# Print all permutations in sorted (lexicographic) order

• Difficulty Level : Hard
• Last Updated : 12 Jul, 2022

Given a string, print all permutations of it in sorted order. For example, if the input string is “ABC”, then output should be “ABC, ACB, BAC, BCA, CAB, CBA”.

We have discussed a program to print all permutations in this post, but here we must print the permutations in increasing order.

Algorithm to print the permutations lexicographic-ally:
Step 1. Sort the given string in non-decreasing order and print it. The first permutation is always the string sorted in non-decreasing order.
Step 2. Start generating next higher permutation. Do it until next higher permutation is not possible. If we reach a permutation where all characters are sorted and in non-increasing order, then that permutation is the last permutation.

Steps to generate the next higher permutation:
Step 1. Take the previously printed permutation and find the rightmost character in it, which is smaller than its next character. Let us call this character as ‘first character’.
Step 2. Now find the ceiling of the ‘first character’. Ceiling is the smallest character on right of ‘first character’, which is greater than ‘first character’. Let us call the ceil character as ‘second character’.
Step 3. Swap the two characters found in above 2 steps.
Step 4. Sort the substring (in non-decreasing order) after the original index of ‘first character’.

Approach:

1. Let us consider the string “ABCDEF”. Let the previously printed permutation be “DCFEBA”.

2. The next permutation in sorted order should be “DEABCF”.

3. Let us understand above steps to find next permutation. The ‘first character’ will be ‘C’. The ‘second character’ will be ‘E’. After swapping these two, we get “DEFCBA”.

4. The final step is to sort the substring after the character original index of ‘first character’. Finally, we get “DEABCF”.

Following is the implementation of the algorithm.

## C++

 `// C++ Program to print all permutations``// of a string in sorted order.``#include ``using` `namespace` `std;` `/* Following function is needed for library function qsort(). Refer``http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */``int` `compare (``const` `void` `*a, ``const` `void` `* b)``{ ``return` `( *(``char` `*)a - *(``char` `*)b ); }` `// A utility function two swap two characters a and b``void` `swap (``char``* a, ``char``* b)``{``    ``char` `t = *a;``    ``*a = *b;``    ``*b = t;``}` `// This function finds the index of the smallest character``// which is greater than 'first' and is present in str[l..h]``int` `findCeil (``char` `str[], ``char` `first, ``int` `l, ``int` `h)``{``    ``// initialize index of ceiling element``    ``int` `ceilIndex = l;` `    ``// Now iterate through rest of the elements and find``    ``// the smallest character greater than 'first'``    ``for` `(``int` `i = l+1; i <= h; i++)``    ``if` `(str[i] > first && str[i] < str[ceilIndex])``            ``ceilIndex = i;` `    ``return` `ceilIndex;``}` `// Print all permutations of str in sorted order``void` `sortedPermutations ( ``char` `str[] )``{``    ``// Get size of string``    ``int` `size = ``strlen``(str);` `    ``// Sort the string in increasing order``    ``qsort``( str, size, ``sizeof``( str[0] ), compare );` `    ``// Print permutations one by one``    ``bool` `isFinished = ``false``;``    ``while` `( ! isFinished )``    ``{``        ``// print this permutation``        ``cout << str << endl;` `        ``// Find the rightmost character which is``        ``// smaller than its next character.``        ``// Let us call it 'first char'``        ``int` `i;``        ``for` `( i = size - 2; i >= 0; --i )``        ``if` `(str[i] < str[i+1])``            ``break``;` `        ``// If there is no such character, all are``        ``// sorted in decreasing order, means we``        ``// just printed the last permutation and we are done.``        ``if` `( i == -1 )``            ``isFinished = ``true``;``        ``else``        ``{``            ``// Find the ceil of 'first char' in``            ``// right of first character.``            ``// Ceil of a character is the smallest``            ``// character greater than it``            ``int` `ceilIndex = findCeil( str, str[i], i + 1, size - 1 );` `            ``// Swap first and second characters``            ``swap( &str[i], &str[ceilIndex] );` `            ``// Sort the string on right of 'first char'``            ``qsort``( str + i + 1, size - i - 1, ``sizeof``(str[0]), compare );``        ``}``    ``}``}` `// Driver program to test above function``int` `main()``{``    ``char` `str[] = ``"ABCD"``;``    ``sortedPermutations( str );``    ``return` `0;``}` `// This is code is contributed by rathbhupendra`

