Generate permutations with only adjacent swaps allowed

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, 
   1. Swap it with the next character. Because each character can be swapped atmost once, we’ll move to the position (i+2). 
   2. 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.

Implementation:

C++

// CPP program to generate permutations with only
// one swap allowed.
#include <cstring>
#include <iostream>
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

<?php
// PHP program to generate permutations
// with only one swap allowed.

function findPermutations($str, $index, $n)
{
    if ($index >= $n || ($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

<script>

      // JavaScript program to generate
      // permutations with only
      // one swap allowed.
      function findPermutations(str, index, n) {
        if (index >= n || index + 1 >= n) {
          document.write(str.join("") + "<br>");
          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);
      }

      function swap(arr, index) {
        var temp = arr[index];
        arr[index] = arr[index + 1];
        arr[index + 1] = temp;
      }
      // Driver code

      var str = "12345".split("");
      var n = str.length;
      findPermutations(str, 0, n);

</script>

Output

12345
12354
12435
13245
13254
21345
21354
21435

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