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Next highest int from digits

From Rosetta Code
Task
Next highest int from digits
You are encouraged to solve this task according to the task description, using any language you may know.

Given a zero or positive integer, the task is to generate the next largest integer using only the given digits*1.

  •   Numbers will not be padded to the left with zeroes.
  •   Use all given digits, with their given multiplicity. (If a digit appears twice in the input number, it should appear twice in the result).
  •   If there is no next highest integer return zero.


*1   Alternatively phrased as:   "Find the smallest integer larger than the (positive or zero) integer   N
which can be obtained by reordering the (base ten) digits of   N".


Algorithm 1
  1.   Generate all the permutations of the digits and sort into numeric order.
  2.   Find the number in the list.
  3.   Return the next highest number from the list.


The above could prove slow and memory hungry for numbers with large numbers of digits, but should be easy to reason about its correctness.


Algorithm 2
  1.   Scan right-to-left through the digits of the number until you find a digit with a larger digit somewhere to the right of it.
  2.   Exchange that digit with the digit on the right that is both more than it, and closest to it.
  3.   Order the digits to the right of this position, after the swap; lowest-to-highest, left-to-right. (I.e. so they form the lowest numerical representation)


E.g.:

    n = 12453
<scan>
    12_4_53
<swap>
    12_5_43
<order-right>
    12_5_34

    return: 12534

This second algorithm is faster and more memory efficient, but implementations may be harder to test.

One method of testing, (as used in developing the task),   is to compare results from both algorithms for random numbers generated from a range that the first algorithm can handle.


Task requirements

Calculate the next highest int from the digits of the following numbers:

  •   0
  •   9
  •   12
  •   21
  •   12453
  •   738440
  •   45072010
  •   95322020


Optional stretch goal
  •   9589776899767587796600



AutoHotkey[edit]

Permutation Version[edit]

Next_highest_int(num){
Arr := []
for i, v in permute(num)
Arr[v] := true
for n, v in Arr
if found
return n
else if (n = num)
found := true
return 0
}
permute(str, k:=0, l:=1){
static res:=[]
r := StrLen(str)
k := k ? k : r
if (l = r)
return SubStr(str, 1, k)
i := l
while (i <= r){
str := swap(str, l, i)
x := permute(str, k, l+1)
if (x<>"")
res.push(x)
str := swap(str, l, i++)
}
if (l=1)
return x:=res, res := []
}
swap(str, l, i){
x := StrSplit(str), var := x[l], x[l] := x[i], x[i] := var
Loop, % x.count()
res .= x[A_Index]
return res
}
Examples:
MsgBox % "" Next_highest_int(0)
. "`n" Next_highest_int(9)
. "`n" Next_highest_int(12)
. "`n" Next_highest_int(21)
. "`n" Next_highest_int(12453)
. "`n" Next_highest_int(738440)
. "`n" Next_highest_int(45072010)
. "`n" Next_highest_int(95322020)
Output:
0
0
21
0
12534
740348
45072100
95322200

Scanning Version[edit]

Next_highest_int(num){
Loop % StrLen(num){
i := A_Index
if (left := SubStr(num, 0-i, 1)) < (right := SubStr(num, 1-i, 1))
break
}
if !(left < right)
return 0
x := StrLen(num) - i
num := swap(num, x, x+1)
Rdigits := rSort(SubStr(num, 1-i))
return SubStr(num,1, StrLen(num)-StrLen(Rdigits)) . Rdigits
}
swap(str, l, i){
x := StrSplit(str), var := x[l], x[l] := x[i], x[i] := var
Loop, % x.count()
res .= x[A_Index]
return res
}
rSort(num){
Arr := []
for i, v in StrSplit(num)
Arr[v, i] := v
for i, obj in Arr
for k, v in obj
res .= v
return res
}
Examples:
MsgBox % "" Next_highest_int(0)
. "`n" Next_highest_int(9)
. "`n" Next_highest_int(12)
. "`n" Next_highest_int(21)
. "`n" Next_highest_int(12453)
. "`n" Next_highest_int(738440)
. "`n" Next_highest_int(45072010)
. "`n" Next_highest_int(95322020)
. "`n" Next_highest_int("9589776899767587796600") ; pass long numbers as text (between quotes)
Output:
0
0
21
0
12534
780344
45072100
95322200
9589776899767587900667

C[edit]

#include <stdbool.h>
#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
 
void swap(char* str, int i, int j) {
char c = str[i];
str[i] = str[j];
str[j] = c;
}
 
void reverse(char* str, int i, int j) {
for (; i < j; ++i, --j)
swap(str, i, j);
}
 
bool next_permutation(char* str) {
int len = strlen(str);
if (len < 2)
return false;
for (int i = len - 1; i > 0; ) {
int j = i, k;
if (str[--i] < str[j]) {
k = len;
while (str[i] >= str[--k]) {}
swap(str, i, k);
reverse(str, j, len - 1);
return true;
}
}
return false;
}
 
uint32_t next_highest_int(uint32_t n) {
char str[16];
snprintf(str, sizeof(str), "%u", n);
if (!next_permutation(str))
return 0;
return strtoul(str, 0, 10);
}
 
int main() {
uint32_t numbers[] = {0, 9, 12, 21, 12453, 738440, 45072010, 95322020};
const int count = sizeof(numbers)/sizeof(int);
for (int i = 0; i < count; ++i)
printf("%d -> %d\n", numbers[i], next_highest_int(numbers[i]));
// Last one is too large to convert to an integer
const char big[] = "9589776899767587796600";
char next[sizeof(big)];
memcpy(next, big, sizeof(big));
next_permutation(next);
printf("%s -> %s\n", big, next);
return 0;
}
Output:
0 -> 0
9 -> 0
12 -> 21
21 -> 0
12453 -> 12534
738440 -> 740348
45072010 -> 45072100
95322020 -> 95322200
9589776899767587796600 -> 9589776899767587900667

C++[edit]

Translation of: Factor
Library: GMP

This solution makes use of std::next_permutation, which is essentially the same as Algorithm 2.

