Problem:
Score : 300 points

Problem Statement

There are N stones arranged in a row. Every stone is painted white or black.
A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is ., and the stone is black if the character is #.

Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone.
Find the minimum number of stones that needs to be recolored.

Constraints

1 \leq N \leq 2\times 10^5

S is a string of length N consisting of . and #.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the minimum number of stones that needs to be recolored.

Sample Input 1

3
#.#

Sample Output 1

1

It is enough to change the color of the first stone to white.

Sample Input 2

5
#.##.

Sample Output 2

2

Sample Input 3

9
.........

Sample Output 3

0

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main{
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        
        int N = sc.nextInt();
        String S = sc.next();
        sc.close();

        int[] numB = new int[N]; //i番目までに黒が何個あるか
        for(int i = 0; i < N; i++){
            if(i == 0){
                if(S.charAt(i) == '#')
                    numB[i] = 1;
            } else {
                if(S.charAt(i) == '#')
                    numB[i] = numB[i-1]+1;
                else
                    numB[i] = numB[i-1];
            }
        }
        int n = numB[N-1];
        int min = numB[N-1]; //全部白に変える
        for(int i = 0; i < N-1; i++){
            if(i == 0 ){
                if(S.charAt(0) == '#'){
                    int cnt = 0;
                    cnt += (numB[i+1]-1);
                    cnt += ((N-i-1) - (n-numB[i+1]));
                    min = Math.min(min, cnt);
                }
            } else {
                if(numB[i] > numB[i-1]) {
                    int cnt = 0;
                    cnt += (numB[i+1]-1);
                    cnt += ((N-i-1) - (n-numB[i+1]));
                    min = Math.min(min, cnt);
                }
            }
        }
        System.out.println(min);

    }
}