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;
typedef long long ll;
const double EPS = 1e-9;
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef pair<int, int> pi;
typedef pair<ll, ll> pll;
typedef map<int, int> mi;
typedef set<int> si;
#define VV(T) vector<vector< T > >
#define dump(x)  cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) << endl

#define rep(i, n) REP(i, 0, n)                              // 0, 1, ..., n-1
#define REP(i, x, n) for(int i = x; i < n; i++)             // x, x + 1, ..., n-1
#define FOREACH(x,a) for(auto& (x) : (a) )

#define ALL(v) (v).begin() , (v).end()
#define RALL(v) (v).rbegin(), (v).rend()

#define pb push_back
#define pu push
#define mp make_pair
#define fi first
#define sc second

#define COUT(x) cout << (x) << endl
#define VECCIN(x) for(auto&youso_: (x) )cin>>youso_
#define VECCOUT(x) for(auto&youso_: (x) )cout<<youso_<<" ";cout<<endl

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; }


int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N; cin >> N;
    string s; cin >> s;
    int ans = 0;
    int ans1 = 0;
    int ans2 = 0;
    for(int i = 0; i < N - 1; i++) {
        if(s[i] == '#' && s[i + 1] == '.') {
            s[i + 1] = '#';
            ans1++;
        }
    }
    for(int i = 0; i < N - 1; i++) {
        if(s[i] == '#' && s[i + 1] == '.') {
            s[i] = '.';
            ans2++;
        }
    }
    ans = min(ans1, ans2);
    if(ans1 == 0) ans = ans2;
    else if(ans2 == 0) ans = ans1;
    COUT(ans);
}