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.

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