Hello coders, today we are going to solve Day 10: Binary Number HackerRank Solution in C++, Java and Python.

Objective
Today, we’re working with binary numbers.
Task
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.
Example
n = 125
The binary representation of 12510 is 11111012. In base 10, there are 5 and 1 consecutive ones in two groups. Print the maximum, 5.
Input Format
A single integer, n.
Constraints
- 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
5
Sample Output 1
1
Sample Input 2
13
Sample Output 2
2
Explanation
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.
Solution – Day 10: Binary Numbers
C++
#include <iostream> using namespace std; int main() { int n; cin >> n; int sum = 0; int max = 0; while (n > 0) { if (n % 2 == 1) { sum++; if (sum > max) max = sum; } else sum = 0; n = n / 2; } cout << max; return 0; }
Java
import java.util.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); // convert to binary and split into strings of consecutive ones String[] groupings = Integer.toBinaryString(n).split("0"); int max = 0; for(String s : groupings){ if(max < s.length()){ max = s.length(); } } System.out.println(max); } }
Python
num = int(input()) result = 0 maximum = 0 while num > 0: if num % 2 == 1: result += 1 if result > maximum: maximum = result else: result = 0 num //= 2 print(maximum)
Disclaimer: The above Problem (Day 10: Binary Numbers) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.