{
m := First_Missing_Positive(obj)
output .= m ", "
}
MsgBox % Trim(output, ", ")
return
- Output:
3, 2, 1, 1, 1
AWK
# syntax: GAWK -f FIND_FIRST_MISSING_POSITIVE.AWK
BEGIN {
PROCINFO["sorted_in"] = "@ind_num_asc"
nums = "[1,2,0],[3,4,-1,1],[7,8,9,11,12]"
printf("%s : ",nums)
n = split(nums,arr1,"],?") - 1
for (i=1; i<=n; i++) {
gsub(/[\[\]]/,"",arr1[i])
split(arr1[i],arr2,",")
for (j in arr2) {
arr3[arr2[j]]++
}
for (j in arr3) {
if (j <= 0) {
continue
}
if (!(1 in arr3)) {
ans = 1
break
}
if (!(j+1 in arr3)) {
ans = j + 1
break
}
}
printf("%d ",ans)
delete arr3
}
printf("\n")
exit(0)
}
- Output:
[1,2,0],[3,4,-1,1],[7,8,9,11,12] : 3 2 1
BASIC
10 DEFINT A-Z: DIM D(100)
20 READ X
30 FOR A=1 TO X
40 READ N
50 FOR I=1 TO N
60 READ D(I)
70 PRINT D(I);
80 NEXT I
90 PRINT "==>";
100 M=1
110 FOR I=1 TO N
120 IF M<D(I) THEN M=D(I)
130 NEXT I
140 FOR I=1 TO M+1
150 FOR J=1 TO N
160 IF D(J)=I GOTO 200
170 NEXT J
180 PRINT I
190 GOTO 210
200 NEXT I
210 NEXT A
220 DATA 3
230 DATA 3, 1,2,0
240 DATA 4, 3,4,-1,1
250 DATA 5, 7,8,9,11,12
- Output:
1 2 0 ==> 3 3 4 -1 1 ==> 2 7 8 9 11 12 ==> 1
BCPL
get "libhdr"
let max(v, n) = valof
$( let x = !v
for i=1 to n-1
if x<v!i do x := v!i
resultis x
$)
let contains(v, n, el) = valof
$( for i=0 to n-1
if v!i=el resultis true
resultis false
$)
let fmp(v, n) = valof
for i=1 to max(v,n)+1
unless contains(v,n,i) resultis i
let show(len, v) be
$( for i=0 to len-1 do writef("%N ", v!i)
writef("==> %N*N", fmp(v, len))
$)
let start() be
$( show(3, table 1,2,0)
show(4, table 3,4,-1,1)
show(5, table 7,8,9,11,12)
$)- Output:
1 2 0 ==> 3 3 4 -1 1 ==> 2 7 8 9 11 12 ==> 1
BQN
FMP ← ⊢(⊑(¬∊˜ )/⊢)1+(↕1+⌈´)
FMP¨ ⟨⟨1,2,0⟩,⟨3,4,¯1,1⟩,⟨7,8,9,11,12⟩⟩
- Output:
⟨ 3 2 1 ⟩
CLU
contains = proc [T, U: type] (needle: T, haystack: U) returns (bool)
where T has equal: proctype (T,T) returns (bool),
U has elements: itertype (U) yields (T)
for item: T in U$elements(haystack) do
if item = needle then return(true) end
end
return(false)
end contains
fmp = proc [T: type] (list: T) returns (int)
where T has elements: itertype (T) yields (int)
n: int := 1
while contains[int, T](n, list) do
n := n + 1
end
return(n)
end fmp
start_up = proc ()
si = sequence[int]
ssi = sequence[si]
po: stream := stream$primary_output()
tests: ssi := ssi$[si$[1,2,0], si$[3,4,-1,1], si$[7,8,9,11,12]]
for test: si in ssi$elements(tests) do
for i: int in si$elements(test) do
stream$puts(po, int$unparse(i) || " ")
end
stream$putl(po, "==> " || int$unparse(fmp[si](test)))
end
end start_up- Output:
