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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <limits.h>

#define REP(i,n) for(i=0;i<n;i++)

int iin() {
    int i;
    scanf("%d", &i);
    return i;
}

long long llin() {
    long long i;
    scanf("%lld", &i);
    return i;
}

double din() {
    double i;
    scanf("%lf", &i);
    return i;
}

char cin() {
    char c;
    scanf("%c", &c);
    return c;
}

void stin(char *s) {
    scanf("%s", s);
}

void iout(int i) {
    printf("%d\n", i);
}

void llout(long long i) {
    printf("%lld\n", i);
}

void dout(double i) {
    printf("%lf\n", i);
}

void cout(char c) {
    printf("%c\n", c);
}

void stout(char *s) {
    printf("%s\n", s);
}

int dfs(int v, int n, int *visited, int table[n][n]) {
    int all_visited = 1, i;

    REP(i,n) {
        if (!visited[i]) {
            all_visited = 0;
            break;
        }
    }

    if (all_visited) {
        return 1;
    }

    int count = 0;

    REP(i,n) {
        if (!table[v][i]) continue;
        if (visited[i]) continue;

        visited[i] = 1;
        count += dfs(i, n, visited, table);
        visited[i] = 0;
    }

    return count;
}

int main() {
    int n = iin();
    char s[n];
    stin(s);
    int i, b = 0, count = 0, fw = 0, lb = 0;
    REP(i,n) {
        if (b) {
            if (s[i] == '.') count++;
        } else {
            if (s[i] == '#') b = 1;
            else fw++;
        }
    }
    i = n - 1;
    while(1) {
        if (s[i] == '.' || i < 0) break;
        lb++;
        i--;
    }
    if (n == fw || n == lb) {
        iout(0);
        return 0;
    }
    iout((n - fw - count - lb > count) ? count : n - fw - count - lb);
    return 0;
}