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 endl "\n"
#define all(x) x.begin(),x.end()
#define pb push_back
#define ff first 
#define ss second 
#define pii pair<int,int>
#define vi vector<int>
#define vc vector<char>
#define vii vector<pii>
#define vvi vector<vi>
#define fori(i,n) for(int i=1;i<=n;++i)
#define forn(i,n) for(int i=0;i<n;++i)
#define sz(x) int(x.size())
#define error(x) cout<<#x<<" "<<x<<endl
#define IOS  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)

const int N=200*1000;
const int mod=1e9+7;

int min3(int a,int b,int c){return min(a,min(b,c));}
int max3(int a,int b,int c){return max(a,max(b,c));}
int min4(int a,int b,int c,int d){return min(min(a,b),min(c,d));}
int max4(int a,int b,int c,int d){return max(max(a,b),max(c,d));}

//#define int int64_t
//int a[N+2];
int pref[N+2]={0};
int32_t  main(){
	IOS;
  int ans=0;
  int n;cin>>n;

  string s;cin>>s;
  if(n==1){cout<<"0\n";return 0;}
  memset(pref,0,sizeof(pref));
  if(s[0]=='.')pref[0]=1;
  for(int i=1;i<n;++i){
    if(s[i]=='.')pref[i]=1+pref[i-1];
    else pref[i]=pref[i-1];
  }
  ans=n-pref[n-1];//all black cost 
  for(int i=0;i<n;++i){
    //uptil ith pos white remaining black 
    ans=min(ans,i+1-pref[i]+(pref[n-1]-pref[i]));
  }
  cout<<ans<<endl;

  return 0;
}