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

• Difficulty Level : Easy
• Last Updated : 19 Sep, 2022

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 ``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

 ``

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``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

 ``

Output

`geeksksxxx`

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

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