# Choose two elements from the given array such that their sum is not present in any of the arrays

• Last Updated : 01 Apr, 2021

Given two arrays A[] and B[], the task is to choose two elements X and Y such that X belongs to A[] and Y belongs to B[] and (X + Y) must not be present in any of the array.
Examples:

Input: A[] = {3, 2, 2}, B[] = {1, 5, 7, 7, 9}
Output: 3 9
3 + 9 = 12 and 12 is not present in
any of the given arrays.
Input: A[] = {1, 3, 5, 7}, B[] = {7, 5, 3, 1}
Output: 7 7

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Approach: Choose X as the maximum element from A[] and Y as the maximum element from B[]. Now, it is obvious that (X + Y) will be greater than the maximum of both the arrays i.e. it will not be present in any og the arrays.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to find the numbers from``// the given arrays such that their``// sum is not present in any``// of the given array``void` `findNum(``int` `a[], ``int` `n, ``int` `b[], ``int` `m)``{``    ``// Find the maximum element``    ``// from both the arrays``    ``int` `x = *max_element(a, a + n);``    ``int` `y = *max_element(b, b + m);``    ``cout << x << ``" "` `<< y;``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 3, 2, 2 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(``int``);``    ``int` `b[] = { 1, 5, 7, 7, 9 };``    ``int` `m = ``sizeof``(b) / ``sizeof``(``int``);` `    ``findNum(a, n, b, m);``    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `// find maximum element in an array``static` `int` `max_element(``int` `a[], ``int` `n)``{``    ``int` `m = Integer.MIN_VALUE;``    ` `    ``for``(``int` `i = ``0``; i < n; i++)``        ``m = Math.max(m, a[i]);``    ` `    ``return` `m;``}` `// Function to find the numbers from``// the given arrays such that their``// sum is not present in any``// of the given array``static` `void` `findNum(``int` `a[], ``int` `n,``                    ``int` `b[], ``int` `m)``{``    ``// Find the maximum element``    ``// from both the arrays``    ``int` `x = max_element(a, n);``    ``int` `y = max_element(b, m);``    ``System.out.print(x + ``" "` `+ y);``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `a[] = { ``3``, ``2``, ``2` `};``    ``int` `n = a.length;``    ``int` `b[] = { ``1``, ``5``, ``7``, ``7``, ``9` `};``    ``int` `m = b.length;` `    ``findNum(a, n, b, m);``}``}` `// This code is contributed by Arnub Kundu`

## Python3

 `# Python3 implementation of the approach` `# Function to find the numbers from``# the given arrays such that their``# sum is not present in any``# of the given array``def` `findNum(a, n, b, m) :` `    ``# Find the maximum element``    ``# from both the arrays``    ``x ``=` `max``(a);``    ``y ``=` `max``(b);``    ``print``(x, y);` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``a ``=` `[ ``3``, ``2``, ``2` `];``    ``n ``=` `len``(a);``    ` `    ``b ``=` `[ ``1``, ``5``, ``7``, ``7``, ``9` `];``    ``m ``=` `len``(b);` `    ``findNum(a, n, b, m);` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `    ``// find maximum element in an array``    ``static` `int` `max_element(``int` `[]a, ``int` `n)``    ``{``        ``int` `m = ``int``.MinValue;``        ` `        ``for``(``int` `i = 0; i < n; i++)``            ``m = Math.Max(m, a[i]);``        ` `        ``return` `m;``    ``}``    ` `    ``// Function to find the numbers from``    ``// the given arrays such that their``    ``// sum is not present in any``    ``// of the given array``    ``static` `void` `findNum(``int` `[]a, ``int` `n,``                        ``int` `[]b, ``int` `m)``    ``{``        ``// Find the maximum element``        ``// from both the arrays``        ``int` `x = max_element(a, n);``        ``int` `y = max_element(b, m);``        ``Console.Write(x + ``" "` `+ y);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]a = { 3, 2, 2 };``        ``int` `n = a.Length;``        ``int` `[]b = { 1, 5, 7, 7, 9 };``        ``int` `m = b.Length;``    ` `        ``findNum(a, n, b, m);``    ``}``}` `// This code is contributed by kanugargng`

## Javascript

 ``
Output:
`3 9`

Time Complexity: O(n)

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