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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; i < n; i++)
using namespace std;
using ll = long long;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  string S;
  cin >> N >> S;
  int count0 = 0;
  int count1 = 0;
  for(int i = 0; i < S.size(); i++) {
    if(S[i] == '#') { break; }
    count0++;
  }
  S.erase(S.begin() + 0, S.begin() + count0);
  for(int i = S.size() - 1; i >= 0; i--) {
    if(S[i] == '.') { break; }
    count1++;
  }
  S.erase(S.end() - count1, S.end());
  if(S.size() == 0) {
    cout << 0 << endl;
    return 0;
  }

  vector<int> H;
  int ct = 1;
  for(int i = 1; i < S.size(); i++) {
    if(S[i] != S[i - 1]) {
      H.push_back(ct);
      ct = 1;
    } else {
      ct++;
    }
  }
  H.push_back(ct);
  // for(int i = 0; i < H.size(); i++) {
  // cout << H[i];
  // }
  int ans = 1000000000;
  for(int i = 0; i <= H.size(); i++) {
    int sum = 0;
    bool flag = false;
    for(int j = 0; j < i; j++) {
      if(j % 2 == 0) sum += H[j];
      if(sum > ans) {
        flag = true;
        break;
      }
    }
    for(int j = i; j < H.size(); j++) {
      if(j % 2 != 0) sum += H[j];
      if(sum > ans) {
        flag = true;
        break;
      }
    }
    if(flag) continue;
    ans = min(ans, sum);
  }

  cout << ans << endl;
  return 0;
}