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: Accepted)

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

#define debug(x) cout <<"DEBUG"<<" "<< #x << ":" << x << '\n'

// n回繰り返し
#define rep(i,n) for(int i = 0; i < ((int)(n)); i++)   // 0-indexed昇順
#define rep1(i,n) for(int i = 1; i <= ((int)(n)); i++) // 1-indexed昇順
#define rrep(i,n) for(int i = ((int)(n)-1); i >= 0; i--)  // 0-indexed降順
#define rrep1(i,n) for(int i = ((int)(n)); i >= 1; i--)   // 1-indexed降順
#define all(x) (x).begin(),(x).end()

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vii;  // 2次元配列
typedef vector<ll> vll;
typedef vector<double> vd;
typedef pair<int, int> PII;

template<typename T>
using vec = vector<T>;
template<class T>
using maxheap = std::priority_queue<T>;
template<class T>
using minheap = std::priority_queue<T, std::vector<T>, std::greater<T>>;

template<class T,class U> inline bool chmax(T &a, const U &b){if(a<b){a=b;return 1;}return 0;}
template<class T,class U> inline bool chmin(T &a, const U &b){if(a>b){a=b;return 1;}return 0;}

const ld Pi = std::acos(-1.0L);
const ll infll = (1LL<<62)-1;
const int inf = (1<<30)-1;
const int mod = 1000000007;
// const int mod = 998244353;


/*FUNCs=================================================*/


/*MAIN==================================================*/
signed main(){
    int _StartTime_ = clock();
    cin.tie(nullptr);
    ios::sync_with_stdio(false);   //cin cout 高速化
    // cout << fixed << setprecision(15);

    int n; cin >> n;
    vi wh(n+1), bl(n+1);  // csum
    rep(i,n) {
        char ch; cin >> ch;
        if(ch=='#') bl[i+1] = 1;
        else wh[i+1] = 1;
        bl[i+1] += bl[i];
        wh[i+1] += wh[i];
    }
//    rep(i,n+1) cout << bl[i] << (i==n ? '\n' : ' ');
//    rep(i,n+1) cout << wh[i] << (i==n ? '\n' : ' ');

    int ans = inf;
    for(int bndry = 1; bndry < n; ++bndry) {
        chmin(ans, bl[bndry]-bl[0] + wh[n]-wh[bndry]);
    }
    chmin(ans, *--wh.end());
    chmin(ans, *--bl.end());
    cout << ans << '\n';

//    debug(cnt1);debug(cnt2);
    // printf("ExecutionTime: %d /ms\n",1000*(int)((clock()-_StartTime_)/CLOCKS_PER_SEC));
}