#include <algorithm>
#include <iostream>
#include <sstream>
 
#include <gmpxx.h>
 
using integer = mpz_class;
 
std::string to_string(const integer& n) {
std::ostringstream out;
out << n;
return out.str();
}
 
integer next_highest(const integer& n) {
std::string str(to_string(n));
if (!std::next_permutation(str.begin(), str.end()))
return 0;
return integer(str);
}
 
int main() {
for (integer n : {0, 9, 12, 21, 12453, 738440, 45072010, 95322020})
std::cout << n << " -> " << next_highest(n) << '\n';
integer big("9589776899767587796600");
std::cout << big << " -> " << next_highest(big) << '\n';
return 0;
}
Output:
0 -> 0
9 -> 0
12 -> 21
21 -> 0
12453 -> 12534
738440 -> 740348
45072010 -> 45072100
95322020 -> 95322200
9589776899767587796600 -> 9589776899767587900667

D[edit]

Translation of: Java
import std.algorithm;
import std.array;
import std.conv;
import std.stdio;
 
string next(string s) {
auto sb = appender!string;
auto index = s.length - 1;
 
// Scan right-to-left through the digits of the number until you find a digit with a larger digit somewhere to the right of it.
while (index > 0 && s[index - 1] >= s[index]) {
index--;
}
// Reached beginning. No next number.
if (index == 0) {
return "0";
}
 
// Find digit on the right that is both more than it, and closest to it.
auto index2 = index;
foreach (i; index + 1 .. s.length) {
if (s[i] < s[index2] && s[i] > s[index - 1]) {
index2 = i;
}
}
 
// Found data, now build string
// Beginning of String
if (index > 1) {
sb ~= s[0 .. index - 1];
}
 
// Append found, place next
sb ~= s[index2];
 
// Get remaining characters
auto chars = [cast(dchar) s[index - 1]];
foreach (i; index .. s.length) {
if (i != index2) {
chars ~= s[i];
}
}
 
// Order the digits to the right of this position, after the swap; lowest-to-highest, left-to-right.
chars.sort;
sb ~= chars;
 
return sb.data;
}
 
long factorial(long n) {
long fact = 1;
foreach (num; 2 .. n + 1) {
fact *= num;
}
return fact;
}
 
void testAll(string s) {
writeln("Test all permutations of: ", s);
string sOrig = s;
string sPrev = s;
int count = 1;
 
// Check permutation order. Each is greater than the last
bool orderOk = true;
int[string] uniqueMap = [s: 1];
while (true) {
s = next(s);
if (s == "0") {
break;
}
 
count++;
if (s.to!long < sPrev.to!long) {
orderOk = false;
}
uniqueMap.update(s, {
return 1;
}, (int a) {
return a + 1;
});
sPrev = s;
}
writeln(" Order: OK = ", orderOk);
 
// Test last permutation
auto reverse = sOrig.dup.to!(dchar[]).reverse.to!string;
writefln(" Last permutation: Actual = %s, Expected = %s, OK = %s", sPrev, reverse, sPrev == reverse);
 
// Check permutations unique
bool unique = true;
foreach (k, v; uniqueMap) {
if (v > 1) {
unique = false;
break;
}
}
writeln(" Permutations unique: OK = ", unique);
 
// Check expected count.
int[char] charMap;
foreach (c; sOrig) {
charMap.update(c, {
return 1;
}, (int v) {
return v + 1;
});
}
long permCount = factorial(sOrig.length);
foreach (k, v; charMap) {
permCount /= factorial(v);
}
writefln(" Permutation count: Actual = %d, Expected = %d, OK = %s", count, permCount, count == permCount);
}
 
void main() {
foreach (s; ["0", "9", "12", "21", "12453", "738440", "45072010", "95322020", "9589776899767587796600", "3345333"]) {
writeln(s, " -> ", next(s));
}
 
testAll("12345");
testAll("11122");
}
Output:
0 -> 0
9 -> 0
12 -> 21
21 -> 0
12453 -> 12534
738440 -> 740348
45072010 -> 45072100
95322020 -> 95322200
9589776899767587796600 -> 9589776899767587900667
3345333 -> 3353334
Test all permutations of: 12345
    Order:  OK =  true
    Last permutation:  Actual = 54321, Expected = 54321, OK = true
    Permutations unique:  OK =  true
    Permutation count:  Actual = 120, Expected = 120, OK = true
Test all permutations of: 11122
    Order:  OK =  true
    Last permutation:  Actual = 22111, Expected = 22111, OK = true
    Permutations unique:  OK =  true
    Permutation count:  Actual = 10, Expected = 10, OK = true

Factor[edit]

This uses the next-permutation word from the math.combinatorics vocabulary. next-permutation wraps around and returns the smallest lexicographic permutation after the largest one, so additionally we must check if we're at the largest permutation and return zero if so. See the implementation of next-permutation here.

Works with: Factor version 0.99 2020-01-23
USING: formatting grouping kernel math math.combinatorics
math.parser sequences ;
 
: next-highest ( m -- n )
number>string dup [ >= ] monotonic?
[ drop 0 ] [ next-permutation string>number ] if ;
 
{
0 9 12 21 12453 738440 45072010 95322020
9589776899767587796600
}
[ dup next-highest "%d -> %d\n" printf ] each
Output:
0 -> 0
9 -> 0
12 -> 21
21 -> 0
12453 -> 12534
738440 -> 740348
45072010 -> 45072100
95322020 -> 95322200
9589776899767587796600 -> 9589776899767587900667

Go[edit]

This uses a modified version of the recursive code in the [Permutations#Go] task.

package main
 
import (
"fmt"
"sort"
)
 
func permute(s string) []string {
var res []string
if len(s) == 0 {
return res
}
b := []byte(s)
var rc func(int) // recursive closure
rc = func(np int) {
if np == 1 {
res = append(res, string(b))
return
}
np1 := np - 1
pp := len(b) - np1
rc(np1)
for i := pp; i > 0; i-- {
b[i], b[i-1] = b[i-1], b[i]
rc(np1)
}
w := b[0]
copy(b, b[1:pp+1])
b[pp] = w
}
rc(len(b))
return res
}
 
func algorithm1(nums []string) {
fmt.Println("Algorithm 1")
fmt.Println("-----------")
for _, num := range nums {
perms := permute(num)
le := len(perms)
if le == 0 { // ignore blanks
continue
}
sort.Strings(perms)
ix := sort.SearchStrings(perms, num)
next := ""
if ix < le-1 {
for i := ix + 1; i < le; i++ {
if perms[i] > num {
next = perms[i]
break
}
}
}
if len(next) > 0 {
fmt.Printf("%29s -> %s\n", commatize(num), commatize(next))
} else {
fmt.Printf("%29s -> 0\n", commatize(num))
}
}
fmt.Println()
}
 
