Related Articles
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 notbool check(int a[], int n){    for (int i = 0; i < n; i++)        if (a[i])            return true;     return false;} // Driver codeint 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 approachclass 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 notdef check(a, n):    for i in range(n):        if (a[i]):            return True     return False # Driver codeif __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 approachusing 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



## Javascript


Output:
YES

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes

My Personal Notes arrow_drop_up