Open In App
Related Articles

Non-overlapping sum of two sets

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Report issue
Report

Given two arrays A[] and B[] of size n. It is given that both array individually contains distinct elements. We need to find the sum of all elements that are not common.

Examples: 

Input : A[] = {1, 5, 3, 8}
        B[] = {5, 4, 6, 7}
Output : 29
1 + 3 + 4 + 6 + 7 + 8 = 29

Input : A[] = {1, 5, 3, 8}
        B[] = {5, 1, 8, 3}
Output : 0
All elements are common.

Brute Force Method: One simple approach is that for each element in A[] check whether it is present in B[], if it is present in then add it to the result. Similarly, traverse B[] and for every element that is not present in B, add it to result. 
Time Complexity: O(n2).
Auxiliary Space: O(1), As constant extra space is used.

Hashing concept: Create an empty hash and insert elements of both arrays into it. Now traverse hash table and add all those elements whose count is 1. (As per the question, both arrays individually have distinct elements)

Below is the implementation of the above approach:

C++

// CPP program to find Non-overlapping sum
#include <bits/stdc++.h>
using namespace std;
 
 
// function for calculating
// Non-overlapping sum of two array
int findSum(int A[], int B[], int n)
{
    // Insert elements of both arrays
    unordered_map<int, int> hash;   
    for (int i = 0; i < n; i++) {
        hash[A[i]]++;
        hash[B[i]]++;
    }
 
    // calculate non-overlapped sum
    int sum = 0;
    for (auto x: hash)
        if (x.second == 1)
            sum += x.first;
     
    return sum;
}
 
// driver code
int main()
{
    int A[] = { 5, 4, 9, 2, 3 };
    int B[] = { 2, 8, 7, 6, 3 };
     
    // size of array
    int n = sizeof(A) / sizeof(A[0]);
 
    // function call
    cout << findSum(A, B, n);
    return 0;
}

                    

Java

// Java program to find Non-overlapping sum
import java.io.*;
import java.util.*;
 
class GFG
{
 
    // function for calculating
    // Non-overlapping sum of two array
    static int findSum(int[] A, int[] B, int n)
    {
        // Insert elements of both arrays
        HashMap<Integer, Integer> hash = new HashMap<>();
        for (int i = 0; i < n; i++)
        {
            if (hash.containsKey(A[i]))
                hash.put(A[i], 1 + hash.get(A[i]));
            else
                hash.put(A[i], 1);
 
            if (hash.containsKey(B[i]))
                hash.put(B[i], 1 + hash.get(B[i]));
            else
                hash.put(B[i], 1);
        }
 
        // calculate non-overlapped sum
        int sum = 0;
        for (Map.Entry entry : hash.entrySet())
        {
            if (Integer.parseInt((entry.getValue()).toString()) == 1)
                sum += Integer.parseInt((entry.getKey()).toString());
        }
 
        return sum;
 
    }
 
    // Driver code
    public static void main(String args[])
    {
        int[] A = { 5, 4, 9, 2, 3 };
        int[] B = { 2, 8, 7, 6, 3 };
 
        // size of array
        int n = A.length;
 
        // function call
        System.out.println(findSum(A, B, n));
    }
}
 
// This code is contributed by rachana soma

                    

Python3

# Python3 program to find Non-overlapping sum
from collections import defaultdict
 
# Function for calculating
# Non-overlapping sum of two array
def findSum(A, B, n):
 
    # Insert elements of both arrays
    Hash = defaultdict(lambda:0)
    for i in range(0, n):
        Hash[A[i]] += 1
        Hash[B[i]] += 1
 
    # calculate non-overlapped sum
    Sum = 0
    for x in Hash:
        if Hash[x] == 1:
            Sum += x
     
    return Sum
 
# Driver code
if __name__ == "__main__":
 
    A = [5, 4, 9, 2, 3]
    B = [2, 8, 7, 6, 3]
     
    # size of array
    n = len(A)
 
    # Function call
    print(findSum(A, B, n))
     
# This code is contributed
# by Rituraj Jain

                    

C#

// C# program to find Non-overlapping sum
using System;
using System.Collections.Generic;
     
class GFG
{
 
    // function for calculating
    // Non-overlapping sum of two array
    static int findSum(int[] A, int[] B, int n)
    {
        // Insert elements of both arrays
        Dictionary<int, int> hash = new Dictionary<int, int>();
        for (int i = 0; i < n; i++)
        {
            if (hash.ContainsKey(A[i]))
            {
                var v = hash[A[i]];
                hash.Remove(A[i]);
                hash.Add(A[i], 1 + v);
            }
            else
                hash.Add(A[i], 1);
 
            if (hash.ContainsKey(B[i]))
            {
                var v = hash[B[i]];
                hash.Remove(B[i]);
                hash.Add(B[i], 1 + v);
            }
            else
                hash.Add(B[i], 1);
        }
 
