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;` `}` |

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## 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` |

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## 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` |

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**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.

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