Move all occurrence of letter ‘x’ from the string s to the end using Recursion

Given a string s, our task is to move all the occurrence of letter x to the end of the string s using recursion.
Note: If there are only letter x in the given string then return the string unaltered.

Examples:

Input: s= “geekxsforgexxeksxx” 
Output: geeksforgeeksxxxxx 
Explanation:
All occurrence of letter ‘x’ is moved to the end.

Input: s = “xxxxx” 
Output: xxxxx 
Explanation:
Since there are only letter x in the given string therefore the output is unaltered. 

Approach:
To solve the problem mentioned above we can use Recursion. Traverse in the string and check recursively if the current character is equal to the character ‘x’ or not. If not then print the character otherwise move to the next character until the length of the string s is reached.

Below is the implementation of the above approach:  

C++

// C++ implementation to Move all occurrence of letter ‘x’
// from the string s to the end using Recursion
#include <bits/stdc++.h>
using namespace std;

// Function to move all 'x' in the end
void moveAtEnd(string s, int i, int l)
{
    if (i >= l)
        return;

    // Store current character
    char curr = s[i];

    // Check if current character is not 'x'
    if (curr != 'x')
        cout << curr;

    // recursive function call
    moveAtEnd(s, i + 1, l);

    // Check if current character is 'x'
    if (curr == 'x')
        cout << curr;

    return;
}

// Driver code
int main()
{
    string s = "geekxsforgexxeksxx";

    int l = s.length();

    moveAtEnd(s, 0, l);

    return 0;
}

Java

// Java implementation to Move all occurrence of letter ‘x’
// from the string s to the end using Recursion
import java.util.*;
class GFG{

// Function to move all 'x' in the end
static void moveAtEnd(String s, int i, int l)
{
    if (i >= l)
        return;

    // Store current character
    char curr = s.charAt(i);

    // Check if current character is not 'x'
    if (curr != 'x')
        System.out.print(curr);

    // recursive function call
    moveAtEnd(s, i + 1, l);

    // Check if current character is 'x'
    if (curr == 'x')
        System.out.print(curr);

    return;
}

// Driver code
public static void main(String args[])
{
    String s = "geekxsforgexxeksxx";

    int l = s.length();

    moveAtEnd(s, 0, l);
}
}

// This code is contributed by Code_Mech

Python3

# Python3 implementation to move all
# occurrences of letter ‘x’ from the
# string s to the end using recursion

# Function to move all 'x' in the end
def moveAtEnd(s, i, l):

    if(i >= l):
       return

    # Store current character
    curr = s[i]

    # Check if current character
    # is not 'x'
    if(curr != 'x'):
        print(curr, end = "")

    # Recursive function call
    moveAtEnd(s, i + 1, l)

    # Check if current character is 'x'
    if(curr == 'x'):
        print(curr, end = "")

    return

# Driver code
if __name__ == '__main__':

    s = "geekxsforgexxeksxx"
    l = len(s)

    moveAtEnd(s, 0, l)

# This code is contributed by Shivam Singh

C#

// C# implementation to Move all occurrence of letter ‘x’
// from the string s to the end using Recursion
using System;
class GFG{

// Function to move all 'x' in the end
static void moveAtEnd(string s, int i, int l)
{
    if (i >= l)
        return;

    // Store current character
    char curr = s[i];

    // Check if current character is not 'x'
    if (curr != 'x')
        Console.Write(curr);

    // recursive function call
    moveAtEnd(s, i + 1, l);

    // Check if current character is 'x'
    if (curr == 'x')
        Console.Write(curr);

    return;
}

// Driver code
public static void Main()
{
    string s = "geekxsforgexxeksxx";

    int l = s.Length;

    moveAtEnd(s, 0, l);
}
}

// This code is contributed by Nidhi_Biet

Javascript

<script>

// Javascript implementation to Move
// all occurrence of letter ‘x’ from
// the string s to the end using Recursion

// Function to move all 'x' in the end
function moveAtEnd(s, i, l)
{
    if (i >= l)
        return;

    // Store current character
    let curr = s[i];

    // Check if current character is not 'x'
    if (curr != 'x')
        document.write(curr);

    // Recursive function call
    moveAtEnd(s, i + 1, l);

    // Check if current character is 'x'
    if (curr == 'x')
        document.write(curr);

    return;
}

// Driver code
let s = "geekxsforgexxeksxx";
let l = s.length;

moveAtEnd(s, 0, l);

// This code is contributed by suresh07

</script>

Output

geeksforgeeksxxxxx

Time Complexity: O(n), where n is the length of the given string.

Auxiliary Space: O(n) for call stack

Another Implementation involving swapping of characters:

In this approach, we will be swapping adjacent characters to bring ‘x’ at the end.

