Maximum Bitwise AND value of subsequence of length K

Given an array a of size N and an integer K. The task is to find the maximum bitwise and value of elements of any subsequence of length K. 
Note: a[i] <= 10⁹
Examples:

Input: a[] = {10, 20, 15, 4, 14}, K = 4 
Output: 4 
{20, 15, 4, 14} is the subsequence with highest ‘&’ value. 
Input: a[] = {255, 127, 31, 5, 24, 37, 15}, K = 5 
Output: 8

Naive Approach: A naive approach is to recursively find the bitwise and value of all subsequences of length K and the maximum among all of them will be the answer. 
Efficient Approach: An efficient approach is to solve it using bit properties. Below are the steps to solve the problem: 

• Iterate from the left(initially left = 31 as 2³² > 10⁹ ) till we find > K numbers in the vector temp (initially temp = arr) whose i-th bit is set. Update the new set of numbers to temp array
• If we do not get > K numbers, the & value of any K elements in the temp array will be the maximum & value possible.
• Repeat Step-1 with left re-initialized as first-bit + 1.

Below is the implementation of the above approach: 

C++

// C++ program to find the sum of
// the addition of all possible subsets.
#include <bits/stdc++.h>
using namespace std;

// Function to perform step-1
vector<int> findSubset(vector<int>& temp, int& last, int k)
{
    vector<int> ans;

    // Iterate from left till 0
    // till we get a bit set of K numbers
    for (int i = last; i >= 0; i--) {
        int cnt = 0;

        // Count the numbers whose
        // i-th bit is set
        for (auto it : temp) {
            int bit = it & (1 << i);
            if (bit > 0)
                cnt++;
        }

        // If the array has numbers>=k
        // whose i-th bit is set
        if (cnt >= k) {
            for (auto it : temp) {
                int bit = it & (1 << i);
                if (bit > 0)
                    ans.push_back(it);
            }

            // Update last
            last = i - 1;

            // Return the new set of numbers
            return ans;
        }
    }

    return ans;
}

// Function to find the maximum '&' value
// of K elements in subsequence
int findMaxiumAnd(int a[], int n, int k)
{
    int last = 31;
    // Temporary arrays
    vector<int> temp1, temp2;

    // Initially temp = arr
    for (int i = 0; i < n; i++) {
        temp2.push_back(a[i]);
    }

    // Iterate till we have >=K elements
    while ((int)temp2.size() >= k) {

        // Temp array
        temp1 = temp2;

        // Find new temp array if
        // K elements are there
        temp2 = findSubset(temp2, last, k);
    }

    // Find the & value
    int ans = temp1[0];
    for (int i = 0; i < k; i++)
        ans = ans & temp1[i];

    return ans;
}

// Driver Code
int main()
{
    int a[] = { 255, 127, 31, 5, 24, 37, 15 };
    int n = sizeof(a) / sizeof(a[0]);
    int k = 4;

    cout << findMaxiumAnd(a, n, k);
}

Java

// Java program to find the sum of
// the addition of all possible subsets.
import java.util.*;
class GFG
{
static int last;

// Function to perform step-1
static Vector<Integer>
       findSubset(Vector<Integer> temp, int k)
{
    Vector<Integer> ans = new Vector<Integer>();

    // Iterate from left till 0
    // till we get a bit set of K numbers
    for (int i = last; i >= 0; i--)
    {
        int cnt = 0;

        // Count the numbers whose
        // i-th bit is set
        for (Integer it : temp)
        {
            int bit = it & (1 << i);
            if (bit > 0)
                cnt++;
        }

        // If the array has numbers>=k
        // whose i-th bit is set
        if (cnt >= k)
        {
            for (Integer it : temp)
            {
                int bit = it & (1 << i);
                if (bit > 0)
                    ans.add(it);
            }

            // Update last
            last = i - 1;

            // Return the new set of numbers
            return ans;
        }
    }
    return ans;
}

// Function to find the maximum '&' value
// of K elements in subsequence
static int findMaxiumAnd(int a[], int n, int k)
{
    last = 31;

    // Temporary arrays
    Vector<Integer> temp1 = new Vector<Integer>();
    Vector<Integer> temp2 = new Vector<Integer>();;

    // Initially temp = arr
    for (int i = 0; i < n; i++)
    {
        temp2.add(a[i]);
    }

    // Iterate till we have >=K elements
    while ((int)temp2.size() >= k)
    {

        // Temp array
        temp1 = temp2;

        // Find new temp array if
        // K elements are there
        temp2 = findSubset(temp2, k);
    }

    // Find the & value
    int ans = temp1.get(0);
    for (int i = 0; i < k; i++)
        ans = ans & temp1.get(i);

    return ans;
}

// Driver Code
public static void main(String[] args)
{
    int a[] = { 255, 127, 31, 5, 24, 37, 15 };
    int n = a.length;
    int k = 4;

    System.out.println(findMaxiumAnd(a, n, k));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the sum of
# the addition of all possible subsets.
last = 31

# Function to perform step-1
def findSubset(temp, k):
    global last
    ans = []

    # Iterate from left till 0
    # till we get a bit set of K numbers
    for i in range(last, -1, -1):
        cnt = 0

        # Count the numbers whose
        # i-th bit is set
        for it in temp:
            bit = it & (1 << i)
            if (bit > 0):
                cnt += 1

