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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<cmath>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>

using namespace std;

typedef long long ll;

typedef std::pair<int, int> ipair;
bool lessPair(const ipair& l, const ipair& r){return l.second < r.second;}
bool morePair(const ipair& l, const ipair& r){return l.second > r.second;}

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

const ll MOD = 1e9 + 7;
void add(long long &a, long long b) { a += b; if (a >= MOD) a -= MOD; }

int main() {
   int N;
   cin >> N;
   string S;
   cin >> S;
   int ans = N;
   int kuro[N + 1];
   int siro[N + 1];
   memset(kuro, 0, sizeof(kuro));
   memset(siro, 0, sizeof(siro));
   for (int i = 0; i < N; i++) {
      if (S[i] == '#') kuro[i + 1] = kuro[i] + 1;
      else kuro[i + 1] = kuro[i];
      if (S[i] == '.') siro[i + 1] = siro[i] + 1;
      else siro[i + 1] = siro[i];
   }
   /*
   for (int i = 0; i <= N; i++) {
      cout << i << " " << kuro[i] << " " << siro[i] << endl;
   }
   */
   for (int i = 1; i < N; i++) {
      if (S[i - 1] == '#' && S[i] == '.') {
         ans = min(ans, kuro[i] + siro[N] - siro[i]);
      }
   }
   ans = min(ans, kuro[N]);
   ans = min(ans, siro[N]);
   cout << ans << endl;
   return 0;
}