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Modify a binary array to Bitwise AND of all elements as 1
• Difficulty Level : Basic
• Last Updated : 25 May, 2021

Given an array, a[] consists of only 0 and 1. The task is to check if it is possible to transform the array such that the AND value between every pair of indices is 1. The only operation allowed is to:

• Take two indices i and j and replace the a[i] and a[j] with a[i] | a[j] where ‘|’ means bitwise OR operation.

If it is possible, then the output is “YES”, otherwise the output is “NO”.

Examples:

```Input:  arr[] = {0, 1, 0, 0, 1}
Output: Yes
Choose these pair of indices (0, 1), (1, 2), (3, 4).

Input: arr[] = {0, 0, 0}
Output: No ```

Approach: The main observation is, if the array consists of at least one 1, then the answer will be YES, otherwise the output will be NO because OR with 1 will give us 1, as the array consists of only 0 and 1.
If there is at least one 1, then we will choose all indices with a 0 value and replace them with an OR value with the index having 1 and the OR value will always be 1.
After all operations, the array will consist of only 1 and the AND value between any pair of indices will be 1 as (1 AND 1)=1.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `// Function to check if it is possible or not``bool` `check(``int` `a[], ``int` `n)``{``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(a[i])``            ``return` `true``;` `    ``return` `false``;``}` `// Driver code``int` `main()``{` `    ``int` `a[] = { 0, 1, 0, 1 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);` `    ``check(a, n) ? cout << ``"YES\n"``                ``: cout << ``"NO\n"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG``{``    ` `    ``// Function to check if it is possible or not``    ``static` `boolean` `check(``int` `a[], ``int` `n)``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``if` `(a[i] == ``1``)``                ``return` `true``;``    ` `        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `a[] = { ``0``, ``1``, ``0``, ``1` `};``        ``int` `n = a.length;``    ` `        ``if``(check(a, n) == ``true` `)``            ``System.out.println(``"YES\n"``) ;``        ``else``            ``System.out.println(``"NO\n"``);``    ``}``}` `// This code is contributed by Ryuga`

## Python3

 `# Python 3 implementation of the``# above approach` `# Function to check if it is``# possible or not``def` `check(a, n):``    ``for` `i ``in` `range``(n):``        ``if` `(a[i]):``            ``return` `True` `    ``return` `False` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``a ``=` `[``0``, ``1``, ``0``, ``1``]``    ``n ``=` `len``(a)``    ` `    ``if``(check(a, n)):``        ``print``(``"YES"``)``    ``else``:``        ``print``(``"NO"``)``        ` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG``{``    ` `    ``// Function to check if it is possible or not``    ``static` `bool` `check(``int` `[]a, ``int` `n)``    ``{``        ``for` `(``int` `i = 0; i < n; i++)``            ``if` `(a[i] == 1)``                ``return` `true``;``    ` `        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]a = { 0, 1, 0, 1 };``        ``int` `n = a.Length;``    ` `        ``if``(check(a, n) == ``true` `)``            ``Console.Write(``"YES\n"``) ;``        ``else``            ``Console.Write(``"NO\n"``);``    ``}``}` `// This code is contributed``// by Akanksha Rai`

## PHP

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## Javascript

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Output:
`YES`

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