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 <vector>
#include <stack>
#include <unordered_set>
#include <numeric>
#include <iterator>
#include <unordered_map>
#include <array>
#include <map>
#include <cstring>
#include <iomanip>
#include <string>
#include <algorithm>
#include <cmath>
#include <ctgmath>
#include <set>
#include <queue>
#include <fstream>
#include <ostream>
#include <random>
#include <chrono>
using namespace std;

signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	int n;
	string s;
	cin >> n >> s;

	string t = s;
	vector<pair<int,char>> comps;

	int curr = 1;
	for (int i = 1; i < n; i++)
	{
		if (s[i] != s[i - 1])
		{
			comps.push_back({ curr, s[i-1]});
			curr = 0;
		}
		curr++;
	}
	comps.push_back({ curr, s[n - 1]});

	int cnt = 0;

	for (int i = 0; i < comps.size() - 1; i++)
	{
		if (comps[i].second == '#' && comps[i + 1].second == '.')
		{
			//option 1
			int opt1, opt2;
			opt1 = opt2 = 0;

			if (comps[i].first < comps[i + 1].first)
			{
				cnt += comps[i].first;
				comps[i].second = '#';
			}
			else
			{
				cnt += comps[i + 1].first;
				comps[i + 1].second = '.';
			}


		}



	}
	cout << cnt << "\n";
}