Replace all occurrences of character X with character Y in given string
Last Updated :
05 Apr, 2023
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;
}
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);
}
}
# 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)
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);
}
}
// 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:
- Get the string str, character X, and Y.
- 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 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 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# 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
<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>
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.
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