func algorithm2(nums []string) {
fmt.Println("Algorithm 2")
fmt.Println("-----------")
outer:
for _, num := range nums {
b := []byte(num)
le := len(b)
if le == 0 { // ignore blanks
continue
}
max := num[le-1]
mi := le - 1
for i := le - 2; i >= 0; i-- {
if b[i] < max {
min := max - b[i]
for j := mi + 1; j < le; j++ {
min2 := b[j] - b[i]
if min2 > 0 && min2 < min {
min = min2
mi = j
}
}
b[i], b[mi] = b[mi], b[i]
c := (b[i+1:])
sort.Slice(c, func(i, j int) bool {
return c[i] < c[j]
})
next := string(b[0:i+1]) + string(c)
fmt.Printf("%29s -> %s\n", commatize(num), commatize(next))
continue outer
} else if b[i] > max {
max = num[i]
mi = i
}
}
fmt.Printf("%29s -> 0\n", commatize(num))
}
}
 
func commatize(s string) string {
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
return s
}
 
func main() {
nums := []string{"0", "9", "12", "21", "12453", "738440", "45072010", "95322020", "9589776899767587796600"}
algorithm1(nums[:len(nums)-1]) // exclude the last one
algorithm2(nums) // include the last one
}
Output:
Algorithm 1
-----------
                            0 -> 0
                            9 -> 0
                           12 -> 21
                           21 -> 0
                       12,453 -> 12,534
                      738,440 -> 740,348
                   45,072,010 -> 45,072,100
                   95,322,020 -> 95,322,200

Algorithm 2
-----------
                            0 -> 0
                            9 -> 0
                           12 -> 21
                           21 -> 0
                       12,453 -> 12,534
                      738,440 -> 740,348
                   45,072,010 -> 45,072,100
                   95,322,020 -> 95,322,200
9,589,776,899,767,587,796,600 -> 9,589,776,899,767,587,900,667

Haskell[edit]

Permutations[edit]

Defining a list of all (if any) digit-shuffle successors of a positive integer, in terms of permutations.

Translation of: Python

(Generator version)

import Data.List (nub, permutations, sort)
 
digitShuffleSuccessors :: Integer -> [Integer]
digitShuffleSuccessors n =
(fmap . (+) <*> (nub . sort . concatMap go . permutations . show)) n
where
go ds
| 0 >= delta = []
| otherwise = [delta]
where
delta = (read ds :: Integer) - n
 
--------------------------- TEST ---------------------------
main :: IO ()
main =
putStrLn $
fTable
"Taking up to 5 digit-shuffle successors of a positive integer:\n"
show
(\xs ->
let harvest = take 5 xs
in rjust
12
' '
(show (length harvest) <> " of " <> show (length xs) <> ": ") <>
show harvest)
digitShuffleSuccessors
[0, 9, 12, 21, 12453, 738440, 45072010, 95322020]
 
------------------------- DISPLAY --------------------------
fTable :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String
fTable s xShow fxShow f xs =
unlines $
s : fmap (((<>) . rjust w ' ' . xShow) <*> ((" -> " <>) . fxShow . f)) xs
where
w = maximum (length . xShow <$> xs)
 
rjust :: Int -> Char -> String -> String
rjust n c = drop . length <*> (replicate n c <>)
Output:
Taking up to 5 digit-shuffle successors of a positive integer:

       0 ->     0 of 0: []
       9 ->     0 of 0: []
      12 ->     1 of 1: [21]
      21 ->     0 of 0: []
   12453 ->   5 of 116: [12534,12543,13245,13254,13425]
  738440 ->    5 of 96: [740348,740384,740438,740483,740834]
45072010 ->  5 of 1861: [45072100,45100027,45100072,45100207,45100270]
95322020 ->     1 of 1: [95322200]

Minimal digit-swaps[edit]

Defining a lazily-evaluated list of all digit-shuffle successors, this time in terms of minimal digit swaps (rather than the full set of permutations).

(The digit-swap approach makes it feasible to obtain successors of this kind for much larger numbers)

import Data.List (unfoldr)
import Data.Bool (bool)
 
------------------- MINIMAL DIGIT-SWAPS --------------------
digitShuffleSuccessors
:: Integral b
=> b -> [b]
digitShuffleSuccessors n =
let go = minimalSwap . splitBy (>)
in unDigits <$>
unfoldr
((\x -> bool (Just (((,) <*> go . reverse) x)) Nothing) <*> null)
(go (reversedDigits n))
 
minimalSwap
:: Ord a
=> ([a], [a]) -> [a]
minimalSwap ([], x:y:xs) = reverse (y : x : xs)
minimalSwap ([], xs) = []
minimalSwap (_, []) = []
minimalSwap (reversedSuffix, pivot:prefix) =
let (less, h:more) = break (> pivot) reversedSuffix
in reverse (h : prefix) ++ less ++ pivot : more
 
 
--------------------------- TEST ---------------------------
main :: IO ()
main = do
putStrLn $
fTable
"Taking up to 5 digit-shuffle successors of a positive integer:\n"
show
(\xs ->
let harvest = take 5 xs
in rjust
12
' '
(show (length harvest) <> " of " <> show (length xs) ++ ": ") ++
show harvest)
digitShuffleSuccessors
[0, 9, 12, 21, 12453, 738440, 45072010, 95322020]
putStrLn $
fTable
"Taking up to 10 digit-shuffle successors of a larger integer:\n"
show
(('\n' :) . unlines . fmap ((" " <>) . show))
(take 10 . digitShuffleSuccessors)
[9589776899767587796600]
 
 
------------------------- GENERIC --------------------------
reversedDigits
:: Integral a
=> a -> [a]
reversedDigits 0 = [0]
reversedDigits n = go n
where
go 0 = []
go x = rem x 10 : go (quot x 10)
 
splitBy :: (a -> a -> Bool) -> [a] -> ([a], [a])
splitBy f xs = go $ break (uncurry f) $ zip xs (tail xs)
where
go (ys, zs)
| null ys = ([], xs)
| otherwise = (fst (head ys) : map snd ys, map snd zs)
 
unDigits
:: (Foldable t, Num a)
=> t a -> a
unDigits = foldl (\a b -> 10 * a + b) 0
 
 
------------------------- DISPLAY --------------------------
fTable :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String
fTable s xShow fxShow f xs =
unlines $
s : fmap (((<>) . rjust w ' ' . xShow) <*> ((" -> " <>) . fxShow . f)) xs
where
w = maximum (length . xShow <$> xs)
 
rjust :: Int -> Char -> String -> String
rjust n c = drop . length <*> (replicate n c <>)
Output:
Taking up to 5 digit-shuffle successors of a positive integer:

