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 = 3Output: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 = 4Output: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

**Approach:**

- Run two loops – one for the first array and second for the second array.
- 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.
- Traverse the set and return the Nth element in the set.

Below is the implementation of the above approach:

## C++

`// 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;` `}` |

## Java

`// Java program to find N'th element in a set formed` `// by sum of two arrays` `import` `java.util.*;` `class` `GFG ` `{` `// Function to calculate the set of sums` `static` `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` ` ` `SortedSet<Integer> s = ` `new` `TreeSet<Integer>();` ` ` `for` `(` `int` `i = ` `0` `; i < size1; i++)` ` ` `for` `(` `int` `j = ` `0` `; j < size2; j++)` ` ` `s.add(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` ` ` ` ` `return` `(` `int` `)s.toArray()[ N-` `1` `];` `}` `// Driver code` `public` `static` `void` `main(String[] args) ` `{` ` ` `int` `arr1[] = {` `1` `, ` `2` `};` ` ` `int` `size1 = arr1.length;` ` ` `int` `arr2[] = {` `3` `, ` `4` `};` ` ` `int` `size2 = arr2.length;` ` ` `int` `N = ` `3` `;` ` ` `int` `res = calculateSetOfSum(arr1, size1, arr2, size2, N);` ` ` `if` `(res == -` `1` `)` ` ` `System.out.println(` `"N'th term doesn't exists in set"` `);` ` ` `else` ` ` `System.out.println(` `"N'th element in the set of sums is "` ` ` `+res);` `}` `}` `// This code is contributed by 29AjayKumar` |

## C#

`// C# program to find N'th element in ` `// a set formed by sum of two arrays` `using` `System;` `using` `System.Linq;` `using` `System.Collections.Generic; ` ` ` `class` `GFG ` `{` `// Function to calculate the set of sums` `static` `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` ` ` `HashSet<` `int` `> s = ` `new` `HashSet<` `int` `>();` ` ` `for` `(` `int` `i = 0; i < size1; i++)` ` ` `for` `(` `int` `j = 0; j < size2; j++)` ` ` `s.Add(arr1[i] + arr2[j]);` ` ` `// If set has less than N elements` ` ` `if` `(s.Count < N)` ` ` `return` `-1;` ` ` `// Find N'tb item in set and return it` ` ` `int` `[]last = s.ToArray();` ` ` `return` `last[s.Count - 1];` `}` `// Driver code` `public` `static` `void` `Main(String[] args) ` `{` ` ` `int` `[]arr1 = {1, 2};` ` ` `int` `size1 = arr1.Length;` ` ` `int` `[]arr2 = {3, 4};` ` ` `int` `size2 = arr2.Length;` ` ` `int` `N = 3;` ` ` `int` `res = calculateSetOfSum(arr1, size1, ` ` ` `arr2, size2, N);` ` ` `if` `(res == -1)` ` ` `Console.WriteLine(` `"N'th term doesn't exists in set"` `);` ` ` `else` ` ` `Console.WriteLine(` `"N'th element in the set"` `+` ` ` `" of sums is "` `+ res);` `}` `}` `// This code is contributed by Rajput-Ji` |

**Output**

N'th element in the set of sums is 6

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

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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.