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: 10Explanation: The element which is common in both arrays is 5. Therefore, the overlapping sum will be (5+5) = 10Input: 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

`// 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

`# 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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