       0 ->     0 of 0: []
       9 ->     0 of 0: []
      12 ->     1 of 1: [21]
      21 ->     0 of 0: []
   12453 ->   5 of 116: [12534,12543,13245,13254,13425]
  738440 ->    5 of 96: [740348,740384,740438,740483,740834]
45072010 ->  5 of 1861: [45072100,45100027,45100072,45100207,45100270]
95322020 ->     1 of 1: [95322200]

Taking up to 10 digit-shuffle successors of a larger integer:

9589776899767587796600 -> 
    9589776899767587900667
    9589776899767587900676
    9589776899767587900766
    9589776899767587906067
    9589776899767587906076
    9589776899767587906607
    9589776899767587906670
    9589776899767587906706
    9589776899767587906760
    9589776899767587907066

J[edit]

   permutations=: A.&i.~ !
   ordered_numbers_from_digits=: [: /:~ ({~ [email protected]#)&.":
   next_highest=: (>:@i:~ { 0 ,~ ]) ordered_numbers_from_digits

   (,. next_highest)&>0 9 12 21 12453 738440 45072010 95322020
       0        0
       9        0
      12       21
      21        0
   12453    12534
  738440   740348
45072010 45072100
95322020 95322200

Java[edit]

Additional testing is performed, including a number with all unique digits and a number with duplicate digits. Included test of all permutations, that the order and correct number of permutations is achieved, and that each permutation is different than all others. If a library is not used, then this testing will provide a better proof of correctness.

 
import java.math.BigInteger;
import java.text.NumberFormat;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
 
public class NextHighestIntFromDigits {
 
public static void main(String[] args) {
for ( String s : new String[] {"0", "9", "12", "21", "12453", "738440", "45072010", "95322020", "9589776899767587796600", "3345333"} ) {
System.out.printf("%s -> %s%n", format(s), format(next(s)));
}
testAll("12345");
testAll("11122");
}
 
private static NumberFormat FORMAT = NumberFormat.getNumberInstance();
 
private static String format(String s) {
return FORMAT.format(new BigInteger(s));
}
 
private static void testAll(String s) {
System.out.printf("Test all permutations of:  %s%n", s);
String sOrig = s;
String sPrev = s;
int count = 1;
 
// Check permutation order. Each is greater than the last
boolean orderOk = true;
Map <String,Integer> uniqueMap = new HashMap<>();
uniqueMap.put(s, 1);
while ( (s = next(s)).compareTo("0") != 0 ) {
count++;
if ( Long.parseLong(s) < Long.parseLong(sPrev) ) {
orderOk = false;
}
uniqueMap.merge(s, 1, (v1, v2) -> v1 + v2);
sPrev = s;
}
System.out.printf(" Order: OK =  %b%n", orderOk);
 
// Test last permutation
String reverse = new StringBuilder(sOrig).reverse().toString();
System.out.printf(" Last permutation: Actual = %s, Expected = %s, OK = %b%n", sPrev, reverse, sPrev.compareTo(reverse) == 0);
 
// Check permutations unique
boolean unique = true;
for ( String key : uniqueMap.keySet() ) {
if ( uniqueMap.get(key) > 1 ) {
unique = false;
}
}
System.out.printf(" Permutations unique: OK =  %b%n", unique);
 
// Check expected count.
Map<Character,Integer> charMap = new HashMap<>();
for ( char c : sOrig.toCharArray() ) {
charMap.merge(c, 1, (v1, v2) -> v1 + v2);
}
long permCount = factorial(sOrig.length());
for ( char c : charMap.keySet() ) {
permCount /= factorial(charMap.get(c));
}
System.out.printf(" Permutation count: Actual = %d, Expected = %d, OK = %b%n", count, permCount, count == permCount);
 
 
}
 
private static long factorial(long n) {
long fact = 1;
for (long num = 2 ; num <= n ; num++ ) {
fact *= num;
}
return fact;
}
 
private static String next(String s) {
StringBuilder sb = new StringBuilder();
int index = s.length()-1;
// Scan right-to-left through the digits of the number until you find a digit with a larger digit somewhere to the right of it.
while ( index > 0 && s.charAt(index-1) >= s.charAt(index)) {
index--;
}
// Reached beginning. No next number.
if ( index == 0 ) {
return "0";
}
 
// Find digit on the right that is both more than it, and closest to it.
int index2 = index;
for ( int i = index + 1 ; i < s.length() ; i++ ) {
if ( s.charAt(i) < s.charAt(index2) && s.charAt(i) > s.charAt(index-1) ) {
index2 = i;
}
}
 
// Found data, now build string
// Beginning of String
if ( index > 1 ) {
sb.append(s.subSequence(0, index-1));
}
 
// Append found, place next
sb.append(s.charAt(index2));
 
// Get remaining characters
List<Character> chars = new ArrayList<>();
chars.add(s.charAt(index-1));
for ( int i = index ; i < s.length() ; i++ ) {
if ( i != index2 ) {
chars.add(s.charAt(i));
}
}
 
// Order the digits to the right of this position, after the swap; lowest-to-highest, left-to-right.
Collections.sort(chars);
for ( char c : chars ) {
sb.append(c);
}
return sb.toString();
}
}
 
Output:
0 -> 0
9 -> 0
12 -> 21
21 -> 0
12,453 -> 12,534
738,440 -> 740,348
45,072,010 -> 45,072,100
95,322,020 -> 95,322,200
9,589,776,899,767,587,796,600 -> 9,589,776,899,767,587,900,667
3,345,333 -> 3,353,334
Test all permutations of:  12345
    Order:  OK =  true
    Last permutation:  Actual = 54321, Expected = 54321, OK = true
    Permutations unique:  OK =  true
    Permutation count:  Actual = 120, Expected = 120, OK = true
Test all permutations of:  11122
    Order:  OK =  true
    Last permutation:  Actual = 22111, Expected = 22111, OK = true
    Permutations unique:  OK =  true
    Permutation count:  Actual = 10, Expected = 10, OK = true


