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 :
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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