1 2 0 ==> 3 3 4 -1 1 ==> 2 7 8 9 11 12 ==> 1
F#
// Find first missing positive. Nigel Galloway: February 15., 2021
let fN g=let g=0::g|>List.filter((<) -1)|>List.sort|>List.distinct
match g|>List.pairwise|>List.tryFind(fun(n,g)->g>n+1) with Some(n,_)->n+1 |_->List.max g+1
[[1;2;0];[3;4;-1;1];[7;8;9;11;12]]|>List.iter(fN>>printf "%d "); printfn ""
- Output:
3 2 1
Factor
USING: formatting fry hash-sets kernel math sequences sets ;
: first-missing ( seq -- n )
>hash-set 1 swap '[ dup _ in? ] [ 1 + ] while ;
{ { 1 2 0 } { 3 4 1 1 } { 7 8 9 11 12 } { 1 2 3 4 5 }
{ -6 -5 -2 -1 } { 5 -5 } { -2 } { 1 } { } }
[ dup first-missing "%u ==> %d\n" printf ] each
- Output:
{ 1 2 0 } ==> 3
{ 3 4 1 1 } ==> 2
{ 7 8 9 11 12 } ==> 1
{ 1 2 3 4 5 } ==> 6
{ -6 -5 -2 -1 } ==> 1
{ 5 -5 } ==> 1
{ -2 } ==> 1
{ 1 } ==> 2
{ } ==> 1
FreeBASIC
function is_in( n() as integer, k as uinteger ) as boolean
for i as uinteger = 1 to ubound(n)
if n(i) = k then return true
next i
return false
end function
function fmp( n() as integer ) as integer
dim as uinteger i = 1
while is_in( n(), i )
i+=1
wend
return i
end function
dim as integer a(1 to 3) = {1, 2, 0}
dim as integer b(1 to 4) = {3, 4, -1, 1}
dim as integer c(1 to 5) = {7, 8, 9, 11, 12}
print fmp(a())
print fmp(b())
print fmp(c())- Output:
3 2 1
Go
package main
import (
"fmt"
"sort"
)
func firstMissingPositive(a []int) int {
var b []int
for _, e := range a {
if e > 0 {
b = append(b, e)
}
}
sort.Ints(b)
le := len(b)
if le == 0 || b[0] > 1 {
return 1
}
for i := 1; i < le; i++ {
if b[i]-b[i-1] > 1 {
return b[i-1] + 1
}
}
return b[le-1] + 1
}
func main() {
fmt.Println("The first missing positive integers for the following arrays are:\n")
aa := [][]int{
{1, 2, 0}, {3, 4, -1, 1}, {7, 8, 9, 11, 12}, {1, 2, 3, 4, 5},
{-6, -5, -2, -1}, {5, -5}, {-2}, {1}, {}}
for _, a := range aa {
fmt.Println(a, "->", firstMissingPositive(a))
}
}
- Output:
The first missing positive integers for the following arrays are: [1 2 0] -> 3 [3 4 -1 1] -> 2 [7 8 9 11 12] -> 1 [1 2 3 4 5] -> 6 [-6 -5 -2 -1] -> 1 [5 -5] -> 1 [-2] -> 1 [1] -> 2 [] -> 1
Haskell
import Data.List (sort)
task :: Integral a => [a] -> a
task = go . (0 :) . sort . filter (> 0)
where
go [x] = succ x
go (w : xs@(x : _))
| x == succ w = go xs
| otherwise = succ w
main :: IO ()
main =
print $
map
task
[[1, 2, 0], [3, 4, -1, 1], [7, 8, 9, 11, 12]]
- Output:
[3,2,1]
Or simply as a filter over an infinite list:
---------- FIRST MISSING POSITIVE NATURAL NUMBER ---------
firstGap :: [Int] -> Int
firstGap xs = head $ filter (`notElem` xs) [1 ..]
