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 <bits/stdc++.h>
using namespace std;
#define ll long long 
#define $ template <tyepename T>

int main(){
    ios::sync_with_stdio(false);
    int n;
    string st;
    cin >> n >> st;
    bool find=false;
    int res1=0, res2=0;
    int aux=0;
    int result=0;
    int result2=0;
    bool start=false;
    for(int i=n-1; i>=0; i--){
        if(st[i]=='#')
            res2++;
        else
            res1++;
        //cout << res1 << " " << res2 << endl;
        if(find && i!=0){
            if(st[i-1]=='.'){
                if(start){
                    result+=res2;
                    result2+=res1;
                }
                else{
                    result+=res1;
                    result2+=res2;
                    if(res1>=res2)
                        start=true;

                    
                }
                res1=0;
                res2=0;         
                find=false;
                //cout << "result1 " << result << " " << result2 <<endl;
            }
        }
        else{
            if(st[i-1]=='#')
                find=true;
        }
    }
    //cout << res1 << " " << res2 << endl;
    result+=res2;
    result2+=res1;
    //cout << "result1 " << result << " " << result2 <<endl;
    cout << min(result, result2) << endl;
}