## C

 `// Program to print all permutations of a string in sorted order.``#include ``#include ``#include ` `/* Following function is needed for library function qsort(). Refer``http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */``int` `compare (``const` `void` `*a, ``const` `void` `* b)``{ ``return` `( *(``char` `*)a - *(``char` `*)b ); }` `// A utility function two swap two characters a and b``void` `swap (``char``* a, ``char``* b)``{``    ``char` `t = *a;``    ``*a = *b;``    ``*b = t;``}` `// This function finds the index of the smallest character``// which is greater than 'first' and is present in str[l..h]``int` `findCeil (``char` `str[], ``char` `first, ``int` `l, ``int` `h)``{``    ``// initialize index of ceiling element``    ``int` `ceilIndex = l;` `    ``// Now iterate through rest of the elements and find``    ``// the smallest character greater than 'first'``    ``for` `(``int` `i = l+1; i <= h; i++)``    ``if` `(str[i] > first && str[i] < str[ceilIndex])``            ``ceilIndex = i;` `    ``return` `ceilIndex;``}` `// Print all permutations of str in sorted order``void` `sortedPermutations ( ``char` `str[] )``{``    ``// Get size of string``    ``int` `size = ``strlen``(str);` `    ``// Sort the string in increasing order``    ``qsort``( str, size, ``sizeof``( str[0] ), compare );` `    ``// Print permutations one by one``    ``bool` `isFinished = ``false``;``    ``while` `( ! isFinished )``    ``{``        ``// print this permutation``        ``printf` `(``"%s \n"``, str);` `        ``// Find the rightmost character which is smaller than its next``        ``// character. Let us call it 'first char'``        ``int` `i;``        ``for` `( i = size - 2; i >= 0; --i )``        ``if` `(str[i] < str[i+1])``            ``break``;` `        ``// If there is no such character, all are sorted in decreasing order,``        ``// means we just printed the last permutation and we are done.``        ``if` `( i == -1 )``            ``isFinished = ``true``;``        ``else``        ``{``            ``// Find the ceil of 'first char' in right of first character.``            ``// Ceil of a character is the smallest character greater than it``            ``int` `ceilIndex = findCeil( str, str[i], i + 1, size - 1 );` `            ``// Swap first and second characters``            ``swap( &str[i], &str[ceilIndex] );` `            ``// Sort the string on right of 'first char'``            ``qsort``( str + i + 1, size - i - 1, ``sizeof``(str[0]), compare );``        ``}``    ``}``}` `// Driver program to test above function``int` `main()``{``    ``char` `str[] = ``"ABCD"``;``    ``sortedPermutations( str );``    ``return` `0;``}`

## Java

 `// Java Program to print all permutations``// of a string in sorted order.` `import` `java.util.*;` `class` `GFG``{` `  ``// This function finds the index of the smallest``  ``// character which is greater than 'first' and is``  ``// present in str[l..h]``  ``static` `int` `findCeil(``char``[] str, ``char` `first, ``int` `l,``                      ``int` `h)``  ``{``    ``// initialize index of ceiling element``    ``int` `ceilIndex = l;` `    ``// Now iterate through rest of the elements and find``    ``// the smallest character greater than 'first'``    ``for` `(``int` `i = l + ``1``; i <= h; i++)``      ``if` `(str[i] > first && str[i] < str[ceilIndex])``        ``ceilIndex = i;` `    ``return` `ceilIndex;``  ``}` `  ``// Print all permutations of str in sorted order``  ``static` `void` `sortedPermutations(``char``[] str)``  ``{``    ``// Get size of string``    ``int` `size = str.length;` `    ``// Sort the string in increasing order``    ``Arrays.sort(str);` `    ``// Print permutations one by one``    ``boolean` `isFinished = ``false``;``    ``while` `(!isFinished) {``      ``// print this permutation``      ``System.out.println(String.valueOf(str));` `      ``// Find the rightmost character which is``      ``// smaller than its next character.``      ``// Let us call it 'first char'``      ``int` `i;``      ``for` `(i = size - ``2``; i >= ``0``; --i)``        ``if` `(str[i] < str[i + ``1``])``          ``break``;` `      ``// If there is no such character, all are``      ``// sorted in decreasing order, means we``      ``// just printed the last permutation and we are``      ``// done.``      ``if` `(i == -``1``)``        ``isFinished = ``true``;``      ``else` `{``        ``// Find the ceil of 'first char' in``        ``// right of first character.``        ``// Ceil of a character is the smallest``        ``// character greater than it``        ``int` `ceilIndex = findCeil(str, str[i], i + ``1``,``                                 ``size - ``1``);` `        ``// Swap first and second characters``        ``char` `temp = str[i];``        ``str[i] = str[ceilIndex];``        ``str[ceilIndex] = temp;` `        ``// Sort the string on right of 'first char'``        ``Arrays.sort(str, i + ``1``, size);``      ``}``    ``}``  ``}` `  ``// Driver program to test above function``  ``public` `static` `void` `main(String[] args)``  ``{``    ``char``[] str = { ``'A'``, ``'B'``, ``'C'``, ``'D'` `};``    ``sortedPermutations(str);``  ``}``}` `// This is code is contributed by phasing17`

