In this article, we will be solving HackerRank Day 10: Binary Numbers problem, part of 30 Days of Code. Majorly, we will use Python3 and Java15 to solve the problem.


Today, we’re working with binary numbers. Check out the Tutorial tab for learning materials and an instructional video!


Given a base-10 integer, n, convert it to binary (base-2). Then find and print the base-10 integer denoting the maximum number of consecutive 1‘s in n‘s binary representation. When working with different bases, it is common to show the base as a subscript.


n = 125

The binary representation of 12510 is 11111012. In base 10, there are 5 and 1 consecutive ones on two groups. Print the maximum, 5.

Input Format

A single integer, n.


  • 1 <= n <= 106

Output Format

Print a single base-10 integer that denotes the maximum number of consecutive 1’s in the binary representation of n.

Sample Input 1


Sample Output 1


Sample Input 2


Sample Output 2



Sample Case 1:

The binary representation of 510 is 1012, so the maximum number of consecutive 1’s is 1.

Sample Case 2:

The binary representation of 1310 is 11012, so the maximum number of consecutive 1’s is 2.

You can find all the source code on my GitHub profile:

Solution Day 10: Binary Numbers in Python3


import math
import os
import random
import re
import sys

if __name__ == '__main__':
    n = int(input().strip())
    numbers = str(bin(n)[2:]).split('0')
    lenghts = [len(num) for num in numbers]

Solution Day 10: Binary Numbers in Java15

import java.math.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import static;
import static;

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(;

        int n = Integer.parseInt(bufferedReader.readLine().trim());
        char[] binary = Integer.toBinaryString(n).toCharArray();
        int tmpCount = 0;
        int maxCount = 0;
        for(int i = 0; i < binary.length; i++){
            tmpCount = (binary[i] == '0') ? 0 : tmpCount + 1; 
            if(tmpCount > maxCount){
                maxCount = tmpCount;
HackerRank Day 10: Binary Numbers Solution | Output – 30 Days of Code

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