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Replace all occurrences of substring

Last Updated : 11 May, 2025
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Given three strings s, s1, and s2 of lengths n, m, and k respectively, the task is to modify the string s by replacing all the substrings s1 with the string s2 in the string s.

Examples:

Input: s = "abababa", s1 = "aba", s2 = "a"
Output: aba
Explanation: Change the substrings s[0, 2] and s[4, 6] to the string s2 modifies the string s to "aba".

Input: s = "geeksforgeeks", s1 = "eek", s2 = "ok"
Output: goksforgoks
Explanation: Change the substrings s[1, 3] and s[9, 11] to the string

Table of Content

[Naive Approach] Using Nested Loop – O(n*m) Time and O(n) Space

The simplest approach to solve the given problem is to traverse the string s and when any string s1 is found as a substring in the string s then replace it by s2.

C++ 
#include <iostream>
#include <string>
using namespace std;

string replaceSubstr(string s, string s1, string s2)
{
    string ans = "";
    int n = s.length(), m = s1.length();

    for (int i = 0; i < n; i++)
    {
        if (i + m <= n && s.substr(i, m) == s1)
        {
            ans += s2;

            // Skip the characters of s1
            i += m - 1;
        }
        else
        {
            ans += s[i];
        }
    }
    return ans;
}

int main()
{
    string s = "abababa", s1 = "aba", s2 = "a";
    cout << replaceSubstr(s, s1, s2) << endl;
    return 0;
}
Java 
public class GFG {
    public static String replaceSubstr(String s, String s1,
                                       String s2)
    {
        StringBuilder ans = new StringBuilder();
        int n = s.length(), m = s1.length();

        for (int i = 0; i < n; i++) {
            if (i + m <= n
                && s.substring(i, i + m).equals(s1)) {
                ans.append(s2);

                // Skip the characters of s1
                i += m - 1;
            }
            else {
                ans.append(s.charAt(i));
            }
        }
        return ans.toString();
    }

    public static void main(String[] args)
    {
        String s = "abababa", s1 = "aba", s2 = "a";
        System.out.println(replaceSubstr(s, s1, s2));
    }
}
Python 
def replaceSubstr(s, s1, s2):
    ans = ""
    n, m = len(s), len(s1)

    i = 0
    while i < n:
        if i + m <= n and s[i:i + m] == s1:
            ans += s2

            # Skip the characters of s1
            i += m - 1
        else:
            ans += s[i]
        i += 1
    return ans


s = "abababa"
s1 = "aba"
s2 = "a"
print(replaceSubstr(s, s1, s2))
C# 
using System;

class GFG {
    static string ReplaceSubstr(string s, string s1,
                                string s2)
    {
        string ans = "";
        int n = s.Length, m = s1.Length;

        for (int i = 0; i < n; i++) {
            if (i + m <= n && s.Substring(i, m) == s1) {
                ans += s2;

                // Skip the characters of s1
                i += m - 1;
            }
            else {
                ans += s[i];
            }
        }
        return ans;
    }

    static void Main()
    {
        string s = "abababa", s1 = "aba", s2 = "a";
        Console.WriteLine(ReplaceSubstr(s, s1, s2));
    }
}
JavaScript 
function replaceSubstr(s, s1, s2)
{
    let ans = "";
    let n = s.length, m = s1.length;

    for (let i = 0; i < n; i++) {
        if (i + m <= n && s.substring(i, i + m) === s1) {
            ans += s2;

            // Skip the characters of s1
            i += m - 1;
        }
        else {
            ans += s[i];
        }
    }
    return ans;
}

let s = "abababa";
let s1 = "aba";
let s2 = "a";
console.log(replaceSubstr(s, s1, s2));

Output
aba

[Expected Approach] Using KMP Algorithm– O(n+m) Time and O(m) Space

The above approach can also be optimized by creating the longest proper prefix and suffix array for the string s1 and then perform the KMP Algorithm to find the occurrences of the string s1 in the string s.

Algorithm:

  • Create a array lps[] that stores the longest proper prefix and suffix for each character and fill this array using the KMP algorithm for the string s1.
  • Initialize two variables say, i and j as 0 to store the position of current character in s and s1 respectively.
  • Initialize a array found[] to store all the starting indexes from which string s1 occurs in s.
  • Iterate over the characters of the string s using variable i and perform the following steps:
    • If s[i] is equal to s1[j], then increment i and j by 1.
    • If j is equal to the length of s1, then add the value (i - j) to the array found and update j as lps[j - 1].
    • Otherwise, if the value of s[i] is not equal to s1[j], then if j is equal to 0, then increment the value of i by 1. Otherwise, update j as lps[j - 1].
  • Initialize a variable say, prev as 0 to store the last changed index and an empty string ans to store the resultant string after replacing all the initial appearances of s1 by s2 in s.
  • Traverse the array found[] and if the value of found[i] is greater than prev, then add the string s2 in place of s1 in ans.
  • After completing the above steps, print the string ans as the result.
C++ 
#include <iostream>
#include <vector>
using namespace std;

