HackerRank Day 29: Bitwise AND 30 days of code solution

Today we are going to solve HackerRank Day 29: Bitwise AND 30 days of code solutions in C, C++, Java, Javascript & Python.

Objective

Welcome to the last day! Today, we’re discussing bitwise operations.

Task

Given set S = {1, 2, 3, . . . ,N}. Find two integers, A and B (where A < B), from set S such that the value of A&B is the maximum possible and also less than a given integer, K. In this case, & represents the bitwise AND operator.

Function Description
Complete the bitwiseAnd function in the editor below.

bitwiseAnd has the following paramter(s):
– int N: the maximum integer to consider
– int K: the limit of the result, inclusive

Returns
– int: the maximum value of A&B within the limit.

Input Format

The first line contains an integer, T, the number of test cases.
Each of the T subsequent lines defines a test case as 2 space-separated integers, N and K, respectively.

Constraints

• 1 <= T <= 10³
• 2 <= N <= 10³
• 2 <= K <= N

Sample Input

STDIN   Function
-----   --------
3       T = 3
5 2     N = 5, K = 2
8 5     N = 8, K = 5
2 2     N = 8, K = 5

Sample Output

1
4
0

HackerRank Day 29: Bitwise AND 30 days of code solutions

Bitwise AND Solution in C

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int maxvalue(int n, int k) {
    int i, j;
    int res = 0, max_res = 0;
    for (i = 1; i <= n; i++) {
        for (j = i+1; j <=n ; j++) {
            int val = i &j;
            if (val > max_res && val < k) {
                max_res = val;
            }
        }
    }
    return max_res;
}
int main(){
    int t; 
    int a0;
    scanf("%d",&t);
    for(a0 = 0; a0 < t; a0++){
        int n; 
        int k; 
        scanf("%d %d",&n,&k);
        printf("%d\n", maxvalue(n, k));
    }
    return 0;
}

Bitwise AND Solution in C++

#include <iostream>
#include <vector>
using namespace std;


int main(){
    int ncases, n, k, max = 0, tmp = 0;
    vector<int> range;
    range.reserve(1000);
    cin >> ncases;
    for(int i = 0; i < ncases; ++i){
        cin >> n >> k;
        for(int j = 0; j < n; ++j)
            range.push_back(j + 1);
        for(int x = 0; x < range.size() - 1; ++x){
            for(int y = x + 1; y < range.size(); ++y){
                tmp = range[x] & range[y];
                if(tmp < k)
                    max = (tmp > max ? tmp : max);
            }
        }
        cout << max << '\n';
        range.clear();
        max = 0;
    }
}

Bitwise AND Solution in Java

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

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner in = new Scanner(System.in);
        int q = in.nextInt();
        for (int i = 0; i < q; i++) {
            int n = in.nextInt();
            int k = in.nextInt();

            int maxed = 0;
            for (int b = 2; b <= n; b++) {
                for (int a = 1; a < b; a++) {
                    if (a == b) continue;
                    int ab = a&b;
                    if (ab > maxed && ab < k) maxed = ab;
                }
            }
            System.out.println(maxed);
        }
    }

Bitwise AND Solution in Python 2

#!/bin/python

import sys


t = int(raw_input().strip())
for a0 in xrange(t):
    n,k = raw_input().strip().split(' ')
    n,k = [int(n),int(k)]
    maxValue = 0
    check = 0
    for i in range(n-1,0,-1):
        for j in range(n,i,-1):
            tmp = i&j
            if ((tmp > maxValue)&(tmp < k)):
                maxValue = tmp
                if (maxValue + 1 == k):
                    check = 1
                    break
        if (check == 1):
            break
    print maxValue

Bitwise AND Solution in Python 3

#!/bin/python3

import math
import os
import random
import re
import sys

def max_bit(n,k):
    maximum = 0
    for i in range(1,n+1):
        for j in range(1,i):
            h = i & j
            if maximum < h < k:
                maximum = h
            if maximum == k-1:
                return maximum
    return maximum

if __name__ == '__main__':
    t = int(input())

    for t_itr in range(t):
        nk = input().split()

        n = int(nk[0])

        k = int(nk[1])
        
        print(max_bit(n,k))

Bitwise AND Solution in Javascript

process.stdin.resume();
process.stdin.setEncoding('ascii');

var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;

process.stdin.on('data', function (data) {
    input_stdin += data;
});

process.stdin.on('end', function () {
    input_stdin_array = input_stdin.split("\n");
    main();    
});

function readLine() {
    return input_stdin_array[input_currentline++];
}

/////////////// ignore above this line ////////////////////

function main() {
    var t = parseInt(readLine());
    for(var a0 = 0; a0 < t; a0++){
        var n_temp = readLine().split(' ');
        var n = parseInt(n_temp[0]);
        var k = parseInt(n_temp[1]);
        function findMaxPoss(arr) {
            var res = 0;
            for(var i = 0; i < arr.length; i++){
                for(var j = i + 1; j < arr.length; j++){
                    var ans = arr[i] & arr[j];
                    if((ans > res) && (ans < k)){
                        res = ans;
                    }
                }
            }
            return res;
        }
        console.log(findMaxPoss(range(n)));
    }
    function range(n){
        return Array.apply(null, Array(n)).map(function (_, i) {return i + 1;});
    }
}

Read More: 30 Days of Code HackerRank Solutions List – Day 0 to Day 29