HackerRank Beautiful Binary String Solution

In this post, we will solve HackerRank Beautiful Binary String Problem Solution.

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.

nction 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.

Output Format

Print the minimum number of steps needed to make the string beautiful.

Sample Input 0

STDIN       Function
-----       --------
7           length of string n = 7
0101010     b = '0101010'

Sample Output 0

Explanation 0:
In this sample, b = “0101010”
The figure below shows a way to get rid of each instance of “010”:

Make the string beautiful by changing 2 characters (b[2] and b[5]).

Sample Input 1

5
01100

Sample Output 1

Sample Case 1:
In this sample b = “01100”
Explanation 1
The substring “010” does not occur in b, so the string is already beautiful in 0 moves.

Sample Input 2

10
0100101010

Sample Output 2

Explanation 2

In this sample b = “0100101010” One solution is to change the values of b[2], b[5] and b[9] to form a beautiful string.

Table of Contents

Beautiful Binary String C Solution

#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("%d\n", 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;
}

Beautiful Binary String C++ Solution

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    string s;
    int n,ans=0;
    cin>>n>>s;
    for(int i = 1;i<n-1;i++)
        if(s[i-1]=='0'&&s[i]=='1'&&s[i+1]=='0') s[i+1]='1', ans++;
    cout<<ans;
    return 0;
}

Beautiful Binary String C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
class Solution {

    static void Main(String[] args) {
        int n = Convert.ToInt32(Console.ReadLine());
        string B = Console.ReadLine();
        int count = 0;

    for (int i = 0; i < n-2; i++){
        if (B[i] != '0'){
            continue;
        }
        if (i+2 < n && B.Substring(i, 3) == "010"){
            count++;
            i +=2;
        }
    }
    Console.WriteLine(count);
    }
}

Beautiful Binary String Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

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

    public static int beautifulBinaryString(String b) {
    // Write your code here
        Pattern p = Pattern.compile("010");
        return (int)p.matcher(b).results().count();    
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int n = Integer.parseInt(bufferedReader.readLine().trim());

        String b = bufferedReader.readLine();

        int result = Result.beautifulBinaryString(b);

        bufferedWriter.write(String.valueOf(result));
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Beautiful Binary String JavaScript Solution

'use strict';

function main() {
    input = input.split('\n').slice(1)[0];
    let steps = 0;
    if (input.length >= 3) {
        for (let i = 0; i < input.length - 2; i++) {
            if (input[i] + input[i + 1] + input[i + 2] === '010') {
                steps++;
                i += 2;
            }
        }
    }
    process.stdout.write(steps + '');
}

process.stdin.resume();
process.stdin.setEncoding('ascii');

let input = "";

process.stdin.on('data', function (data) {
    input += data;
});

process.stdin.on('end', main);

if (process.argv[2] === 'test') {
    process.stdin.pause();
    input = `
    7
    0101010
    `.replace(/^\s+/mg, "").trim();
    process.stdout.write(`Input:\n${input}\n\nOutput:\n`);
    main();
}

Beautiful Binary String Python Solution

#!/bin/python3

import sys


n = int(input().strip())
B = input().strip()
bb = [i for i in B]
strn = "010"
c = 0
while True:
    a = B.find(strn)
    #print(a)
    if a >= 0:
        bb[a+2] = "1"
        #print(bb)
        B = ''.join(bb)
        #print(B)
        c += 1
    else:
        break
print(c)