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)

/*
 * Created by KeigoOgawa
 */


#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <cassert>
#include <cmath>

#define INF (int)1e8
#define EPS 1e-10
#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define RFOR(i, a, b) for (ll i = (b)-1; i >= (a); i--)
#define REP(i, n) for (ll i = 0; i < (n); i++)
#define RREP(i, n) for (ll i = (n) - 1; i >= 0; i--)
#define MIN(a, b) (a > b ? b : a)
#define MAX(a, b) (a > b ? a : b)
#define debug(x) cerr << #x << ": " << x << endl
#define all(a) (a).begin(), (a).end()

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vi> vii;
typedef pair<int, int> PII;

// vector
template<typename T> ostream& operator<<(ostream& s, const vector<T>& v) {
    int len = v.size();
    for (int i = 0; i < len; ++i) {
        s << v[i]; if (i < len - 1) s << "\t";
    }
    return s;
}

// 2 dimentional vector
template<typename T> ostream& operator<<(ostream& s, const vector< vector<T> >& vv) {
    int len = vv.size();
    for (int i = 0; i < len; ++i) {
        s << vv[i] << endl;
    }
    return s;
}


ll N;
string S;

int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);

  cin >> N;
  cin >> S;
  int ans = 0;

  string s = S;
  char prev = s[0];
  for (int i = 1; i < N; i++) {
    if (prev == '#' && s[i] == '.')
    {
        ans++;
        s[i] = '#';
    }
    prev = s[i];
  }

  s = S;
  prev = s[N - 1];
  int ansr = 0;
  for (int i = N - 2; i >= 0; i--)
  {
      if (prev == '.' && s[i] == '#')
      {
          ansr++;
          s[i] = '.';
      }
      prev = s[i];
  }

  ans = min(ans, ansr);

  cout << ans << endl;

    return 0;
}