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: Runtime Error)

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();
		sc = null;
		int i;
		int[] white = new int[10000+1];
		int[] black = new int[10000+1];
		int widx=0,bidx=0;
		if(s.charAt(0)=='#') {
			white[widx++]=0;
		}
		char pre = '\0';
		int preidx = 0;
		for(i=0;i<n;i++) {
			if(pre == '#' && s.charAt(i) == '.'){
				black[bidx++]=i-preidx;
				preidx = i;
			} else if(pre == '.' && s.charAt(i) == '#') {
				white[widx++]=i-preidx;
				preidx = i;
			}
			pre = s.charAt(i);
		}
		
		if(s.charAt(n-1)=='.') {
			white[widx++]=i-preidx;
			black[bidx++]=0;
		} else {
			black[bidx++]=i-preidx;
		}
		int min=200000;
		int sum=0;
		int j;
		for(i=0; i< bidx; i++) {
			sum = 0;
			for(j=0;j<i;j++) {
				sum+=black[j];
			}
			for(++j;j<widx;j++) {
				sum+=white[j];
			}
			min = Math.min(min, sum);
		}
		System.out.println(min);
		
	}
}