Hackerrank Beautiful Binary String problem solution

Hackerrank Beautiful Binary String problem solution – In this Hackerrank Beautiful Binary String problem, Alice has a binary string. She thinks a binary string is beautiful if and only if it doesn’t contain the substring “010”.

In one step, Alice can change a 0 to a 1 or vice versa. Count and print the minimum number of steps needed to make Alice see the string as beautiful.

Example

b=010

She can change any one element and have a beautiful string.

Function Description

Complete the beautifulBinaryString function in the editor below.

beautifulBinaryString has the following parameter(s):

• string b: a string of binary digits

Returns

• int: the minimum moves required

Input Format

The first line contains an integer n, the length of binary string.
The second line contains a single binary string b.

Hackerrank Beautiful Binary String problem solution in Python.

#!/bin/python3

import math
import os
import random
import re
import sys

#
# Complete the 'beautifulBinaryString' function below.
#
# The function is expected to return an INTEGER.
# The function accepts STRING b as parameter.
#

def beautifulBinaryString(b):
    # Write your code here
    return(b.count('010'))

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

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

    b = input()

    result = beautifulBinaryString(b)

    fptr.write(str(result) + 'n')

    fptr.close()

Beautiful Binary String problem solution in Java.

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

public class Solution {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        sc.nextLine();
        String input = sc.nextLine();
        int answer = 0;
        int i = 0;
        while (i < n - 2){
            if (input.substring(i, i + 3).equals("010")){
                answer++;
                i += 3;
                continue;
            }
            i++;
        }
        System.out.println(answer);
    }
}

Problem solution in C++ programming.

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;


int main(){
    int n;
    cin >> n;
    string B;
    cin >> B;
    
    int cnt = 0;
    for (int i = 0; i < B.size(); ++i) {
        if (B.substr(i, 3) == "010") {
            ++cnt;
            i+=2;
        }
    }
    cout << cnt << endl;
    return 0;
}

Problem solution in C programming.

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

int main(){
    int n; 
    scanf("%d",&n);
    char* B = (char *)malloc(10240 * sizeof(char));
    scanf("%s",B);
    printf("%dn", countSubstring(B, "010"));
    return 0;
}
 
int countSubstring(const char *str, const char *sub)
{
    int length = strlen(sub);
    if (length == 0) return 0;
    int count = 0;
    for (str = strstr(str, sub); str; str = strstr(str + length, sub))
        ++count;
    return count;
}

Problem solution in JavaScript programming.

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() {
    let n = parseInt(readLine());
    let B = readLine();
    
    let index = B.indexOf('010');
    let i = 0;
    while(index > -1){
        i++;
        B = B.split('')
        B[index+2] = '1'
        B = B.join('')
        index = B.indexOf('010');
    }
    console.log(i);
   
}