JavaScript[edit]

const compose = (...fn) => (...x) => fn.reduce((a, b) => c => a(b(c)))(...x);
const toString = x => x + '';
const reverse = x => Array.from(x).reduce((p, c) => [c, ...p], []);
const minBiggerThanN = (arr, n) => arr.filter(e => e > n).sort()[0];
const remEl = (arr, e) => {
const r = arr.indexOf(e);
return arr.filter((e,i) => i !== r);
}
 
const nextHighest = itr => {
const seen = [];
let result = 0;
for (const [i,v] of itr.entries()) {
const n = +v;
if (Math.max(n, ...seen) !== n) {
const right = itr.slice(i + 1);
const swap = minBiggerThanN(seen, n);
const rem = remEl(seen, swap);
const rest = [n, ...rem].sort();
result = [...reverse(right), swap, ...rest].join('');
break;
} else {
seen.push(n);
}
}
return result;
};
 
const check = compose(nextHighest, reverse, toString);
 
const test = v => {
console.log(v, '=>', check(v));
}
 
test(0);
test(9);
test(12);
test(21);
test(12453);
test(738440);
test(45072010);
test(95322020);
test('9589776899767587796600');
 
Output:
0 => 0
9 => 0
12 => 21
21 => 0
12453 => 12534
738440 => 740348
45072010 => 45072100
95322020 => 95322200
9589776899767587796600 => 9589776899767587900667

Julia[edit]

using Combinatorics, BenchmarkTools
 
asint(dig) = foldl((i, j) -> 10i + Int128(j), dig)
 
"""
Algorithm 1(A)
Generate all the permutations of the digits and sort into numeric order.
Find the number in the list.
Return the next highest number from the list.
"""
function nexthighest_1A(N)
n = Int128(abs(N))
dig = digits(n)
perms = unique(sort([asint(arr) for arr in permutations(digits(n))]))
length(perms) < 2 && return 0
((N > 0 && perms[end] == n) || (N < 0 && perms[1] == n)) && return 0
pos = findfirst(x -> x == n, perms)
ret = N > 0 ? perms[pos + 1] : -perms[pos - 1]
return ret == N ? 0 : ret
end
 
"""
Algorithm 1(B)
Iterate through the permutations of the digits of a number and get the permutation that
represents the integer having a minimum distance above the given number.
Return the number plus the minimum distance. Does not store all the permutations.
This saves memory versus algorithm 1A, but we still go through all permutations (slow).
"""
function nexthighest_1B(N)
n = Int128(abs(N))
dig = reverse(digits(n))
length(dig) < 2 && return 0
mindelta = n
for perm in permutations(dig)
if (perm[1] != 0) && ((N > 0 && perm > dig) || (N < 0 && perm < dig))
delta = abs(asint(perm) - n)
if delta < mindelta
mindelta = delta
end
end
end
return mindelta < n ? N + mindelta : 0
end
 
"""
Algorithm 2
Scan right-to-left through the digits of the number until you find a digit with a larger digit somewhere to the right of it.
Exchange that digit with the digit on the right that is both more than it, and closest to it.
Order the digits to the right of this position, after the swap; lowest-to-highest, left-to-right.
Very fast, as it does not need to run through all the permutations of digits.
"""
function nexthighest_2(N)
n = Int128(abs(N))
dig, ret = digits(n), N
length(dig) < 2 && return 0
for (i, d) in enumerate(dig)
if N > 0 && i > 1
rdig = dig[1:i-1]
if (j = findfirst(x -> x > d, rdig)) != nothing
dig[i], dig[j] = dig[j], dig[i]
arr = (i == 2) ? dig : [sort(dig[1:i-1], rev=true); dig[i:end]]
ret = asint(reverse(arr))
break
end
elseif N < 0 && i > 1
rdig = dig[1:i-1]
if (j = findfirst(x -> x < d, rdig)) != nothing
dig[i], dig[j] = dig[j], dig[i]
arr = (i == 2) ? dig : [sort(dig[1:i-1]); dig[i:end]]
ret = -asint(reverse(arr))
break
end
end
end
return ret == N ? 0 : ret
end
 
println(" N 1A 1B 2\n", "="^98)
for n in [0, 9, 12, 21, -453, -8888, 12453, 738440, 45072010, 95322020, -592491602, 9589776899767587796600]
println(rpad(n, 25), abs(n) > typemax(Int) ? " "^50 : rpad(nexthighest_1A(n), 25) *
rpad(nexthighest_1B(n), 25), nexthighest_2(n))
end
 
const n = 7384440
@btime nexthighest_1A(n)
println(" for method 1A and n $n.")
@btime nexthighest_1B(n)
println(" for method 1B and n $n.")
@btime nexthighest_2(n)
println(" for method 2 and n $n.")
 
Output:
 N                       1A                       1B                       2
==================================================================================================
0                        0                        0                        0
9                        0                        0                        0
12                       21                       21                       21
21                       0                        0                        0
-453                     -435                     -435                     -435
-8888                    0                        0                        0
12453                    12534                    12534                    12534
738440                   740348                   740348                   740348
45072010                 45072100                 45072100                 45072100
95322020                 95322200                 95322200                 95322200
-592491602               -592491260               -592491260               -592491260
9589776899767587796600                                                     9589776899767587900667
  4.027 ms (40364 allocations: 2.43 MiB)
 for method 1A and n 7384440.
  1.237 ms (28804 allocations: 1.92 MiB)
 for method 1B and n 7384440.
  1.260 μs (14 allocations: 1.36 KiB)
 for method 2 and n 7384440.                      

Kotlin[edit]

Translation of: Java
import java.math.BigInteger
import java.text.NumberFormat
 
fun main() {
for (s in arrayOf(
"0",
"9",
"12",
"21",
"12453",
"738440",
"45072010",
"95322020",
"9589776899767587796600",
"3345333"
)) {
println("${format(s)} -> ${format(next(s))}")
}
testAll("12345")
testAll("11122")
}
 
private val FORMAT = NumberFormat.getNumberInstance()
private fun format(s: String): String {
return FORMAT.format(BigInteger(s))
}
 
private fun testAll(str: String) {
var s = str
println("Test all permutations of: $s")
val sOrig = s
var sPrev = s
var count = 1
 
// Check permutation order. Each is greater than the last
var orderOk = true
val uniqueMap: MutableMap<String, Int> = HashMap()
uniqueMap[s] = 1
while (next(s).also { s = it }.compareTo("0") != 0) {
count++
if (s.toLong() < sPrev.toLong()) {
orderOk = false
}
uniqueMap.merge(s, 1) { a: Int?, b: Int? -> Integer.sum(a!!, b!!) }
sPrev = s
}
println(" Order: OK = $orderOk")
 
