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 int ll;
#define rep( i, n, m ) for(int i = ( n ); i < ( m ); i++)
#define rep_d( i, n, m ) for(int i = ( n ) - 1; i >= ( m ); i--)
#define sort_asc( X ) sort(( X ).begin(), ( X ).end())
#define sort_desc( X ) sort(( X ).begin(), ( X ).end(), greater <>())

template <class T> bool chmax( T& a, T b );
template <class T> bool chmin( T& a, T b );
int* eratosthenes( int N );
bool bit_search( int pattern, int N );
template <class T> void cumulative_sum( T array, size_t N );

const ll INF = 1LL << 60;
const int k_mod = 1e9 + 7;

template <typename T, typename K> struct P {
	T a; K b;
};
typedef struct {
	vector <int> to;
	bool is_visited = false;
} node;
typedef struct {
	int to;
	int cost;
	P <int, int> change;
} edge;

/*
 * @fn modpow
 * @brief O(logn): a^n % mod
 * @param a: 基数
 * @param n: 指数
 * @param mod: 法
 */
ll modpow( ll a, ll n, ll mod ) {
	ll res = 1;
	while (n > 0){
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}
int main( void ){
	int N; cin >> N;
	string S; cin >> S;
	int cnt[2] = {};
	for(auto s:S){
		cnt[0] += (s == '.');
		cnt[1] += !(s == '.');
	}
	cout << min(cnt[0], cnt[1]) << endl;
}

int* eratosthenes( int N ){
	int* prime_array = new int[ N + 1 ];
	int lim = N;
	prime_array[ 0 ] = 0;
	prime_array[ 1 ] = 0;
	rep( i, 2, N + 1 ){
		prime_array[ i ] = 1;
	}
	rep( i, 2, lim ){
		if (prime_array[ i ] == 0)
			continue;
		lim = N / i;
		for(int j = i * 2; j < N + 1; j += i){
			prime_array[ j ] = 0;
		}
	}
	return prime_array;
}
template <class T> void cumulative_sum( T array, size_t N ) {
	rep( i, 1, N + 1 ){
		array[ i ] += array[ i - 1 ];
	}
}
template <class T> bool chmax( T& a, T b ) {
	if (a < b){
		a = b; return true;
	}
	return false;
}
template <class T> bool chmin( T& a, T b ) {
	if (a > b){
		a = b; return true;
	}
	return false;
}
bool bit_search( int pattern, int N ){
	int cnt = 0;
	rep( bit, 0, N ){
		if (pattern & ( 1 << bit )){
			cnt++;
		}
	}
	return true;
}