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  1. Prepare
  2. Tutorials
  3. 30 Days of Code
  4. Day 10: Binary Numbers
  5. Discussions

Day 10: Binary Numbers

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2837 Discussions

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  • romellhamilton14
     + 0 comments

    For C#

    int n = Convert.ToInt32(Console.ReadLine().Trim());

    string binary = "";

    int max = 0;

    int sum = 0;

    int remainder = 0;

    for (int i = n; i > 0; i = i / 2)

    { remainder = i % 2; sum += remainder;

    if (remainder  == 1 &&sum > max)
    max = sum;

else if (remainder == 0)
    sum = 0;

binary += remainder.ToString();

    } binary = new string(binary.Reverse().ToArray());

    Console.WriteLine(max);

  • Kuladeeppappu
     + 0 comments
    if __name__ == '__main__':
    n = int(input().strip())
    k=bin(n)
    k=k.replace("0b", "")
   
    count=0
    max=0
    for i in k:
        if i=='1':
            count=count +1
           
        else:
            count=0
        if count>max:
            max=count     
    print(max)

  • Venom001
     + 0 comments

    I have completed this solution in Python 3 with a bigginer freindly logic

    n = int(input().strip())

binary = bin(n)[2:]   # remove '0b'
max_ones = 0
current = 0

for bit in binary:
    if bit == '1':
        current += 1
        max_ones = max(max_ones, current)
    else:
        current = 0

print(max_ones)

  • ilambrev
     + 0 comments
    import java.io.*;
import java.util.*;

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(bufferedReader.readLine().trim());
        bufferedReader.close();

        System.out.println(Arrays.stream(Integer.toBinaryString(n).split("0+"))
                .map(String::length)
                .max(Comparator.comparingInt(l -> l)).orElse(0));
    }
}

  • abumohd_zalrahm1
     + 0 comments

    A simple one line solution for Python 3

    #!/bin/python3
if __name__ == '__main__':
    n = int(input().strip())
# Solution
    print(max([*map(len, bin(n)[2:].split("0"))]))