// Test last permutation
val reverse = StringBuilder(sOrig).reverse().toString()
println(" Last permutation: Actual = $sPrev, Expected = $reverse, OK = ${sPrev.compareTo(reverse) == 0}")
 
// Check permutations unique
var unique = true
for (key in uniqueMap.keys) {
if (uniqueMap[key]!! > 1) {
unique = false
}
}
println(" Permutations unique: OK = $unique")
 
// Check expected count.
val charMap: MutableMap<Char, Int> = HashMap()
for (c in sOrig.toCharArray()) {
charMap.merge(c, 1) { a: Int?, b: Int? -> Integer.sum(a!!, b!!) }
}
var permCount = factorial(sOrig.length.toLong())
for (c in charMap.keys) {
permCount /= factorial(charMap[c]!!.toLong())
}
println(" Permutation count: Actual = $count, Expected = $permCount, OK = ${count.toLong() == permCount}")
}
 
private fun factorial(n: Long): Long {
var fact: Long = 1
for (num in 2..n) {
fact *= num
}
return fact
}
 
private fun next(s: String): String {
val sb = StringBuilder()
var index = s.length - 1
// Scan right-to-left through the digits of the number until you find a digit with a larger digit somewhere to the right of it.
while (index > 0 && s[index - 1] >= s[index]) {
index--
}
// Reached beginning. No next number.
if (index == 0) {
return "0"
}
 
// Find digit on the right that is both more than it, and closest to it.
var index2 = index
for (i in index + 1 until s.length) {
if (s[i] < s[index2] && s[i] > s[index - 1]) {
index2 = i
}
}
 
// Found data, now build string
// Beginning of String
if (index > 1) {
sb.append(s.subSequence(0, index - 1))
}
 
// Append found, place next
sb.append(s[index2])
 
// Get remaining characters
val chars: MutableList<Char> = ArrayList()
chars.add(s[index - 1])
for (i in index until s.length) {
if (i != index2) {
chars.add(s[i])
}
}
 
// Order the digits to the right of this position, after the swap; lowest-to-highest, left-to-right.
chars.sort()
for (c in chars) {
sb.append(c)
}
return sb.toString()
}
Output:
0 -> 0
9 -> 0
12 -> 21
21 -> 0
12,453 -> 12,534
738,440 -> 740,348
45,072,010 -> 45,072,100
95,322,020 -> 95,322,200
9,589,776,899,767,587,796,600 -> 9,589,776,899,767,587,900,667
3,345,333 -> 3,353,334
Test all permutations of:  12345
    Order:  OK =  true
    Last permutation:  Actual = 54321, Expected = 54321, OK = true
    Permutations unique:  OK =  true
    Permutation count:  Actual = 120, Expected = 120, OK = true
Test all permutations of:  11122
    Order:  OK =  true
    Last permutation:  Actual = 22111, Expected = 22111, OK = true
    Permutations unique:  OK =  true
    Permutation count:  Actual = 10, Expected = 10, OK = true

Perl[edit]

Translation of: Raku
use strict;
use warnings;
use feature 'say';
use bigint;
use List::Util 'first';
 
sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
 
sub next_greatest_index {
my($str) = @_;
my @i = reverse split //, $str;
@i-1 - (1 + first { $i[$_] > $i[$_+1] } 0 .. @i-1);
}
 
sub next_greatest_integer {
my($num) = @_;
my $numr;
return 0 if length $num < 2;
return ($numr = 0 + reverse $num) > $num ? $numr : 0 if length $num == 2;
return 0 unless my $i = next_greatest_index( $num ) // 0;
my $digit = substr($num, $i, 1);
my @rest = sort split '', substr($num, $i);
my $next = first { $rest[$_] > $digit } 1..@rest;
join '', substr($num, 0, $i), (splice(@rest, $next, 1)), @rest;
}
 
say 'Next largest integer able to be made from these digits, or zero if no larger exists:';
 
for (0, 9, 12, 21, 12453, 738440, 45072010, 95322020, 9589776899767587796600, 3345333) {
printf "%30s ->  %s\n", comma($_), comma next_greatest_integer $_;
}
Output:
                             0  ->  0
                             9  ->  0
                            12  ->  21
                            21  ->  0
                        12,453  ->  12,534
                       738,440  ->  740,348
                    45,072,010  ->  45,072,100
                    95,322,020  ->  95,322,200
 9,589,776,899,767,587,796,600  ->  9,589,776,899,767,587,900,667
                     3,345,333  ->  3,353,334

Phix[edit]

algorithm 1[edit]

function nigh(string n)
sequence p = repeat("",factorial(length(n)))
for i=1 to length(p) do
p[i] = permute(i,n)
end for
p = sort(p)
integer k = rfind(n,p)
return iff(k=length(p)?"0",p[k+1])
end function
 
constant tests = {"0","9","12","21","12453",
"738440","45072010","95322020"}
-- (crashes on) "9589776899767587796600"}
atom t0 = time()
for i=1 to length(tests) do
string t = tests[i]
printf(1,"%22s => %s\n",{t,nigh(t)})
end for
?elapsed(time()-t0)
Output:
                     0 => 0
                     9 => 0
                    12 => 21
                    21 => 0
                 12453 => 12534
                738440 => 740348
              45072010 => 45072100
              95322020 => 95322200
"0.2s"

algorithm 2[edit]

function nigh(string n)
integer hi = n[$]
for i=length(n)-1 to 1 by -1 do
integer ni = n[i]
if ni<hi then
string sr = sort(n[i..$])
integer k = rfind(ni,sr)+1
n[i] = sr[k]
sr[k..k] = ""
n[i+1..$] = sr
return n
end if
hi = max(hi,ni)
end for
return "0"
end function
 
constant tests = {"0","9","12","21","12453",
"738440","45072010","95322020",
"9589776899767587796600"}
atom t0 = time()
for i=1 to length(tests) do
string t = tests[i]
printf(1,"%22s => %s\n",{t,nigh(t)})
end for
?elapsed(time()-t0)
Output:
                     0 => 0
                     9 => 0
                    12 => 21
                    21 => 0
                 12453 => 12534
                738440 => 740348
              45072010 => 45072100
              95322020 => 95322200
9589776899767587796600 => 9589776899767587900667
"0s"

Python[edit]

Python: Algorithm 2[edit]

Like Algorithm 2, but digit order is reversed for easier indexing, then reversed on return.

def closest_more_than(n, lst):
"(index of) closest int from lst, to n that is also > n"
large = max(lst) + 1
return lst.index(min(lst, key=lambda x: (large if x <= n else x)))
 
def nexthigh(n):
"Return nxt highest number from n's digits using scan & re-order"
assert n == int(abs(n)), "n >= 0"
this = list(int(digit) for digit in str(int(n)))[::-1]
mx = this[0]
for i, digit in enumerate(this[1:], 1):
if digit < mx:
mx_index = closest_more_than(digit, this[:i + 1])
this[mx_index], this[i] = this[i], this[mx_index]
this[:i] = sorted(this[:i], reverse=True)
return int(''.join(str(d) for d in this[::-1]))
elif digit > mx:
mx, mx_index = digit, i
return 0
 
 
if __name__ == '__main__':
for x in [0, 9, 12, 21, 12453, 738440, 45072010, 95322020,
9589776899767587796600]:
print(f"{x:>12_d} -> {nexthigh(x):>12_d}")
Output:

Note underscores are used in integer representations to aid in comparisons.