## Python3

 `# Python3 Program to print all permutations``# of a string in sorted order.` `# The following function is needed for the sort method`  `# This function finds the index of the smallest character``# which is greater than 'first' and is present in str[l..h]``def` `findCeil(str_, first, l, h):` `    ``# initialize index of ceiling element``    ``ceilIndex ``=` `l` `    ``# Now iterate through rest of the elements and find``    ``# the smallest character greater than 'first'``    ``for` `i ``in` `range``(l ``+` `1``, h ``+` `1``):``        ``if` `(str_[i] > first ``and` `str_[i] < str_[ceilIndex]):``            ``ceilIndex ``=` `i` `    ``return` `ceilIndex`  `# Print all permutations of str in sorted order``def` `sortedPermutations(str_):` `    ``# Get size of string``    ``size ``=` `len``(str_)` `    ``# Sort the string in increasing order``    ``str_ ``=` `sorted``(str_)` `    ``# Print permutations one by one``    ``isFinished ``=` `False``    ``while` `(isFinished ``is` `False``):` `        ``# print this permutation``        ``print``("".join(str_))` `        ``# Find the rightmost character which is``        ``# smaller than its next character.``        ``# Let us call it 'first char'``        ``i ``=` `size ``-` `2``        ``while` `i >``=` `0``:``            ``if` `(str_[i] < str_[i``+``1``]):``                ``break``            ``i ``-``=` `1` `        ``# If there is no such character, all are``        ``# sorted in decreasing order, means we``        ``# just printed the last permutation and we are done.``        ``if` `(i ``=``=` `-``1``):``            ``isFinished ``=` `True``        ``else``:` `            ``# Find the ceil of 'first char' in``            ``# right of first character.``            ``# Ceil of a character is the smallest``            ``# character greater than it``            ``ceilIndex ``=` `findCeil(str_, str_[i], i ``+` `1``, size ``-` `1``)` `            ``# Swap first and second characters``            ``temp ``=` `str_[i]``            ``str_[i] ``=` `str_[ceilIndex]``            ``str_[ceilIndex] ``=` `temp` `            ``# Sort the string on right of 'first char'``            ``str_ ``=` `str_[:i ``+` `1``] ``+` `sorted``(str_[i ``+` `1``:])`  `# Driver program to test above function``str_ ``=` `"ABCD"``sortedPermutations(str_)` `# This code is contributed by phasing17`

