Beautiful Binary String HackerRank Solution

Beautiful Binary String HackerRank Solution
Source : https://www.hackerrank.com/challenges/beautiful-binary-string

Alice has a binary string. She thinks a binary string is beautiful if and only if it doesn't contain the substring .

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

For example, if Alice's string is she can change any one element and have a beautiful string.

Function Description

Complete the beautifulBinaryString function in the editor below. It should return an integer representing the minimum moves required.

beautifulBinaryString has the following parameter(s):

• b: a string of binary digits

Input Format

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

Constraints

• .

Output Format

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

Sample Input 0

7
0101010

Sample Output 0

2  

Explanation 0:

In this sample,

The figure below shows a way to get rid of each instance of :

Because we were able to make the string beautiful by changing characters ( and ), we print .

Sample Input 1

501100

Sample Output 1

0

Sample Case 1:

In this sample

Explanation 1

The substring does not occur in , so the string is already beautiful and we print .

Sample Input 2

100100101010

Sample Output 2

3

Explanation 2

In this sample

One solution is to change the values of to form a beautiful string.

Source : https://www.hackerrank.com/challenges/beautiful-binary-string

Solution

	// Karthikalapati.blogspot.com
	import java.util.Scanner;
	// Idea: - For each 010, we can flip it to 011, and count that as one "step".
	// - An easy way to count the number of steps is to change each 010 to 01 instead,
	// and to compare the length of the resulting string with the original string
	public class Solution {
	static int minSteps(String B) {
	return B.length() - B.replace("010","01").length();
	}
	public static void main(String[] args) {
	Scanner scan = new Scanner(System.in);
	scan.nextInt(); // we don't need "n"
	String B = scan.next();
	scan.close();
	System.out.println(minSteps(B));
	}
	}