HackerRank Beautiful Binary String Solution

Hello Programmers, In this post, you will learn how to solve HackerRank Beautiful Binary String Solution. This problem is a part of the HackerRank Algorithms Series.

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HackerRank Beautiful Binary String Solution

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

HackerRank Beautiful Binary String Solution

Task

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.

Constraints

  • 1 <= n <= 100
  • b[i] ∈ {0, 1}

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

2

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

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

3

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.

HackerRank Beautiful Binary String Solution

Beautiful Binary String Solution in C

#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);
    int i=0,count=0;
    while(B[i]){
        if(B[i]=='0'&&B[i+1]=='1'&&B[i+2]=='0'){
         B[i+2]='1';
         count++;
        }
        i++;
    }
    printf("%d",count);
    return 0;
}

Beautiful Binary String Solution in Cpp

#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;
char B[105];
int main(){
    int n;
    int ans=0;
    scanf("%d", &n);
    scanf("%s", B);
    for(int i=2; B[i]; i++){
        if(B[i-2] == '0' && B[i-1] == '1' && B[i] == '0') B[i] = '1', ans++;
    }
    printf("%d", ans);
    return 0;
}

Beautiful Binary String Solution in Java

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        String B = in.next();
        
        int i = 0;
        int total = 0;
        while (i < B.length()-2) {
            if (B.substring(i,i+3).equals("010")) {
                total++;
                i+=3;
            } else {
                i++;
            }
        }
        System.out.println(total);
    }
}
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Beautiful Binary String Solution in Python

#!/bin/python
import sys
n = int(raw_input().strip())
B = raw_input().strip()
if '010' not in B:
    print 0
else:
    print B.count('010')

Beautiful Binary String Solution using JavaScript

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() {
    var n = parseInt(readLine());
    var B = readLine();
    console.log((B.match(/010/g)||[]).length);
}

Beautiful Binary String Solution in Scala

object Solution {
  def main(args: Array[String]): Unit = {
    val initTime = System.currentTimeMillis()
    val n = lines.next().toInt
    val binString = lines.next()
    var p1 = 0
    val l = binString.length
    if(binString.length < 2) {
      out.println(0)
    } else {
      var count = 0
      while(p1 < l-2) {
        if(binString(p1) == '0' && binString(p1+1) == '1' && binString(p1+2) == '0') {
          count += 1
          p1 += 3
        } else {
          p1 += 1
        }
      }
      out.println(count)
    }
    printTime(initTime, System.currentTimeMillis())
    out.flush()
  }
  private val file = new java.io.File("C:\\Users\\User\\Desktop\\hackerrank\\input_competition.txt")
  private val lines = {
    if(file.exists()) {
      io.Source.fromFile(file).getLines
    } else {
      io.Source.stdin.getLines()
    }
  }
  private val out = new java.io.PrintWriter(System.out)
  private def printTime(init: Long, end:Long):Unit =
    if(file.exists()) out.println(s"${end-init} ms")
}

Beautiful Binary String Solution in Pascal

uses math;
var   s:ansistring; i,n,ans:longint;
 begin
 readln(n);
 readln(s);
 n:=length(s);
 for i:=2 to n-1 do
 if (s[i]='1')  and (s[i-1]='0' ) and (s[i+1]='0' ) then
  begin
   inc(ans);
   s[i+1]:='1';
  end;
 writeln(ans);
 end.

Disclaimer: This problem (Beautiful Binary String) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

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