## C#

 `// C# Program to print all permutations``// of a string in sorted order.` `using` `System;``using` `System.Collections.Generic;` `class` `GFG {``    ``// This function finds the index of the smallest``    ``// character which is greater than 'first' and is``    ``// present in str[l..h]``    ``static` `int` `findCeil(``char``[] str, ``char` `first, ``int` `l,``                        ``int` `h)``    ``{``        ``// initialize index of ceiling element``        ``int` `ceilIndex = l;` `        ``// Now iterate through rest of the elements and find``        ``// the smallest character greater than 'first'``        ``for` `(``int` `i = l + 1; i <= h; i++)``            ``if` `(str[i] > first && str[i] < str[ceilIndex])``                ``ceilIndex = i;` `        ``return` `ceilIndex;``    ``}` `    ``// Print all permutations of str in sorted order``    ``static` `void` `sortedPermutations(``char``[] str)``    ``{``        ``// Get size of string``        ``int` `size = str.Length;` `        ``// Sort the string in increasing order``        ``Array.Sort(str);` `        ``// Print permutations one by one``        ``bool` `isFinished = ``false``;``        ``while` `(!isFinished) {``            ``// print this permutation``            ``Console.WriteLine(``new` `string``(str));` `            ``// Find the rightmost character which is``            ``// smaller than its next character.``            ``// Let us call it 'first char'``            ``int` `i;``            ``for` `(i = size - 2; i >= 0; --i)``                ``if` `(str[i] < str[i + 1])``                    ``break``;` `            ``// If there is no such character, all are``            ``// sorted in decreasing order, means we``            ``// just printed the last permutation and we are``            ``// done.``            ``if` `(i == -1)``                ``isFinished = ``true``;``            ``else` `{``                ``// Find the ceil of 'first char' in``                ``// right of first character.``                ``// Ceil of a character is the smallest``                ``// character greater than it``                ``int` `ceilIndex = findCeil(str, str[i], i + 1,``                                         ``size - 1);` `                ``// Swap first and second characters``                ``char` `temp = str[i];``                ``str[i] = str[ceilIndex];``                ``str[ceilIndex] = temp;` `                ``// Sort the string on right of 'first char'``                ``Array.Sort(str, i + 1, size - i - 1);``            ``}``        ``}``    ``}` `    ``// Driver program to test above function``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``char``[] str = { ``'A'``, ``'B'``, ``'C'``, ``'D'` `};``        ``sortedPermutations(str);``    ``}``}` `// This is code is contributed by phasing17`

## Javascript

 `// JavaScript Program to print all permutations``// of a string in sorted order.` `/* The following function is needed for the sort method */``function` `compare (a, b)``{``    ``return` `a.charCodeAt(0) - b.charCodeAt(0);``}` `// This function finds the index of the smallest character``// which is greater than 'first' and is present in str[l..h]``function` `findCeil (str, first, l, h)``{``    ``// initialize index of ceiling element``    ``let ceilIndex = l;` `    ``// Now iterate through rest of the elements and find``    ``// the smallest character greater than 'first'``    ``for` `(let i = l+1; i <= h; i++)``    ``if` `(str[i] > first && str[i] < str[ceilIndex])``            ``ceilIndex = i;` `    ``return` `ceilIndex;``}` `// Print all permutations of str in sorted order``function` `sortedPermutations ( str)``{``    ``// Get size of string``    ``let size = str.length;` `    ``// Sort the string in increasing order``    ``str = str.split(``""``);``    ``str.sort(compare);` `    ``// Print permutations one by one``    ``let isFinished = ``false``;``    ``while` `( ! isFinished )``    ``{``        ``// print this permutation``        ``console.log(str.join(``""``));` `        ``// Find the rightmost character which is``        ``// smaller than its next character.``        ``// Let us call it 'first char'``        ``let i;``        ``for` `( i = size - 2; i >= 0; --i )``        ``if` `(str[i] < str[i+1])``            ``break``;` `        ``// If there is no such character, all are``        ``// sorted in decreasing order, means we``        ``// just printed the last permutation and we are done.``        ``if` `( i == -1 )``            ``isFinished = ``true``;``        ``else``        ``{``            ``// Find the ceil of 'first char' in``            ``// right of first character.``            ``// Ceil of a character is the smallest``            ``// character greater than it``            ``let ceilIndex = findCeil( str, str[i], i + 1, size - 1 );` `            ``// Swap first and second characters``            ``let temp = str[i];``            ``str[i] = str[ceilIndex];``            ``str[ceilIndex] = temp;` `            ``// Sort the string on right of 'first char'``            ``str1 = str.slice(i + 1);``            ``str1.sort();``            ``str = str.slice(0, i + 1);``            ``str.push(...str1);``        ``}``    ``}``}` `// Driver program to test above function``let str = ``"ABCD"``;``sortedPermutations( str );` `// This code is contributed by phasing17`