--------------------------- TEST -------------------------
main :: IO ()
main =
(putStrLn . unlines) $
fmap
(\xs -> show xs <> " -> " <> (show . firstGap) xs)
[ [1, 2, 0],
[3, 4, -1, 1],
[7, 8, 9, 11, 12]
]
and if xs were large, it could be defined as a set:
import Data.Set (fromList, notMember)
---------- FIRST MISSING POSITIVE NATURAL NUMBER ---------
firstGap :: [Int] -> Int
firstGap xs = head $ filter (`notMember` seen) [1 ..]
where
seen = fromList xs
- Output:
Same output for notElem and notMember versions above:
[1,2,0] -> 3 [3,4,-1,1] -> 2 [7,8,9,11,12] -> 1
J
The first missing positive can be no larger than 1 plus the length of the list, thus:
fmp=: {{ {.y-.~1+i.1+#y }}S:0
(The {{ }} delimiters on definitions, used here, was introduced in J version 9.2)
Example use:
fmp 1 2 0;3 4 _1 1; 7 8 9 11 12
3 2 1
Also, with this approach:
fmp 'abc'
1
JavaScript
(() => {
"use strict";
// ---------- FIRST MISSING NATURAL NUMBER -----------
// firstGap :: [Int] -> Int
const firstGap = xs => {
const seen = new Set(xs);
return filterGen(
x => !seen.has(x)
)(
enumFrom(1)
)
.next()
.value;
};
// ---------------------- TEST -----------------------
// main :: IO ()
const main = () => [
[1, 2, 0],
[3, 4, -1, 1],
[7, 8, 9, 11, 12]
]
.map(xs => `${show(xs)} -> ${firstGap(xs)}`)
.join("\n");
// --------------------- GENERIC ---------------------
// enumFrom :: Int -> [Int]
const enumFrom = function* (x) {
// A non-finite succession of
// integers, starting with n.
let v = x;
while (true) {
yield v;
v = 1 + v;
}
};
// filterGen :: (a -> Bool) -> Gen [a] -> Gen [a]
const filterGen = p =>
// A stream of values which are drawn
// from a generator, and satisfy p.
xs => {
const go = function* () {
let x = xs.next();
while (!x.done) {
const v = x.value;
if (p(v)) {
yield v;
}
x = xs.next();
}
};
return go(xs);
};
// show :: a -> String
const show = x => JSON.stringify(x);
// MAIN ---
return main();
})();
- Output:
[1,2,0] -> 3 [3,4,-1,1] -> 2 [7,8,9,11,12] -> 1
jq
Works with gojq, the Go implementation of jq
In case the target array is very long, it probably makes sense either to sort it, or to use a hash, for quick lookup. For the sake of illustration, we'll use a hash:
# input: an array of integers
def first_missing_positive:
INDEX(.[]; tostring) as $dict
| first(range(1; infinite)
| . as $n
| select($dict|has($n|tostring)|not) ) ;
def examples:
[1,2,0], [3,4,-1,1], [7,8,9,11,12], [-5, -2, -6, -1];
# The task:
examples
| "\(first_missing_positive) is missing from \(.)"- Output:
3 is missing from [1,2,0] 2 is missing from [3,4,-1,1] 1 is missing from [7,8,9,11,12] 1 is missing from [-5,-2,-6,-1]
Julia
for array in [[1,2,0], [3,4,-1,1], [7,8,9,11,12], [-5, -2, -6, -1]]
for n in 1:typemax(Int)
!(n in array) && (println("$array => $n"); break)
end
end
- Output:
[1, 2, 0] => 3 [3, 4, -1, 1] => 2 [7, 8, 9, 11, 12] => 1 [-5, -2, -6, -1] => 1
Nim
In order to avoid the O(n) search in arrays, we could use an intermediate set built from the sequence. But this is useless with the chosen examples.
