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Concatenated string with uncommon characters of two strings

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  • Difficulty Level : Medium
  • Last Updated : 24 May, 2022

Two strings are given and you have to modify 1st string such that all the common characters of the 2nd strings have to be removed and the uncommon characters of the 2nd string have to be concatenated with uncommon characters of the 1st string.
Examples: 
 

Input : S1 = "aacdb"
        S2 = "gafd"
Output : "cbgf"

Input : S1 = "abcs";
        S2 = "cxzca";
Output : "bsxz"

 

The idea is to use hash map where key is character and value is number of strings in which character is present. If a character is present in one string, then count is 1, else if character is present in both strings, count is 2. 
 

  1. Initialize result as empty string.
  2. Push all characters of 2nd string in map with count as 1.
  3. Traverse first string and append all those characters to result that are not present in map. Characters that are present in map, make count 2.
  4. Traverse second string and append all those characters to result whose count is 1.

 

C++




// C++ program Find concatenated string with
// uncommon characters of given strings
#include <bits/stdc++.h>
using namespace std;
 
string concatenatedString(string s1, string s2)
{
    string res = ""; // result
 
    // store all characters of s2 in map
    unordered_map<char, int> m;
    for (int i = 0; i < s2.size(); i++)
        m[s2[i]] = 1;
 
    // Find characters of s1 that are not
    // present in s2 and append to result
    for (int i = 0; i < s1.size(); i++) {
        if (m.find(s1[i]) == m.end())
            res += s1[i];
        else
            m[s1[i]] = 2;
    }
 
    // Find characters of s2 that are not
    // present in s1.
    for (int i = 0; i < s2.size(); i++)
        if (m[s2[i]] == 1)
            res += s2[i];
    return res;
}
 
/* Driver program to test above function */
int main()
{
    string s1 = "abcs";
    string s2 = "cxzca";
    cout << concatenetedString(s1, s2);
    return 0;
}

Java




// Java program Find concatenated string with
// uncommon characters of given strings
import java.util.*;
import java.lang.*;
import java.io.*;
 
class gfg {
    public static String concatenatedString(String s1, String s2)
    {
        // Result
        String res = "";
        int i;
 
        // creating a hashMap to add characters in string s2
        HashMap<Character, Integer> m = new HashMap<Character, Integer>();
        for (i = 0; i < s2.length(); i++)
            m.put(s2.charAt(i), 1);
 
        // Find characters of s1 that are not
        // present in s2 and append to result
        for (i = 0; i < s1.length(); i++)
            if (!m.containsKey(s1.charAt(i)))
                res += s1.charAt(i);
            else
                m.put(s1.charAt(i), 2);
 
        // Find characters of s2 that are not
        // present in s1.
        for (i = 0; i < s2.length(); i++)
            if (m.get(s2.charAt(i)) == 1)
                res += s2.charAt(i);
 
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s1 = "abcs";
        String s2 = "cxzca";
        System.out.println(concatenatedString(s1, s2));
    }
}
 
/* This code is contributed by Devarshi_Singh*/

Python 3




# Python3 program Find concatenated string
# with uncommon characters of given strings
def concatenatedString(s1, s2):
    res = "" # result
    m = {}
     
    # store all characters of s2 in map
    for i in range(0, len(s2)):
        m[s2[i]] = 1
 
    # Find characters of s1 that are not
    # present in s2 and append to result
    for i in range(0, len(s1)):
        if(not s1[i] in m):
            res = res + s1[i]
        else:
            m[s1[i]] = 2
 
    # Find characters of s2 that are not
    # present in s1.        
    for i in range(0, len(s2)):
        if(m[s2[i]] == 1):
            res = res + s2[i]
 
    return res    
     
# Driver Code
if __name__ == "__main__":
    s1 = "abcs"
    s2 = "cxzca"
    print(concatenetedString(s1, s2))
 
# This code is contributed
# by Sairahul099

C#




// C# program Find concatenated string with
// uncommon characters of given strings
using System;
using System.Collections.Generic;
 
class GFG
{
    public static String concatenatedString(String s1,
                                            String s2)
    {
        // Result
        String res = "";
        int i;
 
        // creating a hashMap to add characters
        // in string s2
        Dictionary<char,
                   int> m = new Dictionary<char,
                                           int>();
        for (i = 0; i < s2.Length; i++)
            if (!m.ContainsKey(s2[i]))
                m.Add(s2[i], 1);
 
        // Find characters of s1 that are not
        // present in s2 and append to result
        for (i = 0; i < s1.Length; i++)
            if (!m.ContainsKey(s1[i]))
                res += s1[i];
            else
                m[s1[i]] = 2;
 
        // Find characters of s2 that are not
        // present in s1.
        for (i = 0; i < s2.Length; i++)
            if (m[s2[i]] == 1)
                res += s2[i];
 
        return res;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String s1 = "abcs";
        String s2 = "cxzca";
        Console.WriteLine(concatenatedString(s1, s2));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript program Find concatenated string with
// uncommon characters of given strings
function concatenatedString(s1, s2)
{
    let res = ""; // result
 
    // store all characters of s2 in map
    let m = new Map();
    for (let i = 0; i < s2.length; i++)
        m.set(s2[i],1);
 
    // Find characters of s1 that are not
    // present in s2 and append to result
    for (let i = 0; i < s1.length; i++) {
        if (m.has(s1[i]) == false)
            res += s1[i];
        else
            m.set(s1[i] , 2);
    }
 
    // Find characters of s2 that are not
    // present in s1.
    for (let i = 0; i < s2.length; i++)
        if (m.get(s2[i]) == 1)
            res += s2[i];
    return res;
}
 
/* Driver program to test above function */
let s1 = "abcs";
let s2 = "cxzca";
document.write(concatenetedString(s1, s2));
 
// This code is contributed by shinjanpatra
</script>

Output: 
 

bsxz

Time Complexity: O(M + N), where M and N represents the size of the given two strings.
Auxiliary Space: O(max(M, N)), where M and N represents the size of the given two strings.

This article is contributed by Harshit Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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