Output:

```ABCD
ABDC
....
....
DCAB
DCBA```

Time Complexity:  O(n2 x n!)

Auxiliary Space: O(n)

We can optimize step 4 of the above algorithm for finding next permutation. Instead of sorting the subarray after the ‘first character’, we can reverse the subarray, because the subarray we get after swapping is always sorted in non-increasing order. This optimization makes the time complexity as O(n x n!)

See following optimized code.

## C++

 `#include ``using` `namespace` `std;` `// An optimized version that uses reverse instead of sort for``// finding the next permutation` `// A utility function to reverse a string str[l..h]``void` `reverse(``char` `str[], ``int` `l, ``int` `h)``{``    ``while` `(l < h)``    ``{``        ``swap(&str[l], &str[h]);``        ``l++;``        ``h--;``    ``}``}` `// Print all permutations of str in sorted order``void` `sortedPermutations ( ``char` `str[] )``{``    ``// Get size of string``    ``int` `size = ``strlen``(str);` `    ``// Sort the string in increasing order``    ``qsort``( str, size, ``sizeof``( str[0] ), compare );` `    ``// Print permutations one by one``    ``bool` `isFinished = ``false``;``    ``while` `( ! isFinished )``    ``{``        ``// print this permutation``        ``cout << str << endl;` `        ``// Find the rightmost character which``        ``// is smaller than its next character.``        ``// Let us call it 'first char'``        ``int` `i;``        ``for` `( i = size - 2; i >= 0; --i )``        ``if` `(str[i] < str[i+1])``            ``break``;` `        ``// If there is no such character, all``        ``// are sorted in decreasing order,``        ``// means we just printed the last``        ``// permutation and we are done.``        ``if` `( i == -1 )``            ``isFinished = ``true``;``        ``else``        ``{``            ``// Find the ceil of 'first char' in``            ``// right of first character.``            ``// Ceil of a character is the``            ``// smallest character greater than it``            ``int` `ceilIndex = findCeil( str, str[i], i + 1, size - 1 );` `            ``// Swap first and second characters``            ``swap( &str[i], &str[ceilIndex] );` `            ``// reverse the string on right of 'first char'``            ``reverse( str, i + 1, size - 1 );``        ``}``    ``}``}` `// This code is contributed by rathbhupendra`

## C

 `// An optimized version that uses reverse instead of sort for``// finding the next permutation` `// A utility function to reverse a string str[l..h]``void` `reverse(``char` `str[], ``int` `l, ``int` `h)``{``   ``while` `(l < h)``   ``{``       ``swap(&str[l], &str[h]);``       ``l++;``       ``h--;``   ``}``}` `// Print all permutations of str in sorted order``void` `sortedPermutations ( ``char` `str[] )``{``    ``// Get size of string``    ``int` `size = ``strlen``(str);` `    ``// Sort the string in increasing order``    ``qsort``( str, size, ``sizeof``( str[0] ), compare );` `    ``// Print permutations one by one``    ``bool` `isFinished = ``false``;``    ``while` `( ! isFinished )``    ``{``        ``// print this permutation``        ``printf` `(``"%s \n"``, str);` `        ``// Find the rightmost character which is smaller than its next``        ``// character. Let us call it 'first char'``        ``int` `i;``        ``for` `( i = size - 2; i >= 0; --i )``           ``if` `(str[i] < str[i+1])``              ``break``;` `        ``// If there is no such character, all are sorted in decreasing order,``        ``// means we just printed the last permutation and we are done.``        ``if` `( i == -1 )``            ``isFinished = ``true``;``        ``else``        ``{``            ``// Find the ceil of 'first char' in right of first character.``            ``// Ceil of a character is the smallest character greater than it``            ``int` `ceilIndex = findCeil( str, str[i], i + 1, size - 1 );` `            ``// Swap first and second characters``            ``swap( &str[i], &str[ceilIndex] );` `            ``// reverse the string on right of 'first char'``            ``reverse( str, i + 1, size - 1 );``        ``}``    ``}``}`

