Given two sorted arrays, we can get a set of sums(add one element from the first and one from second). Find the N’th element in the elements of the formed set considered in sorted order.

** Note: ** Set of sums should have unique elements.

Examples:

Input : arr1[] = {1, 2} arr2[] = {3, 4} N = 3 Output : 6 We get following elements set of sums. 4(1+3), 5(2+3 or 1+4), 6(2+4) Third element in above set is 6. Input : arr1[] = { 1,3, 4, 8, 10} arr2[] = {20, 22, 30, 40} N = 4 Output : 25 We get following elements set of sums. 21(1+20), 23(1+22 or 20+3), 24(20+4), 25(22+3)... Fourth element is 25.

Asked in: Microsoft Interview

1- Run two loops – one for first array and second for second array.

2- Just consider each pair and store their sum in a self-balancing-BST (which is implemented by set and map in C++). We use set in C++ here as we need to only see if elements are present or absent, we don’t need key, value pairs.

3- Traverse the set and return the Nth element in the set.

Below is C++ implementation of above idea.

// C++ program to find N'th element in a set formed // by sum of two arrays #include<bits/stdc++.h> using namespace std; //Function to calculate the set of sums int calculateSetOfSum(int arr1[], int size1, int arr2[], int size2, int N) { // Insert each pair sum into set. Note that a set // stores elements in sorted order and unique elements set<int> s; for (int i=0 ; i < size1; i++) for (int j=0; j < size2; j++) s.insert(arr1[i]+arr2[j]); // If set has less than N elements if (s.size() < N) return -1; // Find N'tb item in set and return it set<int>::iterator it = s.begin(); for (int count=1; count<N; count++) it++; return *it; } // Driver code int main() { int arr1[] = {1, 2}; int size1 = sizeof(arr1) / sizeof(arr1[0]); int arr2[] = {3, 4}; int size2 = sizeof(arr2) / sizeof(arr2[0]); int N = 3; int res = calculateSetOfSum(arr1, size1, arr2, size2, N); if (res == -1) cout << "N'th term doesn't exists in set"; else cout << "N'th element in the set of sums is " << res; return 0; }

Output:

Nth element in the set of sums is 25

Time Complexity: O(mn log (mn)) where m is the size of the first array and n is the size of the second array.

### Asked in: Microsoft

This article is contributed by **Sahil Chhabra (akku)**. 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.