# Minimum value among AND of elements of every subset of an array

• Difficulty Level : Basic
• Last Updated : 16 Nov, 2022

Given an array of integers, the task is to find the AND of all elements of each subset of the array and print the minimum AND value among all those.
Examples:

```Input: arr[] = {1, 2, 3}
Output: 0
AND of all possible subsets
(1 & 2) = 0,
(1 & 3) = 1,
(2 & 3) = 2 and
(1 & 2 & 3) = 0.
Minimum among these is 0.

Input: arr[] = {7, 2}
Output: 2    ```

Approach: The minimum AND value of any subset of the array will be the AND of all the elements of the array. So, the simplest way is to find AND of all elements of sub-array.
Below is the implementation of the above approach:
Implementation:

## C++

 `// C++ program for the above approach``#include``using` `namespace` `std;` `void` `minAND(``int` `arr[], ``int` `n)``{` `    ``int` `s = arr[0];` `    ``// Find AND of whole array``    ``for` `(``int` `i = 1; i < n; i++)``    ``{``        ``s = s & arr[i];``    ``}` `    ``// Print the answer``    ``cout << (s) << endl;``}``    ` `// Driver code``int` `main()``{``    ``int` `arr[] = {1, 2, 3};``    ``int` `n = ``sizeof``(arr)/``sizeof``(``int``);``    ``minAND(arr, n);``}` `// This code has been contributed by Arnab Kundu`

## Java

 `// Java program for the above approach``class` `GFG``{` `    ``static` `void` `minAND(``int``[] arr, ``int` `n)``    ``{` `        ``int` `s = arr[``0``];` `        ``// Find AND of whole array``        ``for` `(``int` `i = ``1``; i < n; i++)``        ``{``            ``s = s & arr[i];``        ``}` `        ``// Print the answer``        ``System.out.println(s);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = {``1``, ``2``, ``3``};``        ``int` `n = arr.length;``        ``minAND(arr, n);``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Python

 `# Python program for the above approach``def` `minAND(arr, n):``    ` `    ``s ``=` `arr[``0``]``    ` `    ``# Find AND of whole array``    ``for` `i ``in` `range``(``1``, n):``        ``s ``=` `s & arr[i]``    ` `    ``# Print the answer``    ``print``(s)` `# Driver code``arr ``=` `[``1``, ``2``, ``3``]``n ``=` `len``(arr)``minAND(arr, n)`

## C#

 `// C# program for the above approach``class` `GFG``{` `    ``static` `void` `minAND(``int``[] arr, ``int` `n)``    ``{` `        ``int` `s = arr[0];` `        ``// Find AND of whole array``        ``for` `(``int` `i = 1; i < n; i++)``        ``{``            ``s = s & arr[i];``        ``}` `        ``// Print the answer``        ``System.Console.WriteLine(s);``    ``}``    ` `    ``// Driver code``    ``static` `void` `Main()``    ``{``        ``int``[] arr = {1, 2, 3};``        ``int` `n = arr.Length;``        ``minAND(arr, n);``    ``}``}` `// This code has been contributed by chandan_jnu`

## PHP

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## Javascript

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Output:

`0`

Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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