Find resultant string after concatenating uncommon characters of given strings

Given two strings S₁ and S_2. The task is to concatenate uncommon characters of the S₂ to S₁  and return the resultant string S_{1 .}

Examples:

Input: S₁ = “aacdb”, S₂ = “gafd”
Output: “cbgf”

Input: S₁ = “abcs”, S₂ = “cxzca”;
Output: “bsxz”

Method 1:  Using Hashmap

Approach:

Below is the idea to solve the problem.

The idea is to use Hashmap where the key is a character and the value is an integer i.e. number of strings in which the character is present. If a character is present in one string, then the count is 1, else if the character is present in both strings, the count is 2. 

Follow the steps below to implement the idea:

• Initialize the result as an empty string.
• Push all characters of S₂ string in map with count as 1.
• Traverse first string and append all those characters to result that are not present in map then make their count 2.
• Traverse second string and append all those characters to result whose count is 1.

Below is the implementation of above approach:

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 << concatenatedString(s1, s2);
    return 0;
}

Java

// Java program Find concatenated string with
// uncommon characters of given strings
import java.io.*;
import java.lang.*;
import java.util.*;

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(concatenatedString(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.

Method 2: Using Set

Use a set instead of a hashmap to store the characters of the second string. This can simplify the code and reduce the space complexity.

Approach:

1. Initialize the result as an empty string.
2. Add all characters of S2 string in a set.
3. Traverse S1 string and append all those characters to result that are not present in the set.
4. Return the result.

Below is the implementation of the updated approach:

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 = "";
    unordered_set<char> s(s2.begin(), s2.end());

    for (int i = 0; i < s1.size(); i++) {
        if (s.find(s1[i]) == s.end())
            res += s1[i];
        else
            s.erase(s1[i]);
    }

    for (const char& c : s)
        res += c;

    return res;
}

/* Driver program to test above function */
int main()
{
    string s1 = "abcs";
    string s2 = "cxzca";
    cout << concatenatedString(s1, s2);
    return 0;
}

Java

import java.util.HashSet;

public class ConcatenatedString {
    public static String concatenatedString(String s1, String s2) {
        String res = "";
        HashSet<Character> s = new HashSet<>();
        for (char c : s2.toCharArray()) {
            s.add(c);
        }

        for (char c : s1.toCharArray()) {
            if (!s.contains(c)) {
                res += c;
            } else {
                s.remove(c);
            }
        }

        for (char c : s) {
            res += c;
        }

        return res;
    }

    public static void main(String[] args) {
        String s1 = "abcs";
        String s2 = "cxzca";
        System.out.println(concatenatedString(s1, s2));
    }
}

C#

using System;
using System.Collections.Generic;

public class Program
{
    public static string ConcatenatedString(string s1, string s2)
    {
        string res = "";
        HashSet<char> s = new HashSet<char>(s2);

        for (int i = 0; i < s1.Length; i++) {
            if (!s.Contains(s1[i]))
                res += s1[i];
            else
                s.Remove(s1[i]);
        }

        foreach (char c in s)
            res += c;

        return res;
    }

    public static void Main()
    {
        string s1 = "abcs";
        string s2 = "cxzca";
        Console.WriteLine(ConcatenatedString(s1, s2));
    }
}

Python3

def concatenatedString(s1: str, s2: str) -> str:
    res = ""
    s = set(s2)

    for c in s1:
        if c not in s:
            res += c
        else:
            s.remove(c)

    res += "".join(s)

    return res

# Example usage
s1 = "abcs"
s2 = "cxzca"
print(concatenatedString(s1, s2)) # Output: bsxz

Javascript

function concatenatedString(s1, s2) {
    let res = "";
    let s = new Set(s2);

    for (let i = 0; i < s1.length; i++) {
        if (!s.has(s1[i]))
            res += s1[i];
        else
            s.delete(s1[i]);
    }

    for (let c of s)
        res += c;

    return res;
}

// Driver program to test above function
let s1 = "abcs";
let s2 = "cxzca";
console.log(concatenatedString(s1, s2));

Output

bszx

Time Complaxity: O(M+N)
Auxiliary Space: O(N)

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.