Algorithm
Problem Name: 30 days of code -
In this HackerRank in 30 Days of Code - problem solution,
Objective
Today, we're working with binary numbers. Check out the Tutorial tab for learning materials and an instructional video!
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
'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 (125)10 is (1111101)2. 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 <= 10**6
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 (5)10 is (101)2, so the maximum number of consecutive 1's is 1.
Sample Case 2:
The binary representation of (13)10 is (1101)1 so the maximum number of consecutive 1's is 2.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <assert.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main() {
long int n;
char bin_str[100];
int i = 0;
scanf("%ld",&n);
int count = 0;
int max_count = count;
while(n>0){
int temp = n%2;
n = n/2;
if(temp == 0)
{
count = 0;
}else{
count++;
}
if(max_count < c ount){
max_count = count;
}
}
printf("%d",max_count>;
}
#2 Code Example with C++ Programming
Code -
C++ Programming
#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;
}
#3 Code Example with C# Programming
Code -
C# Programming
using System;
class Solution
{
static void Main(String[] args)
{
var n = int.Parse(Console.ReadLine());
var sum = 0;
var max = 0;
while (n > 0)
{
if (n % 2 == 1)
{
sum++;
if (sum > max)
max = sum;
}
else sum = 0;
n = n / 2;
}
Console.WriteLine(max);
}
}
#4 Code Example with Golang Programming
Code -
Golang Programming
package main
import (
"fmt"
)
func main() {
var n, reminder, count, max int
fmt.Scan(&n)
for n > 0 {
reminder = n % 2
if reminder == 1 {
count++
} else {
count = 0
}
if count > max {
max = count
}
n /= 2
}
fmt.Println(max)
}
#5 Code Example with Java Programming
Code -
Java Programming
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
in.close();
int sum = 0, max = 0;
while (n > 0) {
if (n % 2 == 1) {
sum++;
if (sum > max) {
max = sum;
}
} else {
sum = 0;
}
n = n / 2;
}
System.out.println(max);
}
}
#6 Code Example with Javascript Programming
Code -
Javascript Programming
function main() {
const n = parseInt(readLine().trim(), 10);
const binary = n.toString(2)
var max_count = 0
const arr = binary.split('0')
arr.map((each) => {
if (each.length > max_count) {
max_count = each.length
}
})
console.log(max_count)
}
#7 Code Example with Python Programming
Code -
Python Programming
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)
#8 Code Example with PHP Programming
Code -
PHP Programming
0 ) {
if ( $n % 2 == 1 ) {
$count++;
if ( $count >= $min ) {
$min = $count;
}
} else {
$count = 0;
}
$n = $n / 2;
}
echo $min;
fclose($handle);
?>