for a in [@[1, 2, 0], @[3, 4, -1, 1], @[7, 8, 9, 11, 12], @[-5, -2, -6, -1]]:
for n in 1..int.high:
if n notin a:
echo a, " → ", n
break
- Output:
@[1, 2, 0] → 3 @[3, 4, -1, 1] → 2 @[7, 8, 9, 11, 12] → 1 @[-5, -2, -6, -1] → 1
Perl
#!/usr/bin/perl -l
use strict;
use warnings;
use List::Util qw( first );
my @tests = ( [1,2,0], [3,4,-1,1], [7,8,9,11,12],
[3, 4, 1, 1], [1, 2, 3, 4, 5], [-6, -5, -2, -1], [5, -5], [-2], [1], []);
for my $test ( @tests )
{
print "[ @$test ] ==> ",
first { not { map { $_ => 1 } @$test }->{$_} } 1 .. @$test + 1;
}
- Output:
[ 1 2 0 ] ==> 3 [ 3 4 -1 1 ] ==> 2 [ 7 8 9 11 12 ] ==> 1 [ 3 4 1 1 ] ==> 2 [ 1 2 3 4 5 ] ==> 6 [ -6 -5 -2 -1 ] ==> 1 [ 5 -5 ] ==> 1 [ -2 ] ==> 1 [ 1 ] ==> 2 [ ] ==> 1
Phix
with javascript_semantics procedure test(sequence s) for missing=1 to length(s)+1 do if not find(missing,s) then printf(1,"%v -> %v\n",{s,missing}) exit end if end for end procedure papply({{1,2,0},{3,4,-1,1},{7,8,9,11,12},{1,2,3,4,5},{-6,-5,-2,-1},{5,-5},{-2},{1},{}} ,test)
- Output:
{1,2,0} -> 3
{3,4,-1,1} -> 2
{7,8,9,11,12} -> 1
{1,2,3,4,5} -> 6
{-6,-5,-2,-1} -> 1
{5,-5} -> 1
{-2} -> 1
{1} -> 2
{} -> 1
Python
'''First missing natural number'''
from itertools import count
# firstGap :: [Int] -> Int
def firstGap(xs):
'''First natural number not found in xs'''
return next(x for x in count(1) if x not in xs)
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''First missing natural number in each list'''
print('\n'.join([
f'{repr(xs)} -> {firstGap(xs)}' for xs in [
[1, 2, 0],
[3, 4, -1, 1],
[7, 8, 9, 11, 12]
]
]))
# MAIN ---
if __name__ == '__main__':
main()
- Output:
[1, 2, 0] -> 3 [3, 4, -1, 1] -> 2 [7, 8, 9, 11, 12] -> 1
QBasic
DECLARE FUNCTION isin (n(), k)
DECLARE FUNCTION fmp (n())
DIM a(3)
FOR x = 1 TO UBOUND(a): READ a(x): NEXT x
DIM b(4)
FOR x = 1 TO UBOUND(b): READ b(x): NEXT x
DIM c(5)
FOR x = 1 TO UBOUND(c): READ c(x): NEXT x
PRINT fmp(a())
PRINT fmp(b())
PRINT fmp(c())
Sleep
END
DATA 1,2,0
DATA 3,4,-1,1
DATA 7,8,9,11,12
FUNCTION fmp (n())
j = 1
WHILE isin(n(), j)
j = j + 1
WEND
fmp = j
END FUNCTION
FUNCTION isin (n(), k)
FOR i = LBOUND(n) TO UBOUND(n)
IF n(i) = k THEN isin = 1
NEXT i
END FUNCTION
- Output:
3 2 1
Quackery
Using a bitmap as a set
Treat a number (BigInt) as a set of integers. Add the positive integers to the set, then find the first positive integer not in the set.
