Find N’th item in a set formed by sum of two arrays

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.


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
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++)

    // 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++)
    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";
        cout << "N'th element in the set of sums is "
             << res;
    return 0;


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

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