        // calculate non-overlapped sum
        int sum = 0;
        foreach(KeyValuePair<int, int> entry in hash)
        {
            if ((entry.Value) == 1)
                sum += entry.Key;
        }
 
        return sum;
 
    }
 
    // Driver code
    public static void Main(String []args)
    {
        int[] A = { 5, 4, 9, 2, 3 };
        int[] B = { 2, 8, 7, 6, 3 };
 
        // size of array
        int n = A.Length;
 
        // function call
        Console.WriteLine(findSum(A, B, n));
    }
}
 
// This code is contributed by 29AjayKumar

                    

Javascript

<script>
 
// JavaScript program to find Non-overlapping sum
 
// function for calculating
// Non-overlapping sum of two array
function findSum(A, B, n) {
    // Insert elements of both arrays
    let hash = new Map();
    for (let i = 0; i < n; i++) {
        if (hash.has(A[i]))
            hash.set(A[i], 1 + hash.get(A[i]));
        else
            hash.set(A[i], 1);
 
        if (hash.has(B[i]))
            hash.set(B[i], 1 + hash.get(B[i]));
        else
            hash.set(B[i], 1);
    }
 
    // calculate non-overlapped sum
    let sum = 0;
    for (let entry of hash) {
        if (parseInt((entry[1]).toString()) == 1)
            sum += parseInt((entry[0]).toString());
    }
 
    return sum;
 
}
 
// Driver code
 
let A = [5, 4, 9, 2, 3];
let B = [2, 8, 7, 6, 3];
 
// size of array
let n = A.length;
 
// function call
document.write(findSum(A, B, n));
 
// This code is contributed by gfgking
 
</script>

                    

Output
39

Time Complexity: O(n), since inserting in an unordered map is amortized constant.
Auxiliary Space: O(n).

Another method: Using set data structure

  • Insert elements of Array A in the set data structure and add into sum
  • Check if B’s elements are there in set if exist then remove current element from set, otherwise add current element to sum
  • Finally, return sum

Below is the implementation of the above approach:

C++

// CPP program to find Non-overlapping sum
#include <bits/stdc++.h>
using namespace std;
 
// function for calculating
// Non-overlapping sum of two array
int findSum(int A[], int B[], int n)
{
    int sum = 0;
 
    // Insert elements of Array A in set
    // and add into sum
    set<int> st;
    for (int i = 0; i < n; i++) {
        st.insert(A[i]);
        sum += A[i];
    }
 
    // Check if B's element are there in set
    // if exist then remove current element from
    // set, otherwise add current element into sum
    for (int i = 0; i < n; i++) {
        if (st.find(B[i]) == st.end()) {
            sum += B[i];
        }
        else {
            sum -= B[i];
        }
    }
 
    // Finally, return sum
    return sum;
}
 
// Driver code
int main()
{
    int A[] = { 5, 4, 9, 2, 3 };
    int B[] = { 2, 8, 7, 6, 3 };
 
    // size of array
    int n = sizeof(A) / sizeof(A[0]);
 
    // function call
    cout << findSum(A, B, n);
    return 0;
}
 
// This code is contributed by hkdass001

                    

Java

// Java program to find Non-overlapping sum
 
import java.io.*;
import java.util.*;
 
class GFG {
   
      // function for calculating
    // Non-overlapping sum of two array
    public static int findSum(int[] A, int[] B, int n) {
        int sum = 0;
 
        // Insert elements of Array A in set
        // and add into sum
        Set<Integer> st = new HashSet<>();
        for (int i = 0; i < n; i++) {
            st.add(A[i]);
            sum += A[i];
        }
 
        // Check if B's element are there in set
        // if exist then remove current element from
        // set, otherwise add current element into sum
        for (int i = 0; i < n; i++) {
            if (!st.contains(B[i])) {
                sum += B[i];
            }
            else {
                sum -= B[i];
            }
        }
 
        // Finally, return sum
        return sum;
    }
   
    public static void main (String[] args) {
        int[] A = { 5, 4, 9, 2, 3 };
        int[] B = { 2, 8, 7, 6, 3 };
 
        // size of array
        int n = A.length;
 
        // function call
        System.out.println(findSum(A, B, n));
    }
}
 
// This code is contributed by lokesh.

