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

• Difficulty Level : Medium
• Last Updated : 08 Jul, 2022

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

Recommended Practice

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``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

## 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 s = ``new` `TreeSet();``    ``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`

## Python3

 `# Python3 program to find N'th``# element in a set formed``# by sum of two arrays` `# Function to calculate the set of sums``def` `calculateSetOfSum(arr1, size1,arr2, size2, N):``    ` `    ``# Insert each pair sum into set.``    ``# Note that a set stores elements``    ``# in sorted order and unique elements``    ``s ``=` `set``()``    ``for` `i ``in` `range``(size1):``        ``for` `j ``in` `range``(size2):``            ``s.add(arr1[i]``+``arr2[j])` `    ``# If set has less than N elements``    ``if` `(``len``(s) < N):``        ``return` `-``1` `    ``# Find N'tb item in set and return it``    ``return` `list``(s)[N ``-` `1``]` `# Driver code``arr1 ``=` `[ ``1``, ``2` `]``size1 ``=` `len``(arr1)``arr2 ``=` `[ ``3``, ``4` `]``size2 ``=` `len``(arr2)``N ``=` `3` `res ``=` `calculateSetOfSum(arr1, size1,``                            ``arr2, size2, N)``if` `(res ``=``=` `-``1``):``    ``print``(``"N'th term doesn't exists in set"``)``else``:``    ``print``(f``"N'th element in the set of sums is {res}"``)`  `# This code is contributed by shinjanpatra`

## 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`

## Javascript

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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.

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