## Java

 `import` `java.util.*;` `// An optimized version that uses reverse instead of sort``// for finding the next permutation``class` `GFG``{``  ` `    ``// A utility function two swap two characters a and b``    ``static` `void` `swap(``char``[] str, ``int` `i, ``int` `j)``    ``{``        ``char` `t = str[i];``        ``str[i] = str[j];``        ``str[j] = t;``    ``}` `    ``// A utility function to reverse a string str[l..h]``    ``static` `void` `reverse(``char` `str[], ``int` `l, ``int` `h)``    ``{``        ``while` `(l < h) {``            ``swap(str, l, h);``            ``l++;``            ``h--;``        ``}``    ``}` `    ``// This function finds the index of the smallest``    ``// character which is greater than 'first' and is``    ``// present in str[l..h]``    ``static` `int` `findCeil(``char` `str[], ``char` `first, ``int` `l,``                        ``int` `h)``    ``{``        ``// initialize index of ceiling element``        ``int` `ceilIndex = l;` `        ``// Now iterate through rest of the elements and find``        ``// the smallest character greater than 'first'``        ``for` `(``int` `i = l + ``1``; i <= h; i++)``            ``if` `(str[i] > first && str[i] < str[ceilIndex])``                ``ceilIndex = i;` `        ``return` `ceilIndex;``    ``}` `    ``// Print all permutations of str in sorted order``    ``static` `void` `sortedPermutations(``char` `str[])``    ``{``        ``// Get size of string``        ``int` `size = str.length;` `        ``// Sort the string in increasing order``        ``Arrays.sort(str);` `        ``// Print permutations one by one``        ``boolean` `isFinished = ``false``;``        ``while` `(!isFinished) {``            ``// print this permutation``            ``System.out.println(str);` `            ``// Find the rightmost character which is``            ``// smaller than its next character.``            ``// Let us call it 'first char'``            ``int` `i;``            ``for` `(i = size - ``2``; i >= ``0``; --i)``                ``if` `(str[i] < str[i + ``1``])``                    ``break``;` `            ``// If there is no such character, all are``            ``// sorted in decreasing order, means we``            ``// just printed the last permutation and we are``            ``// done.``            ``if` `(i == -``1``)``                ``isFinished = ``true``;``            ``else` `{``                ``// Find the ceil of 'first char' in``                ``// right of first character.``                ``// Ceil of a character is the smallest``                ``// character greater than it``                ``int` `ceilIndex = findCeil(str, str[i], i + ``1``,``                                         ``size - ``1``);` `                ``// Swap first and second characters``                ``swap(str, i, ceilIndex);` `                ``// reverse the string on right of 'first``                ``// char'``                ``reverse(str, i + ``1``, size - ``1``);``            ``}``        ``}``    ``}` `    ``// Driver program to test above function``    ``public` `static` `void` `main(String[] args)``    ``{``        ``char` `str[] = ``"ABCD"``.toCharArray();``        ``sortedPermutations(str);``    ``}``}` `// This code is contributed by Swarn Pallav Bhaskar`

## Python3

 `# An optimized version that uses reverse``# instead of sort for finding the next``# permutation` `# A utility function to reverse a``# string str[l..h]``def` `reverse(``str``, l, h):``    ` `    ``while` `(l < h) :``        ``str``[l], ``str``[h] ``=` `str``[h], ``str``[l]``        ``l ``+``=` `1``        ``h ``-``=` `1``        ` `    ``return` `str``    ` `def` `findCeil(``str``, c, k, n):``    ` `    ``ans ``=` `-``1``    ``val ``=` `c``    ` `    ``for` `i ``in` `range``(k, n ``+` `1``):``        ``if` `str``[i] > c ``and` `str``[i] < val:``            ``val ``=` `str``[i]``            ``ans ``=` `i``            ` `    ``return` `ans``            ` `# Print all permutations of str in sorted order``def` `sortedPermutations(``str``):``    ` `    ``# Get size of string``    ``size ``=` `len``(``str``)` `    ``# Sort the string in increasing order``    ``str` `=` `''.join(``sorted``(``str``))` `    ``# Print permutations one by one``    ``isFinished ``=` `False``    ` `    ``while` `(``not` `isFinished):``        ` `        ``# Print this permutation``        ``print``(``str``)` `        ``# Find the rightmost character which``        ``# is smaller than its next character.