// Compute the LPS (Longest Prefix Suffix) array
vector<int> computeLPS(string pattern) {
    int m = pattern.length();
    vector<int> lps(m, 0);
    int len = 0;

    for (int i = 1; i < m; ) {
        if (pattern[i] == pattern[len]) {
            lps[i++] = ++len;
        } else {
            if (len != 0) len = lps[len - 1];
            else lps[i++] = 0;
        }
    }

    return lps;
}

// Function to perform the replace using KMP
string replaceSubstr(string s, string s1, string s2) {
    int n = s.length();
    int m = s1.length();
    vector<int> lps = computeLPS(s1);

    string ans;
    int i = 0, j = 0;

    while (i < n) {
        if (s[i] == s1[j]) {
            i++, j++;
            if (j == m) {
                // Match found; replace
                ans += s2;
                j = 0;  // Reset pattern pointer for next match
            }
        } else {
            if (j != 0) j = lps[j - 1];
            else {
                ans += s[i++];
            }
        }
    }

    // Add remaining unmatched characters
    if (j != 0) ans += s.substr(i - j, j);

    return ans;
}

int main()
{
    string s = "abababa", s1 = "aba", s2 = "a";
    cout << replaceSubstr(s, s1, s2) << endl;
    return 0;
}
Java 
import java.util.*;

public class GfG{

    // Compute the LPS (Longest Prefix Suffix) array
    public static int[] computeLPS(String pattern) {
        int m = pattern.length();
        int[] lps = new int[m];
        int len = 0;

        for (int i = 1; i < m; ) {
            if (pattern.charAt(i) == pattern.charAt(len)) {
                lps[i++] = ++len;
            } else {
                if (len != 0) len = lps[len - 1];
                else lps[i++] = 0;
            }
        }

        return lps;
    }

    // Function to perform the replace using KMP
    public static String replaceSubstr(String s, String s1, String s2) {
        int[] lps = computeLPS(s1);
        StringBuilder ans = new StringBuilder();
        int i = 0, j = 0;
        int n = s.length(), m = s1.length();

        while (i < n) {
            if (s.charAt(i) == s1.charAt(j)) {
                i++; j++;
                if (j == m) {
                    // Match found; replace
                    ans.append(s2);
                    j = 0;  // Reset pattern pointer for next match
                }
            } else {
                if (j != 0) j = lps[j - 1];
                else ans.append(s.charAt(i++));
            }
        }

        // Add remaining unmatched characters
        if (j != 0) ans.append(s.substring(i - j, i));

        return ans.toString();
    }

    public static void main(String[] args) {
        String s = "abababa", s1 = "aba", s2 = "a";
        System.out.println(replaceSubstr(s, s1, s2));
    }
}
Python 
# Compute the LPS (Longest Prefix Suffix) array
def compute_lps(pattern):
    m = len(pattern)
    lps = [0] * m
    length = 0
    i = 1
    while i < m:
        if pattern[i] == pattern[length]:
            length += 1
            lps[i] = length
            i += 1
        else:
            if length != 0:
                length = lps[length - 1]
            else:
                lps[i] = 0
                i += 1
    return lps

# Function to perform the replace using KMP
def replaceSubstr(s, s1, s2):
    n, m = len(s), len(s1)
    lps = compute_lps(s1)
    ans = []
    i = j = 0

    while i < n:
        if s[i] == s1[j]:
            i += 1
            j += 1
            if j == m:
                
                # Match found; replace
                ans.append(s2)
                
                # Reset pattern pointer for next match
                j = 0  
        else:
            if j != 0:
                j = lps[j - 1]
            else:
                ans.append(s[i])
                i += 1

    # Add remaining unmatched characters
    if j != 0:
        ans.append(s[i - j:i])

    return ''.join(ans)


if __name__ == '__main__':
    s = "abababa"
    s1 = "aba"
    s2 = "a";
    print(replaceSubstr(s, s1, s2))
C# 
using System;
using System.Text;

class GfG{

    // Compute the LPS (Longest Prefix Suffix) array
    static int[] ComputeLPS(string pattern) {
        int m = pattern.Length;
        int[] lps = new int[m];
        int len = 0;
        int i = 1;

        while (i < m) {
            if (pattern[i] == pattern[len]) {
                lps[i++] = ++len;
            } else {
                if (len != 0) len = lps[len - 1];
                else lps[i++] = 0;
            }
        }

        return lps;
    }

    // Function to perform the replace using KMP
    static string replaceSubstr(string s, string s1, string s2) {
        int[] lps = ComputeLPS(s1);
        StringBuilder ans = new StringBuilder();
        int i = 0, j = 0;
        int n = s.Length, m = s1.Length;

        while (i < n) {
            if (s[i] == s1[j]) {
                i++; j++;
                if (j == m) {
                    // Match found; replace
                    ans.Append(s2);
                    j = 0;  // Reset pattern pointer for next match
                }
            } else {
                if (j != 0) j = lps[j - 1];
                else ans.Append(s[i++]);
            }
        }