           0 ->            0
           9 ->            0
          12 ->           21
          21 ->            0
      12_453 ->       12_534
     738_440 ->      740_348
  45_072_010 ->   45_072_100
  95_322_020 ->   95_322_200
9_589_776_899_767_587_796_600 -> 9_589_776_899_767_587_900_667

Python: Algorithm 1[edit]

I would not try it on the stretch goal, otherwise results as above.

from itertools import permutations
 
 
def nexthigh(n):
"Return next highest number from n's digits using search of all digit perms"
assert n == int(abs(n)), "n >= 0"
this = tuple(str(int(n)))
perms = sorted(permutations(this))
for perm in perms[perms.index(this):]:
if perm != this:
return int(''.join(perm))
return 0

Python: Generator[edit]

A variant which defines (in terms of a concatMap over permutations), a generator of all digit-shuffle successors for a given integer:

'''Next highest int from digits'''
 
from itertools import chain, islice, permutations, tee
 
 
# --------------- LAZY STREAM OF SUCCESSORS ----------------
 
# digitShuffleSuccessors :: Int -> [Int]
def digitShuffleSuccessors(n):
'''Iterator stream of all digit-shuffle
successors of n, where 0 <= n.
'''

def go(ds):
delta = int(''.join(ds)) - n
return [] if 0 >= delta else [delta]
return map(
add(n),
sorted(
set(concatMap(go)(
permutations(str(n))
))
)
)
 
 
# -------------------------- TEST --------------------------
# main :: IO ()
def main():
'''Taking up to 5 digit-shuffle successors for each:'''
 
def showSuccs(n):
def go(xs):
ys, zs = tee(xs)
harvest = take(n)(ys)
return (
repr(len(harvest)) + ' of ' + (
repr(len(list(zs))) + ': '
)
).rjust(12, ' ') + repr(harvest)
return go
 
print(
fTable(main.__doc__ + '\n')(str)(showSuccs(5))(
digitShuffleSuccessors
)([
0,
9,
12,
21,
12453,
738440,
45072010,
95322020
])
)
 
 
# ------------------------ GENERIC -------------------------
 
# add (+) :: Num a => a -> a -> a
def add(a):
'''Curried addition.'''
def go(b):
return a + b
return go
 
 
# concatMap :: (a -> [b]) -> [a] -> [b]
def concatMap(f):
'''The concatenation of a mapping.
The list monad can be derived by using a function f
which wraps its output in a list, using an empty
list to represent computational failure).
'''

def go(xs):
return chain.from_iterable(map(f, xs))
return go
 
 
# fTable :: String -> (a -> String) ->
# (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function ->
fx display function -> f -> xs -> tabular string.
'''

def gox(xShow):
def gofx(fxShow):
def gof(f):
def goxs(xs):
ys = [xShow(x) for x in xs]
w = max(map(len, ys))
 
def arrowed(x, y):
return y.rjust(w, ' ') + (
' -> ' + fxShow(f(x))
)
return s + '\n' + '\n'.join(
map(arrowed, xs, ys)
)
return goxs
return gof
return gofx
return gox
 
 
# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.
'''

def go(xs):
return (
xs[0:n]
if isinstance(xs, (list, tuple))
else list(islice(xs, n))
)
return go
 
 
# MAIN ---
if __name__ == '__main__':
main()
Output:
Taking up to 5 digit-shuffle successors for each:

       0 ->    0 of 0:  []
       9 ->    0 of 0:  []
      12 ->    1 of 1:  [21]
      21 ->    0 of 0:  []
   12453 ->  5 of 116:  [12534, 12543, 13245, 13254, 13425]
  738440 ->   5 of 96:  [740348, 740384, 740438, 740483, 740834]
45072010 -> 5 of 1861:  [45072100, 45100027, 45100072, 45100207, 45100270]
95322020 ->    1 of 1:  [95322200]

Raku[edit]

(formerly Perl 6)

Works with: Rakudo version 2020.01

Minimal error trapping. Assumes that the passed number is an integer. Handles positive or negative integers, always returns next largest regardless (if possible).

use Lingua::EN::Numbers;
 
sub next-greatest-index ($str, &op = &infix:«<» ) {
my @i = $str.comb;
(1..^@i).first: { &op(@i[$_ - 1], @i[$_]) }, :end, :k;
}
 
multi next-greatest-integer (Int $num where * >= 0) {
return 0 if $num.chars < 2;
return $num.flip > $num ?? $num.flip !! 0 if $num.chars == 2;
return 0 unless my $i = next-greatest-index( $num ) // 0;
my $digit = $num.substr($i, 1);
my @rest = (flat $num.substr($i).comb).sort(+*);
my $next = @rest.first: * > $digit, :k;
$digit = @rest.splice($next,1);
join '', flat $num.substr(0,$i), $digit, @rest;
}
 
multi next-greatest-integer (Int $num where * < 0) {
return 0 if $num.chars < 3;
return $num.abs.flip < -$num ?? -$num.abs.flip !! 0 if $num.chars == 3;
return 0 unless my $i = next-greatest-index( $num, &CORE::infix:«>» ) // 0;
my $digit = $num.substr($i, 1);
my @rest = (flat $num.substr($i).comb).sort(-*);
my $next = @rest.first: * < $digit, :k;
$digit = @rest.splice($next,1);
join '', flat $num.substr(0,$i), $digit, @rest;
}
 