                    

Python3

# python program to find Non-overlapping sum
 
# function for calculating
# Non-overlapping sum of two array
def findSum(A, B, n):
    sum = 0;
 
    # Insert elements of Array A in set
    # and add into sum
    st = set();
    for i in range(0,n):
        st.add(A[i]);
        sum += A[i];
     
    # Check if B's element are there in set
    # if exist then remove current element from
    # set, otherwise add current element into sum
    for i in range (0, n):
        if (B[i] in st):
            sum -= B[i];
        else :
            sum += B[i];
 
    # Finally, return sum
    return sum;
 
# Driver code
A = [ 5, 4, 9, 2, 3 ];
B = [ 2, 8, 7, 6, 3 ];
 
# size of array
n = len(A);
 
# function call
print(findSum(A, B, n));

                    

C#


                    

Javascript

<div id="highlighter_664145" class="syntaxhighlighter nogutter  "><table border="0" cellpadding="0" cellspacing="0"><tbody><tr><td class="code"><div class="container"><div class="line number1 index0 alt2"><code class="comments">// Javascript program to find Non-overlapping sum</code></div><div class="line number2 index1 alt1"> </div><div class="line number3 index2 alt2"><code class="comments">// function for calculating</code></div><div class="line number4 index3 alt1"><code class="comments">// Non-overlapping sum of two array</code></div><div class="line number5 index4 alt2"><code class="keyword">function</code> <code class="plain">findSum(A, B, n)</code></div><div class="line number6 index5 alt1"><code class="plain">{</code></div><div class="line number7 index6 alt2"><code class="undefined spaces">    </code><code class="plain">let sum = 0;</code></div><div class="line number8 index7 alt1"> </div><div class="line number9 index8 alt2"><code class="undefined spaces">    </code><code class="comments">// Insert elements of Array A in set</code></div><div class="line number10 index9 alt1"><code class="undefined spaces">    </code><code class="comments">// and add into sum</code></div><div class="line number11 index10 alt2"><code class="undefined spaces">    </code><code class="plain">let st = </code><code class="keyword">new</code> <code class="plain">Set();</code></div><div class="line number12 index11 alt1"><code class="undefined spaces">    </code><code class="keyword">for</code> <code class="plain">(let i = 0; i < n; i++) {</code></div><div class="line number13 index12 alt2"><code class="undefined spaces">        </code><code class="plain">st.add(A[i]);</code></div><div class="line number14 index13 alt1"><code class="undefined spaces">        </code><code class="plain">sum += A[i];</code></div><div class="line number15 index14 alt2"><code class="undefined spaces">    </code><code class="plain">}</code></div><div class="line number16 index15 alt1"><code class="undefined spaces">    </code> </div><div class="line number17 index16 alt2"><code class="undefined spaces">    </code><code class="comments">// Check if B's element are there in set</code></div><div class="line number18 index17 alt1"><code class="undefined spaces">    </code><code class="comments">// if exist then remove current element from</code></div><div class="line number19 index18 alt2"><code class="undefined spaces">    </code><code class="comments">// set, otherwise add current element into sum</code></div><div class="line number20 index19 alt1"><code class="undefined spaces">    </code><code class="keyword">for</code> <code class="plain">(let i = 0; i < n; i++) {</code></div><div class="line number21 index20 alt2"><code class="undefined spaces">        </code><code class="keyword">if</code> <code class="plain">(!st.has(B[i])) {</code></div><div class="line number22 index21 alt1"><code class="undefined spaces">            </code><code class="plain">sum += B[i];</code></div><div class="line number23 index22 alt2"><code class="undefined spaces">        </code><code class="plain">}</code></div><div class="line number24 index23 alt1"><code class="undefined spaces">        </code><code class="keyword">else</code> <code class="plain">{</code></div><div class="line number25 index24 alt2"><code class="undefined spaces">            </code><code class="plain">sum -= B[i];</code></div><div class="line number26 index25 alt1"><code class="undefined spaces">        </code><code class="plain">}</code></div><div class="line number27 index26 alt2"><code class="undefined spaces">    </code><code class="plain">}</code></div><div class="line number28 index27 alt1"> </div><div class="line number29 index28 alt2"><code class="undefined spaces">    </code><code class="comments">// Finally, return sum</code></div><div class="line number30 index29 alt1"><code class="undefined spaces">    </code><code class="keyword">return</code> <code class="plain">sum;</code></div><div class="line number31 index30 alt2"><code class="plain">}</code></div><div class="line number32 index31 alt1"> </div><div class="line number33 index32 alt2"><code class="comments">// Driver code</code></div><div class="line number34 index33 alt1"><code class="undefined spaces">    </code><code class="plain">let A = [ 5, 4, 9, 2, 3 ];</code></div><div class="line number35 index34 alt2"><code class="undefined spaces">    </code><code class="plain">let B = [ 2, 8, 7, 6, 3 ];</code></div><div class="line number36 index35 alt1"> </div><div class="line number37 index36 alt2"><code class="undefined spaces">    </code><code class="comments">// size of array</code></div><div class="line number38 index37 alt1"><code class="undefined spaces">    </code><code class="plain">let n = A.length;</code></div><div class="line number39 index38 alt2"> </div><div class="line number40 index39 alt1"><code class="undefined spaces">    </code><code class="comments">// function call</code></div><div class="line number41 index40 alt2"><code class="undefined spaces">    </code><code class="plain">document.write(findSum(A, B, n));</code></div></div></td></tr></tbody></table></div>

                    

Output
39

Time Complexity: O(n*log n)
Auxiliary Space: O(n)
 



Last Updated : 20 Jan, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads