# Modify a binary array to Bitwise AND of all elements as 1

Given an array a[] consisting 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The main observation is if the array consists 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 only 0 and 1.
If there is at least one 1 then we will choose all index with 0 value and replace with OR value with the index having 1 and the OR value will be 1 always.
After all operation, 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 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[0]); ` ` `  `    ``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 [tabby title="PHP"]

 ` `

Output:

```YES
```

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