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

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++ implementation of the above approach
#include <bits/stdc++.h>
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 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

# 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# 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
// PHP implementation of the
// above approach

// Function to check if it is 
// possible or not
function check($a, $n)
{
    for ($i = 0; $i < $n; $i++) 
        if ($a[$i]) 
            return true;

    return false;
}

// Driver code
$a = array(0, 1, 0, 1);
$n = sizeof($a);

if(check($a, $n)) 
    echo "YES\n";
else
    echo "NO\n";

// This code is contributed
// by Akanksha Rai
?>

<script>

// Javascript implementation of the above approach

// Function to check if it is possible or not
function check(a, n)
{
    for (var i = 0; i < n; i++) 
        if (a[i]) 
            return true;

    return false;
}

// Driver code
var a = [0, 1, 0, 1 ];
var n = a.length;
check(a, n) ? document.write( "YES")
            : document.write( "NO\n");


</script> 

YES

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.