Below is the implementation of the above technique:

C++

// C++ program for above approach
#include<bits/stdc++.h>
using namespace std;

// Recursive program to bring 'x'
// to the end
void rec(char *a, int i)
{

    // When the string is completed
    // from reverse direction end of recursion
    if(i == 0)
    {
      cout << a << endl;
      return;
    }

    // If the character x is found
    if(a[i] == 'x')
    {

      // Transverse the whole string
      int j = i;
      while(a[j] != '\0' && a[j+1] != '\0')
      {

        // Swap the x so that
        // it moves to the last
        swap(a[j], a[j+1]);
        j++;
      }
    }

    // call to the smaller problem now
    rec(a, i - 1);
}

// Driver Code
int main()
{
    char a[] = {'g', 'e', 'e', 'k', 'x',
            's', 'x', 'x', 'k', 's', '\0'};

    // Size of a
    int n = 10;

    // Call to rec
    rec(a,n-1);
}
/* This code is contributed by Harsh kedia */

Java

// Java program for the
// above approach
import java.util.*;
class Main{

// Recursive program to
// bring 'x' to the end
public static void rec(char a[],
                       int i)
{
  // When the string is completed
  // from reverse direction end
  // of recursion
  if(i == 0)
  {
    System.out.println(a);
    return;
  }

  // If the character x is found
  if(a[i] == 'x')
  {

    // Transverse the whole string
    int j = i;
    while(a[j] != '\0' &&
          a[j + 1] != '\0')
    {
      // Swap the x so that
      // it moves to the last
      char temp = a[j];
      a[j] = a[j + 1];
      a[j + 1] = temp;
      j++;
    }
  }

  // call to the smaller
  // problem now
  rec(a, i - 1);
}

// Driver code
public static void main(String[] args)
{
  char a[] = {'g', 'e', 'e', 'k',
              'x', 's', 'x', 'x',
              'k', 's', '\0'};

  // Size of a
  int n = 10;

  // Call to rec
  rec(a,n-1);
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for above approach

# Recursive program to bring 'x'
# to the end
def rec(a, i):

    # When the string is completed
    # from reverse direction end
    # of recursion
    if (i == 0):
        a.pop()
        print("".join(a))
        return

    # If the character x is found
    if (a[i] == 'x'):

      # Transverse the whole string
      j = i
      while(a[j] != '\0' and
            a[j + 1] != '\0'):

        # Swap the x so that
        # it moves to the last
        (a[j], a[j + 1]) = (a[j + 1], a[j])
        j += 1

    # Call to the smaller problem now
    rec(a, i - 1)

# Driver code
if __name__=="__main__":

    a = [ 'g', 'e', 'e', 'k', 'x',
          's', 'x', 'x', 'k', 's', '\0' ]

    # Size of a
    n = 10

    # Call to rec
    rec(a, n - 1)

# This code is contributed by rutvik_56

C#

// C# program for the
// above approach
using System;
class GFG
{

    // Recursive program to
    // bring 'x' to the end
    static void rec(char[] a, int i)
    {

      // When the string is completed
      // from reverse direction end
      // of recursion
      if(i == 0)
      {
        Console.WriteLine(a);
        return;
      }

      // If the character x is found
      if(a[i] == 'x')
      {

        // Transverse the whole string
        int j = i;
        while(a[j] != '\0' &&
              a[j + 1] != '\0')
        {

          // Swap the x so that
          // it moves to the last
          char temp = a[j];
          a[j] = a[j + 1];
          a[j + 1] = temp;
          j++;
        }
      }

      // call to the smaller
      // problem now
      rec(a, i - 1);
    }

  // Driver code
  static void Main()
  {
      char[] a = {'g', 'e', 'e', 'k',
              'x', 's', 'x', 'x',
              'k', 's', '\0'};

      // Size of a
      int n = 10;

      // Call to rec
      rec(a,n-1);
  }
}

// This code is contributed by divyesh072019

Javascript

<script>
    // Javascript program for the above approach

    // Recursive program to
    // bring 'x' to the end
    function rec(a, i)
    {

      // When the string is completed
      // from reverse direction end
      // of recursion
      if(i == 0)
      {
        document.write(a.join(""));
        return;
      }

      // If the character x is found
      if(a[i] == 'x')
      {

        // Transverse the whole string
        let j = i;
        while(a[j] != '\0' &&
              a[j + 1] != '\0')
        {

          // Swap the x so that
          // it moves to the last
          let temp = a[j];
          a[j] = a[j + 1];
          a[j + 1] = temp;
          j++;
        }
      }

      // call to the smaller
      // problem now
      rec(a, i - 1);
    }

    let a = ['g', 'e', 'e', 'k', 'x', 's', 'x', 'x', 'k', 's', '\0'];

    // Size of a
    let n = 10;

    // Call to rec
    rec(a, n - 1);

    // This code is contributed by decode2207.
</script>

Output

geeksksxxx

Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(N).