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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

/* #pragma */
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

/*alias*/
#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
#define next asdnext
#define prev asdprev
#define itn int

/*struct*/
struct aaa {
	aaa() {
		cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(20);
	};
}aaaaaaa;

/* const */
const int INF = 1001001001;
const long long LINF = 1001001001001001001ll;
const int MOD = 1e9 + 7; //10^9
const double EPS = 1e-9;
const int dx[] = { 1, 1, 0, -1, -1, -1, 0, 1 }, dy[] = { 0, 1, 1, 1, 0, -1, -1, -1 };
//Counterclockwise from the 6o'clock direction
//Take the x-axis downward and the y-axis right

//#define int long long

//g++ -std=c++14 -Wall --pedantic-errors -fsanitize=signed-integer-overflow -fno-sanitize-recover=all -o template template.cpp


signed main()
{
	int n; cin >> n;
	string s; cin >> s;
	int cnt1{};
	int cnt2{};
	for(int i = 0; i < s.size(); ++i){
		if(s[i] == '#'){
			if(s[i+1] == '.'){
				s[i+1] = '#';
				++cnt1;
			}
		}
	}

	for(int i = s.size()-1; i <= 0; --i){
		if(s[i] == '.'){
			if(s[i-1] == '#'){
				s[i-1] = '#';
				++cnt2;
			}
		}
	}

	int cnt = min(cnt1, cnt2);

	return 0;
}