        # If the array has numbers>=k
        # whose i-th bit is set
        if (cnt >= k):
            for it in temp:
                bit = it & (1 << i)
                if (bit > 0):
                    ans.append(it)

            # Update last
            last = i - 1

            # Return the new set of numbers
            return ans

    return ans

# Function to find the maximum '&' value
# of K elements in subsequence
def findMaxiumAnd(a, n, k):
    global last

    # Temporary arrays
    temp1, temp2, = [], []

    # Initially temp = arr
    for i in range(n):
        temp2.append(a[i])

    # Iterate till we have >=K elements
    while len(temp2) >= k:

        # Temp array
        temp1 = temp2

        # Find new temp array if
        # K elements are there
        temp2 = findSubset(temp2, k)

    # Find the & value
    ans = temp1[0]
    for i in range(k):
        ans = ans & temp1[i]

    return ans

# Driver Code
a = [255, 127, 31, 5, 24, 37, 15]
n = len(a)
k = 4

print(findMaxiumAnd(a, n, k))

# This code is contributed by Mohit Kumar

C#

// C# program to find the sum of
// the addition of all possible subsets.
using System;
using System.Collections.Generic;

class GFG
{
static int last;

// Function to perform step-1
static List<int>findSubset(List<int> temp, int k)
{
    List<int> ans = new List<int>();

    // Iterate from left till 0
    // till we get a bit set of K numbers
    for (int i = last; i >= 0; i--)
    {
        int cnt = 0;

        // Count the numbers whose
        // i-th bit is set
        foreach (int it in temp)
        {
            int bit = it & (1 << i);
            if (bit > 0)
                cnt++;
        }

        // If the array has numbers>=k
        // whose i-th bit is set
        if (cnt >= k)
        {
            foreach (int it in temp)
            {
                int bit = it & (1 << i);
                if (bit > 0)
                    ans.Add(it);
            }

            // Update last
            last = i - 1;

            // Return the new set of numbers
            return ans;
        }
    }
    return ans;
}

// Function to find the maximum '&' value
// of K elements in subsequence
static int findMaxiumAnd(int []a, int n, int k)
{
    last = 31;

    // Temporary arrays
    List<int> temp1 = new List<int>();
    List<int> temp2 = new List<int>();;

    // Initially temp = arr
    for (int i = 0; i < n; i++)
    {
        temp2.Add(a[i]);
    }

    // Iterate till we have >=K elements
    while ((int)temp2.Count >= k)
    {

        // Temp array
        temp1 = temp2;

        // Find new temp array if
        // K elements are there
        temp2 = findSubset(temp2, k);
    }

    // Find the & value
    int ans = temp1[0];
    for (int i = 0; i < k; i++)
        ans = ans & temp1[i];

    return ans;
}

// Driver Code
public static void Main(String[] args)
{
    int []a = { 255, 127, 31, 5, 24, 37, 15 };
    int n = a.Length;
    int k = 4;

    Console.WriteLine(findMaxiumAnd(a, n, k));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript program to find the sum of
// the addition of all possible subsets.

// Function to perform step-1
function findSubset(temp,k)
{
    let ans = [];

    // Iterate from left till 0
    // till we get a bit set of K numbers
    for (let i = last; i >= 0; i--)
    {
        let cnt = 0;

        // Count the numbers whose
        // i-th bit is set
        for (let it=0;it< temp.length;it++)
        {
            let bit = temp[it] & (1 << i);
            if (bit > 0)
                cnt++;
        }

        // If the array has numbers>=k
        // whose i-th bit is set
        if (cnt >= k)
        {
            for (let it=0;it< temp.length;it++)
            {
                let bit = temp[it] & (1 << i);
                if (bit > 0)
                    ans.push(temp[it]);
            }

            // Update last
            last = i - 1;

            // Return the new set of numbers
            return ans;
        }
    }
    return ans;
}

// Function to find the maximum '&' value
// of K elements in subsequence
function findMaxiumAnd(a,n,k)
{
    last = 31;

    // Temporary arrays
    let temp1 = [];
    let temp2 = [];

    // Initially temp = arr
    for (let i = 0; i < n; i++)
    {
        temp2.push(a[i]);
    }

    // Iterate till we have >=K elements
    while (temp2.length >= k)
    {

        // Temp array
        temp1 = temp2;

        // Find new temp array if
        // K elements are there
        temp2 = findSubset(temp2, k);
    }

    // Find the & value
    let ans = temp1[0];
    for (let i = 0; i < k; i++)
        ans = ans & temp1[i];

    return ans;
}

// Driver Code
let a=[255, 127, 31, 5, 24, 37, 15 ];
let n = a.length;
let k = 4;
document.write(findMaxiumAnd(a, n, k));



// This code is contributed by unknown2108
</script>

Output:

24

Time Complexity: O(N*N), as we are using a loop to traverse N times and in each traversal we are calling the findSubset function which will cost O (N) time. Where N is the number of elements in the array.

Auxiliary Space: O(N), as we are using extra space for the temp array. Where N is the number of elements in the array.