Find last element in Array formed from bitwise AND of array elements

Given an array A[] of size N, the task is to find the last remaining element in a new array B containing all pairwise bitwise AND of elements from A i.e., B consists of N?(N ? 1) / 2 elements, each of the form A_i & A_j for some 1 ? i < j ? N. And we can perform the following operation any number of times on a new array till there is only one element remaining such as:

• Let X and Y be the current maximum and minimum elements of array B respectively and remove X and Y from array B and insert X|Y into it.

Examples:

Input: A[] = {2, 7, 1}
Output: 3
?Explanation: Array B will be [A₁ & A₂, A₁ & A₃, A₂ & A₃] = [2 & 7, 2 & 1, 7 & 1] = [2, 0, 1].
Then, we do the following operations on B:
Remove 2 and 0 from B and insert 2|0=2 into it. Now, B=[1, 2].
Remove 2 and 1 from B and insert 2|1=3 into it. Now, B=[3].
The last remaining element is thus 3.

Input: A[] = {4, 6, 7, 2}
Output: 6

Approach: The problem can be solved based on the following observation:

Observations:

• The property of bitwise or is that if we are performing X | Y, bit i will be set if atleast one of bit i in X or Y is set.
• This leads us to the most crucial observation of the problem, for every bit i, if bit i is set in atleast one of B₁, B₂, …B_N?(N?1)/2, then bit i will be set in the final remaining element of B when all the operations are performed. We don’t need to worry about how the operations are performed.
• For bit i to be set in atleast one element of B, we need to have atleast 2 elements in A say j and k where A_j and A_k both have bit i set. This sets the bit i in A_j & A_k.
• Now we have the following solution, iterate over every valid bit i, count the number of elements in array A which have bit i set. If this count is greater than 1, the final answer will have bit i set else it will be unset.

Follow the steps mentioned below to implement the idea:

• Create an array of size 32 to store the count of bits set in i^th position.
• Traverse the array and for each array element:
  • Find the positions in which the bit is set.
  • Increment the set bit count for that position by 1.
• Traverse the array storing the set bit count.
  • If the count is at least 2, set that bit in the final answer.
• Return the number formed as the required answer.

Below is the implementation of the above approach:

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find last remaining
// element in a new array
int find(int a[], int n)
{
    int count = 0;
    int b[33] = { 0 };
    for (int bit = 30; bit >= 0; bit--) {
        for (int i = 0; i < n; i++) {
            if (((1 << bit) & (a[i])) != 0) {
                b[bit]++;
            }
        }
    }
    for (int bit = 30; bit >= 0; bit--) {
        if (b[bit] > 1)
            count = count + (1 << bit);
    }
    return count;
}

// Driver Code
int main()
{
    int A[] = { 2, 7, 1 };
    int N = 3;

    // Function call
    cout << (find(A, N));
    return 0;
}

// This code is contributed by Rohit Pradhan

// Java code to implement the approach

import java.io.*;
import java.util.*;

public class GFG {

    // Function to find last remaining
    // element in a new array
    public static int find(int a[], int n)
    {
        int count = 0;
        int b[] = new int[33];
        for (int bit = 30; bit >= 0; bit--) {
            for (int i = 0; i < n; i++) {
                if (((1 << bit) & (a[i])) != 0) {
                    b[bit]++;
                }
            }
        }
        for (int bit = 30; bit >= 0; bit--) {
            if (b[bit] > 1)
                count = count + (1 << bit);
        }
        return count;
    }

    // Driver code
    public static void main(String[] args)
    {
        int A[] = { 2, 7, 1 };
        int N = A.length;

        // Function call
        System.out.println(find(A, N));
    }
}

# Python code for the above approach

# Function to find last remaining element in a new array
def find(a, n):
    count = 0
    b = [0] * 33
    for bit in range(30, -1, -1):
        for i in range(n):
            if(((1 << bit) & (a[i])) is not 0):
                b[bit] += 1

    for bit in range(30, -1, -1):
        if(b[bit] > 1):
            count = count + (1 << bit)

    return count

A = [2, 7, 1]
N = len(A)

# Function call
print(find(A, N))

# This code is contributed by lokesh

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{
    // Function to find last remaining
    // element in a new array
    public static int find(int[] a, int n)
    {
        int count = 0;
        int[] b = new int[33];
        for (int bit = 30; bit >= 0; bit--) {
            for (int i = 0; i < n; i++) {
                if (((1 << bit) & (a[i])) != 0) {
                    b[bit]++;
                }
            }
        }
        for (int bit = 30; bit >= 0; bit--) {
            if (b[bit] > 1)
                count = count + (1 << bit);
        }
        return count;
    }

 // Driver code
 public static void Main(String[] args)
 {
     int[] A = { 2, 7, 1 };
        int N = A.Length;

        // Function call
       Console.WriteLine(find(A, N));
 }
}

// This code is contributed by code_hunt.

<script>
// JS code to implement the approach

    // Function to find last remaining
    // element in a new array
    function find(a, n)
    {
        let count = 0;
        let b = new Array(33);
        for (let i = 0; i < n; i++) {
            b[i]=0;
        }
        for (let bit = 30; bit >= 0; bit--) {
            for (let i = 0; i < n; i++) {
                if (((1 << bit) & (a[i])) != 0) {
                    b[bit]++;
                }
            }
        }
        for (let bit = 30; bit >= 0; bit--) {
            if (b[bit] > 1)
                count = count + (1 << bit);
        }
        return count;
    }

// Driver code

        let A = [ 2, 7, 1 ];
        let N = A.length;

        // Function call
        document.write(find(A, N));

// This code is contributed by sanjoy_62.
</script>

3

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