``        ``# Let us call it 'first char' ``        ``for` `i ``in` `range``(size ``-` `2``, ``-``1``, ``-``1``):``            ``if` `(``str``[i] < ``str``[i ``+` `1``]):``                ``break` `        ``# If there is no such character, all``        ``# are sorted in decreasing order,``        ``# means we just printed the last``        ``# permutation and we are done.``        ``if` `(i ``=``=` `-``1``):``            ``isFinished ``=` `True``        ``else``:``            ` `            ``# Find the ceil of 'first char' in``            ``# right of first character.``            ``# Ceil of a character is the``            ``# smallest character greater than it``            ``ceilIndex ``=` `findCeil(``str``, ``str``[i], i ``+` `1``,``                                           ``size ``-` `1``)` `            ``# Swap first and second characters``            ``str``[i], ``str``[ceilIndex] ``=`  `str``[ceilIndex], ``str``[i]` `            ``# Reverse the string on right of 'first char'``            ``str` `=` `reverse(``str``, i ``+` `1``, size ``-` `1``)``            ` `# This code is contributed by rohan07`

## C#

 `using` `System;` `// An optimized version that uses reverse instead of sort``// for finding the next permutation``public` `class` `GFG {` `    ``// A utility function two swap two characters a and b``    ``static` `void` `swap(``char``[] str, ``int` `i, ``int` `j) {``        ``char` `t = str[i];``        ``str[i] = str[j];``        ``str[j] = t;``    ``}` `    ``// A utility function to reverse a string str[l..h]``    ``static` `void` `reverse(``char` `[]str, ``int` `l, ``int` `h) {``        ``while` `(l < h) {``            ``swap(str, l, h);``            ``l++;``            ``h--;``        ``}``    ``}` `    ``// This function finds the index of the smallest``    ``// character which is greater than 'first' and is``    ``// present in str[l..h]``    ``static` `int` `findCeil(``char` `[]str, ``char` `first, ``int` `l, ``int` `h) {``        ``// initialize index of ceiling element``        ``int` `ceilIndex = l;` `        ``// Now iterate through rest of the elements and find``        ``// the smallest character greater than 'first'``        ``for` `(``int` `i = l + 1; i <= h; i++)``            ``if` `(str[i] > first && str[i] < str[ceilIndex])``                ``ceilIndex = i;` `        ``return` `ceilIndex;``    ``}` `    ``// Print all permutations of str in sorted order``    ``static` `void` `sortedPermutations(``char` `[]str) {``        ``// Get size of string``        ``int` `size = str.Length;` `        ``// Sort the string in increasing order``        ``Array.Sort(str);` `        ``// Print permutations one by one``        ``bool` `isFinished = ``false``;``        ``while` `(!isFinished) {``            ``// print this permutation``            ``Console.WriteLine(str);` `            ``// Find the rightmost character which is``            ``// smaller than its next character.``            ``// Let us call it 'first char'``            ``int` `i;``            ``for` `(i = size - 2; i >= 0; --i)``                ``if` `(str[i] < str[i + 1])``                    ``break``;` `            ``// If there is no such character, all are``            ``// sorted in decreasing order, means we``            ``// just printed the last permutation and we are``            ``// done.``            ``if` `(i == -1)``                ``isFinished = ``true``;``            ``else` `{``                ``// Find the ceil of 'first char' in``                ``// right of first character.``                ``// Ceil of a character is the smallest``                ``// character greater than it``                ``int` `ceilIndex = findCeil(str, str[i], i + 1, size - 1);` `                ``// Swap first and second characters``                ``swap(str, i, ceilIndex);` `                ``// reverse the string on right of 'first``                ``// char'``                ``reverse(str, i + 1, size - 1);``            ``}``        ``}``    ``}` `    ``// Driver program to test above function``    ``public` `static` `void` `Main(String[] args) {``        ``char` `[]str = ``"ABCD"``.ToCharArray();``        ``sortedPermutations(str);``    ``}``}` `// This code contributed by umadevi9616`

## Javascript

 ``

Time Complexity: O(n*n!)

Auxiliary Space: O(n)

The above programs print duplicate permutation when characters are repeated. We can avoid it by keeping track of the previous permutation. While printing, if the current permutation is same as previous permutation, we won’t be printing it.

My Personal Notes arrow_drop_up