Find the overlapping sum of two arrays

Given two arrays A[] and B[] having n unique elements each. The task is to find the overlapping sum of the two arrays. That is the sum of elements which is common in both of the arrays.

Note: Elements in the arrays are unique. That is the array does not contains duplicates.

Examples:

Input : A[] = {1, 5, 3, 8}
        B[] = {5, 4, 6, 7}
Output : 10
Explanation : The element which is common in both arrays is 5.
Therefore, the overlapping sum will be (5+5) = 10

Input : A[] = {1, 5, 3, 8}
        B[] = {5, 1, 8, 3}
Output : 99

Brute Force Method : The simple approach is that for each element in A[] check whether it is present in B[] and if it is present in B[] then add that number two times(once for A[] and once for B[]) to the sum. Repeat this procedure for all elements in the array A[].
Time Complexity: O(n^2).

Efficient Method : An efficient method is to use Hashing. Traverse both of the arrays and insert the elements into a hash table to keep track of the count of elements. Add the elements to sum whose count equals to two.

Below is the implementation of above approach:

CPP

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// CPP program to find overlapping sum
#include <bits/stdc++.h>
using namespace std;
  
// Function for calculating
// overlapping sum of two array
int findSum(int A[], int B[], int n)
{   
    // unordered map to store count of 
    // elements
    unordered_map<int,int> hash;
      
    // insert elements of A[] into
    // unordered_map
    for(int i=0;i<n;i++)
    {
        if(hash.find(A[i])==hash.end())
        {
            hash.insert(make_pair(A[i],1));
        }
        else
        {
            hash[A[i]]++;
        }
    }
      
    // insert elements of B[] into
    // unordered_map
    for(int i=0;i<n;i++)
    {
        if(hash.find(B[i])==hash.end())
        {
            hash.insert(make_pair(B[i],1));
        }
        else
        {
            hash[B[i]]++;
        }
    }
  
    // calculate overlapped sum
    int sum = 0;
    for(auto itr = hash.begin(); itr!=hash.end(); itr++)
    {
        if((itr->second)==2)
        {
            sum += (itr->first)*2;
        }
    }
      
    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;
}

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Python

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# Python3 program to find overlapping sum
  
# Function for calculating
# overlapping sum of two array
def findSum(A, B, n):
      
    # unordered map to store count of
    # elements
    hash=dict()
  
    # insert elements of A into
    # unordered_map
    for i in range(n):
        hash[A[i]] = hash.get(A[i], 0) + 1
  
    # insert elements of B into
    # unordered_map
    for i in range(n):
        hash[B[i]] = hash.get(B[i], 0) + 1
  
  
    # calculate overlapped sum
    sum = 0
  
    for i in hash:
        if hash[i] == 2:
            sum += i * 2
  
    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))
  
# This code is contributed by mohit kumar 29

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Output:

10

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

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : mohit kumar 29



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