[ 0 0 rot
witheach
[ dup 0 > iff
[ bit | ]
else drop ]
[ dip 1+
1 >> dup 1 &
0 = until ]
drop ] is task ( [ --> n )
' [ [ 1 2 0 ] [ 3 4 -1 1 ] [ 7 8 9 11 12 ] ]
witheach [ task echo sp ]- Output:
3 2 1
Using filtering and sorting
Filter out the non-positive integers, and then non-unique elements (after adding zero).
uniquewith is defined at Remove duplicate elements#Quackery and conveniently sorts the nest.
Then hunt for the first item which does not have the same value as its index.
[ dup size [] rot
witheach
[ dup 0 > iff
join
else drop ]
0 join
uniquewith >
witheach
[ i^ != if
[ drop i^
conclude ] ] ] is task ( [ -> n )
' [ [ 1 2 0 ] [ 3 4 -1 1 ] [ 7 8 9 11 12 ] ]
witheach [ task echo sp ]- Output:
3 2 1
Raku
say $_, " ==> ", (1..Inf).first( -> \n { n ∉ $_ } ) for
[ 1, 2, 0], [3, 4, 1, 1], [7, 8, 9, 11, 12], [1, 2, 3, 4, 5], [-6, -5, -2, -1], [5, -5], [-2], [1], []
- Output:
[1 2 0] ==> 3 [3 4 1 1] ==> 2 [7 8 9 11 12] ==> 1 [1 2 3 4 5] ==> 6 [-6 -5 -2 -1] ==> 1 [5 -5] ==> 1 [-2] ==> 1 [1] ==> 2 [] ==> 1
REXX
This REXX version doesn't need to sort the elements of the sets, it uses an associated array.
/*REXX program finds the smallest missing positive integer in a given list of integers. */
parse arg a /*obtain optional arguments from the CL*/
if a='' | a="," then a= '[1,2,0] [3,4,-1,1] [7,8,9,11,12] [1,2,3,4,5]' ,
"[-6,-5,-2,-1] [5,-5] [-2] [1] []" /*maybe use defaults.*/
say 'the smallest missing positive integer for the following sets is:'
say
do j=1 for words(a) /*process each set in a list of sets.*/
set= translate( word(a, j), ,'],[') /*extract a " from " " " " */
#= words(set) /*obtain the number of elements in set.*/
@.= . /*assign default value for set elements*/
do k=1 for #; x= word(set, k) /*obtain a set element (an integer). */
@.x= x /*assign it to a sparse array. */
end /*k*/
do m=1 for # until @.m==. /*now, search for the missing integer. */
end /*m*/
if @.m=='' then m= 1 /*handle the case of a "null" set. */
say right( word(a, j), 40) ' ───► ' m /*show the set and the missing integer.*/
end /*j*/ /*stick a fork in it, we're all done. */
- output when using the default inputs:
the smallest missing positive integer for the following sets is:
[1,2,0] ───► 3
[3,4,-1,1] ───► 2
[7,8,9,11,12] ───► 1
[1,2,3,4,5] ───► 6
[-6,-5,-2,-1] ───► 1
[5,-5] ───► 1
[-2] ───► 1
[1] ───► 2
[] ───► 1
Ring
nums = [[1,2,0], [3,4,-1,1], [7,8,9,11,12], [1,2,3,4,5],
[-6,-5,-2,-1], [5,-5], [-2], [1], []]
for n = 1 to len(nums)
see "the smallest missing positive integer for "
? (arrayToStr(nums[n]) + ": " + fmp(nums[n]))
next
func fmp(ary)
if len(ary) > 0
for m = 1 to max(ary) + 1
if find(ary, m) < 1 return m ok
next ok return 1
func arrayToStr(ary)
res = "[" s = ","
for n = 1 to len(ary)
if n = len(ary) s = "" ok
res += "" + ary[n] + s
next return res + "]"- Output:
the smallest missing positive integer for [1,2,0]: 3 the smallest missing positive integer for [3,4,-1,1]: 2 the smallest missing positive integer for [7,8,9,11,12]: 1 the smallest missing positive integer for [1,2,3,4,5]: 6 the smallest missing positive integer for [-6,-5,-2,-1]: 1 the smallest missing positive integer for [5,-5]: 1 the smallest missing positive integer for [-2]: 1 the smallest missing positive integer for [1]: 2 the smallest missing positive integer for []: 1
Ruby
nums = [1,2,0], [3,4,-1,1], [7,8,9,11,12]
puts nums.map{|ar|(1..).find{|candidate| !ar.include?(candidate) }}.join(", ")
- Output:
3, 2, 1
True BASIC
FUNCTION isin (n(), k)
FOR i = LBOUND(n) TO UBOUND(n)
IF n(i) = k THEN LET isin = 1
NEXT i
END FUNCTION
FUNCTION fmp (n())
LET j = 1
DO WHILE isin(n(), j) = 1
LET j = j + 1
LOOP
LET fmp = j
END FUNCTION
DIM a(3)
FOR x = 1 TO UBOUND(a)
READ a(x)
NEXT x
DIM b(4)
FOR x = 1 TO UBOUND(b)
READ b(x)
NEXT x
DIM c(5)
FOR x = 1 TO UBOUND(c)
READ c(x)
NEXT x
PRINT fmp(a())
PRINT fmp(b())
PRINT fmp(c())
DATA 1,2,0
DATA 3,4,-1,1
DATA 7,8,9,11,12
END
V (Vlang)
fn first_missing_positive(a []int) int {
mut b := []int{}
for e in a {
if e > 0 {
b << e
}
}
b.sort<int>()
le := b.len
if le == 0 || b[0] > 1 {
return 1
}
for i in 1..le {
if b[i]-b[i-1] > 1 {
return b[i-1] + 1
}
}
return b[le-1] + 1
}
fn main() {
println("The first missing positive integers for the following arrays are:\n")
aa := [
[1, 2, 0], [3, 4, -1, 1], [7, 8, 9, 11, 12], [1, 2, 3, 4, 5],
[-6, -5, -2, -1], [5, -5], [-2], [1]]
for a in aa {
println("$a -> ${first_missing_positive(a)}")
}
}- Output:
Same as go entry
Wren
import "/sort" for Sort
var firstMissingPositive = Fn.new { |a|
var b = a.where { |i| i > 0 }.toList
Sort.insertion(b)
if (b.isEmpty || b[0] > 1) return 1
var i = 1
while (i < b.count) {
if (b[i] - b[i-1] > 1) return b[i-1] + 1
i = i + 1
}
return b[-1] + 1
}
System.print("The first missing positive integers for the following arrays are:\n")
var aa = [
[ 1, 2, 0], [3, 4, -1, 1], [7, 8, 9, 11, 12], [1, 2, 3, 4, 5],
[-6, -5, -2, -1], [5, -5], [-2], [1], []
]
for (a in aa) System.print("%(a) -> %(firstMissingPositive.call(a))")- Output:
The first missing positive integers for the following arrays are: [1, 2, 0] -> 3 [3, 4, -1, 1] -> 2 [7, 8, 9, 11, 12] -> 1 [1, 2, 3, 4, 5] -> 6 [-6, -5, -2, -1] -> 1 [5, -5] -> 1 [-2] -> 1 [1] -> 2 [] -> 1
XPL0
proc ShowMissing(Arr, Len);
int Arr, Len, N, N0, I;
[N:= 1;
repeat N0:= N;
for I:= 0 to Len-1 do
if Arr(I) = N then N:= N+1;
until N = N0;
IntOut(0, N); ChOut(0, ^ );
];
int I, Nums;
[for I:= 0 to 2 do
[Nums:= [[1,2,0], [3,4,-1,1], [7,8,9,11,12], [0]];
ShowMissing( Nums(I), (Nums(I+1)-Nums(I))/4 );
];
]- Output:
3 2 1