Replace all occurrences of character X with character Y in given string

Given a string str and two characters X and Y, the task is to write a recursive function to replace all occurrences of character X with character Y.

Examples:

Input: str = abacd, X = a, Y = x 
Output: xbxcd

Input: str = abacd, X = e, Y = y 
Output: abacd  

Iterative Approach: The idea is to iterate over the given string and if any character X is found then replace that character with Y.

Code-

C++

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

// Function to replace all occurrences
// of character c1 with character c2
void replaceCharacter(char* str, char c1, char c2)
{

    int j, n = strlen(str);
    for (int i = j = 0; i < n; i++) {
        if (str[i] != c1) {
            str[j++] = str[i];
        }
        else {
            str[j++] = c2;
        }
    }

    str[j] = '\0';
}

// Driver Code
int main()
{
    // Given string
    char str[] = "abacd";
    char c1 = 'a';
    char c2 = 'x';

    // Function call
    replaceCharacter(str, c1, c2);

    // Print the string
    cout << str;
    return 0;
}

Java

public class Main {
    // Function to replace all occurrences
    // of character c1 with character c2
    static void replaceCharacter(char[] str, char c1, char c2) {
        int j = 0;
        int n = str.length;
        for (int i = 0; i < n; i++) {
            if (str[i] != c1) {
                str[j++] = str[i];
            } else {
                str[j++] = c2;
            }
        }
    }

    public static void main(String[] args) {
        // Given string
        char[] str = "abacd".toCharArray();
        char c1 = 'a';
        char c2 = 'x';

        // Function call
        replaceCharacter(str, c1, c2);

        // Print the string
        System.out.println(str);
    }
}

Python3

# Function to replace all occurrences
# of character c1 with character c2
def replaceCharacter(str, c1, c2):
    n = len(str)
    res = ""
    for i in range(n):
        if str[i] != c1:
            res += str[i]
        else:
            res += c2
    return res

# Driver Code
if __name__ == '__main__':
    # Given string
    str = "abacd"
    c1 = 'a'
    c2 = 'x'

    # Function call
    str = replaceCharacter(str, c1, c2)

    # Print the string
    print(str)

C#

using System;

public class MainClass
{

  // Function to replace all occurrences
  // of character c1 with character c2
  static void ReplaceCharacter(char[] str, char c1, char c2)
  {
    int j = 0;
    int n = str.Length;
    for (int i = 0; i < n; i++)
    {
      if (str[i] != c1)
      {
        str[j++] = str[i];
      }
      else
      {
        str[j++] = c2;
      }
    }
  }

  public static void Main(string[] args)
  {

    // Given string
    char[] str = "abacd".ToCharArray();
    char c1 = 'a';
    char c2 = 'x';

    // Function call
    ReplaceCharacter(str, c1, c2);

    // Print the string
    Console.WriteLine(str);
  }
}

Javascript

// Function to replace all occurrences
// of character c1 with character c2
function replaceCharacter(str, c1, c2) {
    let result = "";
    for (let i = 0; i < str.length; i++) {
        if (str[i] != c1) {
            result += str[i];
        } else {
            result += c2;
        }
    }
    return result;
}

// Given string
let str = "abacd";
let c1 = 'a';
let c2 = 'x';

// Function call
let newStr = replaceCharacter(str, c1, c2);

// Print the string
console.log(newStr);

Output-

xbxcd

Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time.
Auxiliary Space: O(1), as we are not using any extra space.

