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<string.h>

int main(void){
    int N;
    scanf("%d", &N);
    char S[N+1];
    scanf("%s", S);

    int max = strlen(S);
    int ans = max;

    int idx = 1;
    int left = 0;
    int right = 0;
    int i;
    int flag = 1;
    for(i = 0; i < N-1; i++){
        if(S[i] != S[i+1]){flag = 0; break;}
    }

    //これだと(2*10^5)^2で10^10以上で間に合わなさそうなので
    //累積和を使ってO(n)チックにしたい
    /*for(idx = 1; idx <= N-1; idx++){
        for(i = 0; i < idx; i++){if(S[i] == '#')left++;}
        for(i = idx; i < N; i++){if(S[i] == '.')right++;}
        if(ans > (left+right)){ans = left+right;}
        left = 0;
        right = 0;
    }*/

    int lc[N];
    lc[0] = 0;
    for(idx = 0; idx < N-1; idx++){
        if(S[idx] == '#'){lc[idx+1] = lc[idx] + 1;}
        //これ（真下）がないと、何も代入されてない要素ができてしまう
        else{lc[idx+1] = lc[idx];}
        //printf("idx+1:%d, lc[%d]:%d\n", idx+1, idx+1, lc[idx+1]);
    }
    int rc[N];
    rc[N-1] = 0;
    for(idx = N-1; idx > 0; idx--){
        if(S[idx] == '.'){rc[idx-1] = rc[idx] + 1;}
        else{rc[idx-1] = rc[idx];}
        //printf("idx-1:%d, rc[%d]:%d\n", idx-1, idx-1, lc[idx-1]);
    }

    for(i = 0; i < N-1; i++){
        if(ans > lc[i+1] + rc[i]){ans = lc[i+1] + rc[i];}
    }
    
    if(flag == 1){printf("0\n");}
    else{printf("%d\n", ans);}
    return 0;
}