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: Accepted)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define REP(i,a,b) for (int i = a; i < (b); ++i)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define YY cout << "Yes" << endl;
#define NN cout << "No" << endl;
#define ve vector
#define pint pair<int,int>
const int INF = 1000000007;
const long double EPS = 1e-15;
const long double PI = acos(-1);
typedef long long ll;
using namespace std;
using Graph = vector<vector<int>>;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { return (a * b) / gcd(a,b); }
// *max_element(all(d))
// priority_queue<int,vector<int>,greater<int>>
int solve(){
  int n;
  cin >> n;
  string s;
  cin >> s;
  int ans = INF;
  vector<int>sum(n + 1);
  rep(i,n){
    if(s[i] == '#')sum[i + 1] = sum[i] + 1;
    else sum[i + 1] = sum[i];
  }
  rep(i,n + 1)ans = min(ans,sum[i] + n - i - (sum[n] - sum[i]));
  cout << ans << endl;
  return 0; 
}
int main(){
  // ios::sync_with_stdio(false);cin.tie(nullptr);
  // cout << fixed;cout << setprecision(16);
  solve();
  return 0;
}