Recursive Approach: The idea is to recursively traverse the given string and replace the character with value X with Y. Below are the steps:

1. Get the string str, character X, and Y.
2. Recursively Iterate from index 0 to string length.
   • Base Case: If we reach the end of the string the exit from the function. 
     

if(str[0]=='\0')
   return ;

• Recursive Call: If the base case is not met, then check for the character at 0th index if it is X then replace that character with Y and recursively iterate for next character. 
  

if(str[0]==X) 
  str[0] = Y

• Return Statement: At each recursive call(except the base case), return the recursive function for the next iteration. 
  

return recursive_function(str + 1, X, Y)

Below is the recursive implementation:

C++

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

// Function to replace all occurrences
// of character c1 with character c2
void replaceCharacter(char input[],
                    char c1, char c2)
{
    // Base Case
    // If the string is empty
    if (input[0] == '\0') {
        return;
    }

    // If the character at starting
    // of the given string is equal
    // to c1, replace it with c2
    if (input[0] == c1) {
        input[0] = c2;
    }

    // Getting the answer from recursion
    // for the smaller problem
    return replaceCharacter(input + 1,
                            c1, c2);
}

// Driver Code
int main()
{
    // Given string
    char str[] = "abacd";
    char c1 = 'a';
    char c2 = 'x';

    // Function call
    replaceCharacter(str, c1, c2);

    // Print the string
    cout << str;
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
import java.io.*;

class GFG{

// Function to replace all occurrences
// of character c1 with character c2
static String replaceCharacter(String str,
                             char c1, char c2)
{
    // Base Case
    // If the string is empty
    if (str.length() == 1)
    {
        return str;
    }
    char x=str.charAt(0);
    // If the character at starting
    // of the given string is equal
    // to c1, replace it with c2
    if (str.charAt(0) == c1)
    {
        x=c2;
        str = c2+str.substring(1);
    }

    // Getting the answer from recursion
    // for the smaller problem
    return x+replaceCharacter(str.substring(1),
                            c1, c2);
}

// Driver Code
public static void main(String[] args)
{

    // Given string
    String str = "abacd";
    char c1 = 'a';
    char c2 = 'x';

    // Function call
    System.out.println(replaceCharacter(str, c1, c2));
}
}

// This code is contributed by cyrus18

Python3

# Python3 program for the above approach

# Function to replace all occurrences
# of character c1 with character c2
def replaceCharacter(input, c1, c2):

    input = list(str)

    # If the character at starting
    # of the given string is equal
    # to c1, replace it with c2
    for i in range(0, len(str)):
        if (input[i] == c1):
            input[i] = c2;

        # Print the string
        print(input[i], end = "")

# Driver Code

# Given string
str = "abacd"
c1 = 'a'
c2 = 'x'

# Function call
replaceCharacter(str, c1, c2);

# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;

class GFG{

// Function to replace all occurrences
// of character c1 with character c2
static void replaceCharacter(string str,
                             char c1, char c2)
{
    char[] input = str.ToCharArray();

    // If the character at starting
    // of the given string is equal
    // to c1, replace it with c2
    for(int i = 0; i < str.Length; i++)
    {
        if (input[i] == c1)
        {
            input[i] = c2;
        }

        // Print the string
        Console.Write(input[i]);
    }
}

// Driver Code
public static void Main()
{

    // Given string
    string str = "abacd";
    char c1 = 'a';
    char c2 = 'x';

    // Function call
    replaceCharacter(str, c1, c2);
}
}

// This code is contributed by sanjoy_62

Javascript

<script>
// javascript program for the above approach


// Function to replace all occurrences
// of character c1 with character c2
function replaceCharacter(str,c1, c2)
{
    // Base Case
    // If the string is empty
    if (str.length == 1)
    {
        return str;
    }
    var x=str.charAt(0);
    // If the character at starting
    // of the given string is equal
    // to c1, replace it with c2
    if (str.charAt(0) == c1)
    {
        x=c2;
        str = c2+str.substring(1);
    }

    // Getting the answer from recursion
    // for the smaller problem
    return x+replaceCharacter(str.substring(1),
                            c1, c2);
}

// Driver Code
//Given string
var str = "abacd";
var c1 = 'a';
var c2 = 'x';

// Function call
document.write(replaceCharacter(str, c1, c2));

// This code is contributed by 29AjayKumar
</script>

Output

xbxcd

Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time, where N is the length of the string.
Auxiliary Space: O(N), due to recursive stack space.