Maximize Bitwise AND of Array by replacing at most one element

Given an array arr[] containing N positive integers, the task is to maximize bitwise AND of the arr[] by picking at most one element of the arr[] and increment or decrement it by any value.

Examples:

Input: arr[] = {1, 2, 3}
Output: 2
Explanation: Following are the operations performed to maximize the bitwise AND of arr[]
Initially bitwise AND of {1, 2, 3} = 0
Incrementing arr[0] by 1 updates arr[] = {2, 2, 3}. 
Now bitwise AND of {2, 2, 3} is 2 which is maximum possible. 
Therefore, the answer is 2. 

Input: arr[] = {10, 10}
Output: 10
Explanation: Do not perform any operation that would be optimal.  

Approach: The problem statement of this problem says to maximize AND of arr[] by replacing almost one element from arr[]. The brute force method to solve this problem is to take bitwise AND of all the elements in arr[] except one at a time and find an appropriate replacement for that element whose contribution is not added in the whole bitwise AND of arr[] each time. Do this for all the elements in arr[] and find the optimal result. 

Below is the implementation of the above approach.

#include <bits/stdc++.h>
using namespace std;

// Function to find maximum AND by
// Replacing atmost one element by any value
int maxAnd(int n, vector<int> a)
{
    // To Calculate answer
    int ans = 0;

    // Iterating through the array
    for (int i = 0; i < n; i++) {
        int max_and = 0XFFFFFFFF;

        // Checking and for element and
        // Leaving only jth element
        for (int j = 0; j < n; j++) {
            if (i != j) {
                max_and = max_and & a[j];
            }
        }

        // Comparing previous answers
        // and max answers
        ans = max(ans, max_and);
    }
    return ans;
}

// Driver Code
int main()
{
    int N = 3;
    vector<int> arr = { 1, 2, 3 };

    // Function Call
    cout << maxAnd(N, arr) << "\n";
}

import java.io.*;

class GFG{
  
// Function to find maximum AND by
// Replacing atmost one element by any value
static int maxAnd(int n,int[] a)
{
    
    // To Calculate answer
    int ans = 0;

    // Iterating through the array
    for(int i = 0; i < n; i++) 
    {
        int max_and = 0XFFFFFFFF;

        // Checking and for element and
        // Leaving only jth element
        for(int j = 0; j < n; j++)
        {
            if (i != j)
            {
                max_and = max_and & a[j];
            }
        }

        // Comparing previous answers
        // and max answers
        ans = Math.max(ans, max_and);
    }
    return ans;
}

// Driver Code
public static void main(String[] args)
{
    int N = 3;
    int[] arr = { 1, 2, 3 };
    
    // Function Call
    System.out.println( maxAnd(N, arr) );
}
}

// This code is contributed by Potta Lokesh

# Function to find maximum AND by
# Replacing atmost one element by any value
def maxAnd(n, a):
    
    # To Calculate answer
    ans = 0

    # Iterating through the array
    for i in range(0, n):
        max_and = 0XFFFFFFFF

        # Checking and for element and
        # Leaving only jth element
        for j in range(0, n):
            if (i != j):
                max_and = max_and & a[j]

        # Comparing previous answers
        # and max answers
        ans = max(ans, max_and)
        
    return ans

# Driver Code
if __name__ == '__main__':
    
    N = 3
    arr = [ 1, 2, 3 ]
    
    print(maxAnd(N, arr))
    
    # This code is contributed by Samim Hossain Mondal.

using System;

class GFG{
  
// Function to find maximum AND by
// Replacing atmost one element by any value
static int maxAnd(int n,int[] a)
{
    
    // To Calculate answer
    int ans = 0;

    // Iterating through the array
    for(int i = 0; i < n; i++) 
    {
        uint max_and = 0XFFFFFFFF;

        // Checking and for element and
        // Leaving only jth element
        for(int j = 0; j < n; j++)
        {
            if (i != j)
            {
                max_and = Convert.ToUInt32(max_and & a[j]);
            }
        }

        // Comparing previous answers
        // and max answers
        ans = Math.Max(ans, (int)max_and);
    }
    return ans;
}

// Driver Code
public static void Main()
{
    int N = 3;
    int[] arr = { 1, 2, 3 };
    
    // Function Call
    Console.Write( maxAnd(N, arr) );
}
}

// This code is contributed by gfgking

<script>

// Function to find maximum AND by
// Replacing atmost one element by any value
const maxAnd = (n, a) => {
    
    // To Calculate answer
    let ans = 0;

    // Iterating through the array
    for(let i = 0; i < n; i++) 
    {
        let max_and = 0XFFFFFFFF;

        // Checking and for element and
        // Leaving only jth element
        for(let j = 0; j < n; j++)
        {
            if (i != j) 
            {
                max_and = max_and & a[j];
            }
        }

        // Comparing previous answers
        // and max answers
        ans = Math.max(ans, max_and);
    }
    return ans;
}

// Driver Code
let N = 3;
let arr = [ 1, 2, 3 ];

// Function Call
document.write(maxAnd(N, arr));

// This code is contributed by rakeshsahni

</script>

2

Time Complexity: O(N^2) 
Auxiliary Space: O(1)