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# Generate permutations with only adjacent swaps allowed

• Difficulty Level : Medium
• Last Updated : 16 Jun, 2021

Given a string on length N. You can swap only the adjacent elements and each element can be swapped atmost once. Find the no of permutations of the string that can be generated after performing the swaps as mentioned.
Examples:

```Input : 12345
Output : 12345 12354 12435 13245 13254
21345 21354 21435```

Source: Goldman Sachs Interview

Consider any i-th character in the string. There are two possibilities for this character:
1.) Don’t swap it, i.e. don’t do anything with this character and move to the next character.
2.) Swap it. As it can be swapped with its adjacent,
……..a.) Swap it with the next character. Because each character can be swapped atmost once, we’ll move to the position (i+2).
……. b.) Swap it with the previous character – we don’t need to consider this case separately as i-th character is the next character of (i-1)th which is same as the case 2.a.

## C++

 `// CPP program to generate permutations with only``// one swap allowed.``#include ``#include ``using` `namespace` `std;` `void` `findPermutations(``char` `str[], ``int` `index, ``int` `n)``{``    ``if` `(index >= n || (index + 1) >= n) {``        ``cout << str << endl;``        ``return``;``    ``}` `    ``// don't swap the current position``    ``findPermutations(str, index + 1, n);` `    ``// Swap with the next character and``    ``// revert the changes. As explained``    ``// above, swapping with previous is``    ``// is not needed as it anyways happens``    ``// for next character.``    ``swap(str[index], str[index + 1]);``    ``findPermutations(str, index + 2, n);``    ``swap(str[index], str[index + 1]);``}` `// Driver code``int` `main()``{``    ``char` `str[] = { ``"12345"` `};``    ``int` `n = ``strlen``(str);``    ``findPermutations(str, 0, n);``    ``return` `0;``}`

## Java

 `// Java program to generate permutations with only``// one swap allowed.` `class` `GFG {` `    ``static` `void` `findPermutations(``char` `str[], ``int` `index, ``int` `n) {``        ``if` `(index >= n || (index + ``1``) >= n) {``            ``System.out.println(str);``            ``return``;``        ``}` `        ``// don't swap the current position``        ``findPermutations(str, index + ``1``, n);` `        ``// Swap with the next character and``        ``// revert the changes. As explained``        ``// above, swapping with previous is``        ``// is not needed as it anyways happens``        ``// for next character.``        ``swap(str, index);``        ``findPermutations(str, index + ``2``, n);``        ``swap(str, index);``    ``}` `    ``static` `void` `swap(``char` `arr[], ``int` `index) {``        ``char` `temp = arr[index];``        ``arr[index] = arr[index + ``1``];``        ``arr[index + ``1``] = temp;``    ``}``// Driver code` `    ``public` `static` `void` `main(String[] args) {``        ``char` `str[] = ``"12345"``.toCharArray();``        ``int` `n = str.length;``        ``findPermutations(str, ``0``, n);``    ``}``}``// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to generate permutations``# with only one swap allowed.``def` `findPermutations(string, index, n):``    ``if` `index >``=` `n ``or` `(index ``+` `1``) >``=` `n:``        ``print``(''.join(string))``        ``return` `    ``# don't swap the current position``    ``findPermutations(string, index ``+` `1``, n)` `    ``# Swap with the next character and``    ``# revert the changes. As explained``    ``# above, swapping with previous is``    ``# is not needed as it anyways happens``    ``# for next character.``    ``string[index], \``    ``string[index ``+` `1``] ``=` `string[index ``+` `1``], \``                        ``string[index]``                        ` `    ``findPermutations(string, index ``+` `2``, n)``    ` `    ``string[index], \``    ``string[index ``+` `1``] ``=` `string[index ``+` `1``], \``                        ``string[index]` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``string ``=` `list``(``"12345"``)``    ``n ``=` `len``(string)``    ``findPermutations(string, ``0``, n)` `# This code is contributed by``# sanjeev2552`

## C#

 `// C# program to generate permutations with only``// one swap allowed.``using` `System;``public` `class` `GFG {` `    ``static` `void` `findPermutations(``char` `[]str, ``int` `index, ``int` `n) {``        ``if` `(index >= n || (index + 1) >= n) {``            ``Console.WriteLine(str);``            ``return``;``        ``}` `        ``// don't swap the current position``        ``findPermutations(str, index + 1, n);` `        ``// Swap with the next character and``        ``// revert the changes. As explained``        ``// above, swapping with previous is``        ``// is not needed as it anyways happens``        ``// for next character.``        ``swap(str, index);``        ``findPermutations(str, index + 2, n);``        ``swap(str, index);``    ``}` `    ``static` `void` `swap(``char` `[]arr, ``int` `index) {``        ``char` `temp = arr[index];``        ``arr[index] = arr[index + 1];``        ``arr[index + 1] = temp;``    ``}``// Driver code` `    ``public` `static` `void` `Main() {``        ``char` `[]str = ``"12345"``.ToCharArray();``        ``int` `n = str.Length;``        ``findPermutations(str, 0, n);``    ``}``}` `// This code is contributed by Rajput-Ji`

## PHP

 `= ``\$n` `|| (``\$index` `+ 1) >= ``\$n``)``    ``{``        ``echo` `\$str``, ``"\n"``;``        ``return``;``    ``}` `    ``// don't swap the current position``    ``findPermutations(``\$str``, ``\$index` `+ 1, ``\$n``);` `    ``// Swap with the next character and``    ``// revert the changes. As explained``    ``// above, swapping with previous is``    ``// is not needed as it anyways happens``    ``// for next character.``    ``list(``\$str``[``\$index``],``         ``\$str``[``\$index` `+ 1]) = ``array``(``\$str``[``\$index` `+ 1],``                                   ``\$str``[``\$index``]);` `    ``findPermutations(``\$str``, ``\$index` `+ 2, ``\$n``);``    ``list(``\$str``[``\$index``],``         ``\$str``[``\$index` `+ 1]) = ``array``(``\$str``[``\$index` `+ 1],``                                   ``\$str``[``\$index``]);``}` `// Driver code``\$str` `= ``"12345"` `;``\$n` `= ``strlen``(``\$str``);``findPermutations(``\$str``, 0, ``\$n``);` `// This code is contributed by Sach_Code``?>`

## Javascript

 ``

Output:

```12345
12354
12435
13245
13254
21345
21354
21435```

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