Move all occurence 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++

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// C++ implementation to Move all occurence 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;
}

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Java

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// Java implementation to Move all occurence 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

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Python3

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# 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

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C#

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// C# implementation to Move all occurence 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

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Output:

geeksforgeeksxxxxx

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