say "Next largest integer able to be made from these digits, or zero if no larger exists:";
printf "%30s -> %s%s\n", .&comma, .&next-greatest-integer < 0 ?? '' !! ' ', .&next-greatest-integer.&comma for
flat 0, (9, 12, 21, 12453, 738440, 45072010, 95322020, 9589776899767587796600, 3345333,
95897768997675877966000000000000000000000000000000000000000000000000000000000000000000).map: { $_, -$_ };
Output:
Next largest integer able to be made from these digits, or zero if no larger exists:
                             0  ->  0
                             9  ->  0
                            -9  ->  0
                            12  ->  21
                           -12  ->  0
                            21  ->  0
                           -21  -> -12
                        12,453  ->  12,534
                       -12,453  -> -12,435
                       738,440  ->  740,348
                      -738,440  -> -738,404
                    45,072,010  ->  45,072,100
                   -45,072,010  -> -45,072,001
                    95,322,020  ->  95,322,200
                   -95,322,020  -> -95,322,002
 9,589,776,899,767,587,796,600  ->  9,589,776,899,767,587,900,667
-9,589,776,899,767,587,796,600  -> -9,589,776,899,767,587,796,060
                     3,345,333  ->  3,353,334
                    -3,345,333  -> -3,343,533
95,897,768,997,675,877,966,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000  ->  95,897,768,997,675,879,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,667
-95,897,768,997,675,877,966,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000  -> -95,897,768,997,675,877,960,600,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

REXX[edit]

/*REXX program finds the  next highest positive integer  from a list of decimal digits. */
parse arg n /*obtain optional arguments from the CL*/
if n='' | n="," then n= 0 9 12 21 12453 738440 45072010 95322020 /*use the defaults?*/
w= length( commas( word(n, words(n) ) ) ) /*maximum width number (with commas). */
 
do j=1 for words(n); y= word(n, j) /*process each of the supplied numbers.*/
masky= mask(y) /*build a digit mask for a supplied int*/
lim= copies(9, length(y) ) /*construct a LIMIT for the DO loop. */
 
do #=y+1 to lim until mask(#)==masky /*search for a number that might work. */
if verify(y, #) \== 0 then iterate /*does # have all the necessary digits?*/
end /*#*/
 
if #>lim then #= 0 /*if # > lim, then there is no valid #*/
say 'for ' right(commas(y),w) " ─── the next highest integer is: " right(commas(#),w)
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg _; do ?=length(_)-3 to 1 by -3; _= insert(',', _, ?); end; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
mask: parse arg z, $; @.= 0 /* [↓] build an unsorted digit mask. */
do k=1 for length(z); parse var z _ +1 z; @._= @._ + 1
end /*k*/
do m=0 for 10; if @.m==0 then iterate; $= $ || copies(m, @.m)
end /*m*/; return $ /* [↑] build a sorted digit mask.*/
output   when using the default inputs:
for           0  ─── the next highest integer is:           0
for           9  ─── the next highest integer is:           0
for          12  ─── the next highest integer is:          21
for          21  ─── the next highest integer is:           0
for      12,453  ─── the next highest integer is:      12,534
for     738,440  ─── the next highest integer is:     740,348
for  45,072,010  ─── the next highest integer is:  45,072,100
for  95,322,020  ─── the next highest integer is:  95,322,200

Rust[edit]

fn next_permutation<T: PartialOrd>(array: &mut [T]) -> bool {
let len = array.len();
if len < 2 {
return false;
}
let mut i = len - 1;
while i > 0 {
let j = i;
i -= 1;
if array[i] < array[j] {
let mut k = len - 1;
while array[i] >= array[k] {
k -= 1;
}
array.swap(i, k);
array[j..len].reverse();
return true;
}
}
false
}
 
fn next_highest_int(n: u128) -> u128 {
use std::iter::FromIterator;
let mut chars: Vec<char> = n.to_string().chars().collect();
if !next_permutation(&mut chars) {
return 0;
}
String::from_iter(chars).parse::<u128>().unwrap()
}
 
fn main() {
for n in &[0, 9, 12, 21, 12453, 738440, 45072010, 95322020, 9589776899767587796600] {
println!("{} -> {}", n, next_highest_int(*n));
}
}
Output:
0 -> 0
9 -> 0
12 -> 21
21 -> 0
12453 -> 12534
738440 -> 740348
45072010 -> 45072100
95322020 -> 95322200
9589776899767587796600 -> 9589776899767587900667

Sidef[edit]

func next_from_digits(n, b = 10) {
 
var a = n.digits(b).flip
 
while (a.next_permutation) {
with (a.flip.digits2num(b)) { |t|
return t if (t > n)
}
}
 
return 0
}
 
say 'Next largest integer able to be made from these digits, or zero if no larger exists:'
 
for n in (
0, 9, 12, 21, 12453, 738440, 3345333, 45072010,
95322020, 982765431, 9589776899767587796600,
) {
printf("%30s ->  %s\n", n, next_from_digits(n))
}
Output:
Next largest integer able to be made from these digits, or zero if no larger exists:
                             0  ->  0
                             9  ->  0
                            12  ->  21
                            21  ->  0
                         12453  ->  12534
                        738440  ->  740348
                       3345333  ->  3353334
                      45072010  ->  45072100
                      95322020  ->  95322200
                     982765431  ->  983124567
        9589776899767587796600  ->  9589776899767587900667

zkl[edit]

fcn nextHightest(N){	// N is int, BigInt or string -->String.  Algorithm 2
// ds:=N.split().copy(); // mutable, int
ds:=N.toString().split("").apply("toInt").copy(); // handle "234" or BigInt
if(ds.len()<2) return(0);
m:=ds[-1];
foreach i in ([ds.len()-1 .. 0,-1]){
d:=ds[i];
if(d<m){
dz,j,z := ds[i,*], dz.sort().filter1n('>(d)), dz[j];
dz.del(j);
// return( ds[0,i].extend( z, dz.sort() ).concat().toInt() );
return( ds[0,i].extend( z, dz.sort() ).concat() );
}
m=m.max(d);
}
"0"
}
ns:=T(0, 9, 12, 21, 12453, 738440, 45072010, 95322020);
foreach n in (ns){ println("%,d --> %,d".fmt(n,nextHightest(n))) }
 
n:="9589776899767587796600"; // or BigInt(n)
println("%s --> %s".fmt(n,nextHightest(n)));
Output:
0 --> 0
9 --> 0
12 --> 21
21 --> 0
12,453 --> 12,534
738,440 --> 740,348
45,072,010 --> 45,072,100
95,322,020 --> 95,322,200
9589776899767587796600 --> 9589776899767587900667