Remove Characters Present in Other

Last Updated : 8 Sep, 2026

Given two strings s1 and s2, remove every character from s1 that is present in s2. Return the resulting string.

Note: Both strings contain only lowercase English letters, and |s1| > |s2|.

Examples:

Input: s1 = "computer", s2 = "cat"
Output: "ompuer"
Explanation: After removing characters(c, a, t) from string1 we get "ompuer".
Input: s1 = "occurrence", s2 = "car"
Output: "ouene"
Explanation: After removing characters (c, a, r) from string1 we get "ouene".

Try It Yourself

Table of Content

[Naive Approach] Using In-built Search function - O(|s1| × |s2|) Time and O(1) Space
[Expected Approach] Using Frequency Array - O(|s1| + |s2|) Time and O(1) Space

[Naive Approach] Using In-built Search function - O(|s1| × |s2|) Time and O(1) Space

For every character of s1, the idea is to check whether the same character exists in s2 using a built-in search function. If it is found, skip it; otherwise, add it to the result.

Initialize an empty result string.
Traverse each character of s1.
Use the built-in search function to check whether the character exists in s2.
If the character is not found, add it to the result.
Return the result.

C++

#include <bits/stdc++.h>
using namespace std;

// Returns s1 after removing all characters
// that are present in s2.
string removeChars(string &s1, string &s2)
{
    string res;

    // Traverse every character of s1.
    for (char ch : s1)
    {
        // Check whether the current character
        // is present in s2.
        if (s2.find(ch) == string::npos)
        {
            // If not present, keep the character.
            res += ch;
        }
    }

    return res;
}

int main()
{
    string s1 = "occurrence";
    string s2 = "car";

    cout << removeChars(s1, s2) << endl;

    return 0;
}

Java

import java.util.*;

class GFG {

    // Returns s1 after removing all characters
    // that are present in s2.
    static String removeChars(String s1, String s2)
    {
        StringBuilder res = new StringBuilder();

        // Traverse every character of s1.
        for (char ch : s1.toCharArray()) {

            // Check whether the current character
            // is present in s2.
            if (s2.indexOf(ch) == -1) {

                // If not present, keep the character.
                res.append(ch);
            }
        }

        return res.toString();
    }

    public static void main(String[] args)
    {
        String s1 = "occurrence";
        String s2 = "car";

        System.out.println(removeChars(s1, s2));
    }
}

Python

# Returns s1 after removing all characters
# that are present in s2.
def removeChars(s1, s2):
    res = ""

    # Traverse every character of s1.
    for ch in s1:

        # Check whether the current character
        # is present in s2.
        if ch not in s2:

            # If not present, keep the character.
            res += ch

    return res


# Driver Code
if __name__ == "__main__":
    s1 = "occurrence"
    s2 = "car"

    print(removeChars(s1, s2))

C#

using System;

class GFG {

    // Returns s1 after removing all characters
    // that are present in s2.
    static string removeChars(string s1, string s2)
    {
        string res = "";

        // Traverse every character of s1.
        foreach(char ch in s1)
        {
            // Check whether the current character
            // is present in s2.
            if (s2.IndexOf(ch) == -1) {

                // If not present, keep the character.
                res += ch;
            }
        }

        return res;
    }

    static void Main()
    {
        string s1 = "occurrence";
        string s2 = "car";

        Console.WriteLine(removeChars(s1, s2));
    }
}

JavaScript

// Returns s1 after removing all characters
// that are present in s2.
function removeChars(s1, s2)
{
    let res = "";

    // Traverse every character of s1.
    for (let ch of s1) {

        // Check whether the current character
        // is present in s2.
        if (s2.indexOf(ch) === -1) {

            // If not present, keep the character.
            res += ch;
        }
    }

    return res;
}

// Driver Code
let s1 = "occurrence";
let s2 = "car";

console.log(removeChars(s1, s2));

Output

ouene

[Expected Approach] Using Frequency Array - O(|s1| + |s2|) Time and O(1) Space

The idea is to store the frequency of every character of s2 in a frequency array of size 26. Then traverse s1 and keep only those characters whose frequency in s2 is 0.

