Beautiful Binary String


Problem Statement :


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 ]  e  {0, 1 }

Output Format

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



Solution :



title-img


                            Solution in C :

In   C++  :





#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;
}








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);
    }
}








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;
}








In   Python3  :







#!/bin/python3
import re

import sys


n = int(input().strip())
B = input().strip()
print(len(re.findall("010", B)))
                        








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