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: neveropenxxxxx
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 endvoid 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 codeint 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 Recursionimport java.util.*;class GFG{// Function to move all 'x' in the endstatic 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 codepublic 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 enddef 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 codeif __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 Recursionusing System;class GFG{// Function to move all 'x' in the endstatic 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 codepublic 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 endfunction 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 codelet s = "geekxsforgexxeksxx";let l = s.length;moveAtEnd(s, 0, l);// This code is contributed by suresh07</script> |
Output
neveropenxxxxx
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 endvoid 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 Codeint 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 approachimport java.util.*;class Main{ // Recursive program to // bring 'x' to the endpublic 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 codepublic 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 enddef 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 codeif __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 approachusing 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
neveropenksxxx
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(N).
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