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Remove all occurrences of a character in a string | Recursive approach

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  • Difficulty Level : Medium
  • Last Updated : 31 May, 2022

Given string str, the task is to write a recursive program to remove all the occurrences of a character X in the string.

Examples:

Input: str = “geeksforgeeks”, c = ‘e’ 
Output: gksforgks
Input: str = “geeksforgeeks”, c = ‘g’ 
Output: eeksforeeks 
 

Iterative Approach: The iterative approach to this problem can be found in this post.
Recursive Approach: Below are the steps: 
 

  1. Get the string str and character X for which character X is to be removed.
  2. Recursively iterate for all the character in the string: 
    • Base Case: If the length of the string str called recursively is 0 then return the empty string from the function. 
       
if(str.length()==0) {
   return "";
}
  • Recursive Call: If the base case is not met, then check for the character at 0th index if it is X then recursively iterate for the substring removing the first character.
if (str[0] == X) {
        return recursive_function(str.substr(1), X);
}
  • Return Statement: At each recursive call(except the base case and the above condition), return the recursive function for the next iteration including the character at 0th index
     
return str[0] + recursive_function(str.substr(1), X)

Below is the implementation of the above approach:
 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to remove all occurrences
// of a character in the string
string removeCharRecursive(string str,
                           char X)
{
    // Base Case
    if (str.length() == 0) {
        return "";
    }
 
    // Check the first character
    // of the given string
    if (str[0] == X) {
 
        // Pass the rest of the string
        // to recursion Function call
        return removeCharRecursive(str.substr(1), X);
    }
 
    // Add the first character of str
    // and string from recursion
    return str[0]
           + removeCharRecursive(str.substr(1), X);
}
 
// Driver Code
int main()
{
    // Given String
    string str = "geeksforgeeks";
 
    // Given character
    char X = 'e';
 
    // Function Call
    str = removeCharRecursive(str, X);
    cout << str;
    return 0;
}

Java




// Java program for the above approach
class GFG{
     
// Function to remove all occurrences
// of a character in the string
static String removeCharRecursive(String str,
                                  char X)
{
     
    // Base Case
    if (str.length() == 0)
    {
        return "";
    }
 
    // Check the first character
    // of the given string
    if (str.charAt(0) == X)
    {
 
        // Pass the rest of the string
        // to recursion Function call
        return removeCharRecursive(
               str.substring(1), X);
    }
 
    // Add the first character of str
    // and string from recursion
    return str.charAt(0) +
           removeCharRecursive(
           str.substring(1), X);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given String
    String str = "geeksforgeeks";
 
    // Given character
    char X = 'e';
 
    // Function call
    str = removeCharRecursive(str, X);
     
    System.out.println(str);
}
}
 
// This code is contributed by jrishabh99

Python3




# Python3 program for the above approach
 
# Function to remove all occurrences
# of a character in the string
def removeCharRecursive(str, X):
     
    # Base Case
    if (len(str) == 0):
        return ""
     
    # Check the first character
    # of the given string
    if (str[0] == X):
 
        # Pass the rest of the string
        # to recursion Function call
        return removeCharRecursive(str[1:], X)
     
    # Add the first character of str
    # and string from recursion
    return str[0] + removeCharRecursive(str[1:], X)
 
# Driver Code
 
# Given String
str = "geeksforgeeks"
 
# Given character
X = 'e'
 
# Function call
str = removeCharRecursive(str, X)
 
print(str)
 
# This code is contributed by sanjoy_62

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to remove all occurrences
// of a character in the string
static String removeCharRecursive(String str,
                                  char X)
{
     
    // Base Case
    if (str.Length == 0)
    {
        return "";
    }
 
    // Check the first character
    // of the given string
    if (str[0] == X)
    {
 
        // Pass the rest of the string
        // to recursion Function call
        return removeCharRecursive(
            str.Substring(1), X);
    }
 
    // Add the first character of str
    // and string from recursion
    return str[0] + removeCharRecursive(
                    str.Substring(1), X);
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given String
    String str = "geeksforgeeks";
 
    // Given character
    char X = 'e';
 
    // Function call
    str = removeCharRecursive(str, X);
     
    Console.WriteLine(str);
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
 
// javascript program for the above approach
 
     
// Function to remove all occurrences
// of a character in the string
function removeCharRecursive(str,X)
{
     
    // Base Case
    if (str.length == 0)
    {
        return "";
    }
 
    // Check the first character
    // of the given string
    if (str.charAt(0) == X)
    {
 
        // Pass the rest of the string
        // to recursion Function call
        return removeCharRecursive(
               str.substring(1), X);
    }
 
    // Add the first character of str
    // and string from recursion
    return str.charAt(0) +
           removeCharRecursive(
           str.substring(1), X);
}
 
// Driver Code
//Given String
var str = "geeksforgeeks";
 
// Given character
var X = 'e';
 
// Function call
str = removeCharRecursive(str, X);
 
document.write(str);
 
// This code is contributed by 29AjayKumar
</script>

Output: 

gksforgks

 

Time Complexity: O(N), as we are using a recursive calls to traverse N times, where N is the length of the string.

Auxiliary Space: O(1), as we are not using any extra space.
 


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