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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
using static System.Console;

namespace AtCoder_1
{
    class Program
    {
        static void Main(string[] args)
        {
            var n = CinT();
            var s = Cin();
            int white = 0;
            int black = 0;
            int ubBlk = 0;
            bool blacked = false;
            for(int i = n - 1; i > -1; i--)
            {
                if(s[i] == '#')
                {
                    white++;
                    ubBlk = 0;
                }
                if(s[i] == '.')
                {
                    black++;
                    ubBlk++;
                }
                if(i == 0)
                {
                    black -= ubBlk;
                }
            }
            Cout(Math.Min(black, white));


            Cin();
        }

        static int loop(int num, int max)
        {
            if (num > max)
            {
                return 0;
            }
            else if (num < 0)
            {
                return max;
            }
            else
            {
                return num;
            }
        }
        static int GCD(int a, int b)
        {
            if (a < b) { swap(ref a, ref b); }
            if (a % b == 0) { return b; }
            else { return GCD(b, a % b); }
        }

        static void swap<T>(ref T a, ref T b)
        {
            T t = a;
            a = b;
            b = t;
        }

        static readonly int MOD10E9 = (1000000000 + 7);
        static bool isInt(double d) { return (int)d == d; }
        static string Cin() { return ReadLine(); }
        static string[] CinAr() { return ReadLine().Split(' '); }
        static int CinT() { return S2i(Cin()); }
        static int[] CinTAr() { return CinAr().Select((x, i) => { return S2i(x); }).ToArray(); }
        static long CinL() { return S2L(Cin()); }
        static long[] CinLAr() { return CinAr().Select((x, i) => { return S2L(x); }).ToArray(); }
        static double CinD() { return S2D(Cin()); }
        static double[] CinDAr() { return CinAr().Select((x, i) => { return S2D(x); }).ToArray(); }
        static int S2i(string s) { return int.Parse(s); }
        static long S2L(string s) { return long.Parse(s); }
        static double S2D(string s) { return double.Parse(s); }
        static void Cout(object s) { WriteLine(s); }
        static int[] yakusuu(int n)
        {
            if (n == 1) { return new int[] { 1 }; }
            HashSet<int> res = new HashSet<int>();
            for (int i = 1; i < Math.Ceiling(n / 2.0); i++)
            {
                if (n % i == 0)
                {
                    res.Add(i);
                    res.Add(n / i);
                }
            }
            return res.ToArray();
        }
        static void CoutAr<Type>(IEnumerable<Type> types)
        {
            string text = string.Join(" ", types);
            WriteLine(text);
        }

        public static long nCr(int n, int r)
        {
            // naive: return Factorial(n) / (Factorial(r) * Factorial(n - r));
            return nPr(n, r) / Factorial(r);
        }

        public static long nPr(int n, int r)
        {
            // naive: return Factorial(n) / Factorial(n - r);
            return FactorialDivision(n, n - r);
        }

        private static long FactorialDivision(int topFactorial, int divisorFactorial)
        {
            long result = 1;
            for (int i = topFactorial; i > divisorFactorial; i--)
                result *= i;
            return result;
        }

        private static long Factorial(int i)
        {
            if (i <= 1)
                return 1;
            return i * Factorial(i - 1);
        }
    }
}