        // Add remaining unmatched characters
        if (j != 0) ans.Append(s.Substring(i - j, j));

        return ans.ToString();
    }

    static void Main() {
        string s = "abababa", s1 = "aba", s2 = "a";
        Console.WriteLine(replaceSubstr(s, s1, s2));
    }
}
JavaScript 
// Compute the LPS (Longest Prefix Suffix) array
function computeLPS(pattern) {
    const m = pattern.length;
    const lps = Array(m).fill(0);
    let len = 0;

    for (let i = 1; i < m;) {
        if (pattern[i] === pattern[len]) {
            lps[i++] = ++len;
        } else {
            if (len !== 0) len = lps[len - 1];
            else lps[i++] = 0;
        }
    }
    return lps;
}

// Function to perform the replace using KMP
function replaceSubstr(s, s1, s2) {
    const lps = computeLPS(s1);
    let ans = "";
    let i = 0, j = 0;
    const n = s.length, m = s1.length;

    while (i < n) {
        if (s[i] === s1[j]) {
            i++; j++;
            if (j === m) {
                // Match found; replace
                ans += s2;
                j = 0;  // Reset pattern pointer for next match
            }
        } else {
            if (j !== 0) j = lps[j - 1];
            else ans += s[i++];
        }
    }

    // Add remaining unmatched characters
    if (j !== 0) ans += s.slice(i - j, i);
    return ans;
}

// Driver Code
const s = "abababa", s1 = "aba", s2 = "a";
console.log(replaceSubstr(s, s1, s2));

Output
aba

Time Complexity: The time complexity is O(n + m), where n is the length of string s and m is the length of string s1.
The LPS function runs in O(m) time to compute the prefix-suffix array, and the main loop processes string s in O(n) time.
Auxiliary Space: The LPS array requires O(m) space, where m is the length of s1, so the overall space complexity is O(m).

Using Library Methods - O(n*m) time and O(n) space

C++ 
#include <iostream>
#include <string>
using namespace std;

// Function to perform the replace
string replaceSubstr(string s, const string& s1, const string& s2) {
    size_t pos = 0;

    // Replace all non-overlapping occurrences of s1 with s2
    while ((pos = s.find(s1, pos)) != string::npos) {
        s.replace(pos, s1.length(), s2);
        
        // Move past the replaced part
        pos += s2.length();  
    }

    return s;
}

int main() {
    string s = "abababa", s1 = "aba", s2 = "a";
    cout << replaceSubstr(s, s1, s2) << endl;
    return 0;
}
Java 
public class GfG {

    // Function to perform the replace
    static String replaceSubstr(String s, String s1,
                                    String s2) {
        
        // Replace all non-overlapping occurrences
        return s.replace(s1, s2);
    }

    public static void main(String[] args) {
        String s = "abababa", s1 = "aba", s2 = "a";
        System.out.println(replaceSubstr(s, s1, s2));
    }
}
Python 
# Function to perform the replace
def replaceSubstr(s, s1, s2):
    
    # Replace all non-overlapping occurrences
    return s.replace(s1, s2)

if __name__ == "__main__":
    s = "abababa"
    s1 = "aba"
    s2 = "a"
    print(replaceSubstr(s, s1, s2))
C# 
using System;

class GfG{

    // Function to perform the replace 
    static string replaceSubstr(string s, string s1, string s2) {
        
        // Replace all non-overlapping occurrences
        return s.Replace(s1, s2);
    }

    static void Main() {
        string s = "abababa", s1 = "aba", s2 = "a";
        Console.WriteLine(replaceSubstr(s, s1, s2));
    }
}
JavaScript 
// Function to perform the replace 
function replaceSubstr(s, s1, s2) {
  
    // Replace all non-overlapping occurrences
    return s.replaceAll(s1, s2);  // ES2021+

    // Or for older environments:
    // return s.split(s1).join(s2);
}

// Driver Code
const s = "abababa", s1 = "aba", s2 = "a";
console.log(replaceSubstr(s, s1, s2));

Output
aba

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In-place replace multiple occurrences of a pattern

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