Create a frequency array freq[26] initialized with 0.
Traverse s2 and increment the frequency of each character.
Traverse s1.
If freq[ch - 'a'] == 0, add the character to the result.
Return the resulting string.

C++

#include <bits/stdc++.h>
using namespace std;

// Returns s1 after removing all characters
// that are present in s2.
string removeChars(string &s1, string &s2)
{
    // Stores the frequency of each character in s2.
    int freq[26] = {0};

    // Traverse every character of s2.
    for (char ch : s2)
    {
        // Increment the frequency of the current character.
        freq[ch - 'a']++;
    }

    string res;

    // Traverse every character of s1.
    for (char ch : s1)
    {
        // Keep the character only if it is not present in s2.
        if (freq[ch - 'a'] == 0)
        {
            res += ch;
        }
    }

    return res;
}

int main()
{
    string s1 = "occurrence";
    string s2 = "car";

    cout << removeChars(s1, s2) << endl;

    return 0;
}

Java

import java.util.*;

class GFG {

    // Returns s1 after removing all characters
    // that are present in s2.
    static String removeChars(String s1, String s2)
    {
        // Stores the frequency of each character in s2.
        int[] freq = new int[26];

        // Traverse every character of s2.
        for (char ch : s2.toCharArray()) {

            // Increment the frequency of the current
            // character.
            freq[ch - 'a']++;
        }

        StringBuilder res = new StringBuilder();

        // Traverse every character of s1.
        for (char ch : s1.toCharArray()) {

            // Keep the character only if it is not present
            // in s2.
            if (freq[ch - 'a'] == 0) {
                res.append(ch);
            }
        }

        return res.toString();
    }

    public static void main(String[] args)
    {
        String s1 = "occurrence";
        String s2 = "car";

        System.out.println(removeChars(s1, s2));
    }
}

Python

# Returns s1 after removing all characters
# that are present in s2.
def removeChars(s1, s2):

    # Stores the frequency of each character in s2.
    freq = [0] * 26

    # Traverse every character of s2.
    for ch in s2:

        # Increment the frequency of the current character.
        freq[ord(ch) - ord('a')] += 1

    res = ""

    # Traverse every character of s1.
    for ch in s1:

        # Keep the character only if it is not present in s2.
        if freq[ord(ch) - ord('a')] == 0:
            res += ch

    return res


# Driver Code
if __name__ == "__main__":
    s1 = "occurrence"
    s2 = "car"

    print(removeChars(s1, s2))

C#

using System;

class GFG {
    
    // Returns s1 after removing all characters
    // that are present in s2.
    static string removeChars(string s1, string s2)
    {
        // Stores the frequency of each character in s2.
        int[] freq = new int[26];

        // Traverse every character of s2.
        foreach(char ch in s2)
        {
            // Increment the frequency of the current
            // character.
            freq[ch - 'a']++;
        }

        string res = "";

        // Traverse every character of s1.
        foreach(char ch in s1)
        {
            // Keep the character only if it is not present
            // in s2.
            if (freq[ch - 'a'] == 0) {
                res += ch;
            }
        }

        return res;
    }

    static void Main()
    {
        string s1 = "occurrence";
        string s2 = "car";

        Console.WriteLine(removeChars(s1, s2));
    }
}

JavaScript

// Returns s1 after removing all characters
// that are present in s2.
function removeChars(s1, s2)
{
    // Stores the frequency of each character in s2.
    let freq = new Array(26).fill(0);

    // Traverse every character of s2.
    for (let ch of s2) {

        // Increment the frequency of the current character.
        freq[ch.charCodeAt(0) - "a".charCodeAt(0)]++;
    }

    let res = "";

    // Traverse every character of s1.
    for (let ch of s1) {

        // Keep the character only if it is not present in
        // s2.
        if (freq[ch.charCodeAt(0) - "a".charCodeAt(0)]
            === 0) {
            res += ch;
        }
    }

    return res;
}


// Driver Code
let s1 = "occurrence";
let s2 = "car";

console.log(removeChars(s1, s2));

Output

ouene

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