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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

int min(int a,int b){
  if(a>b) return b;
  return a;
}

int main(void){
  int n,i,tmp,ans=100000;
  scanf("%d",&n);
  
  char s[n];
  int a[n],b[n];
  for(i=0;i<n;i++){
    a[i]=0;
    b[i]=0;
  }
  
  scanf("%s",s);
  
  if(s[0]=='#') a[0]=1;
  if(s[n-1]=='.') b[n-1]=1;
  
  for(i=1;i<n;i++){
    if(s[i]=='#') a[i]=a[i-1]+1;
    else a[i]=a[i-1];
    if(s[n-1-i]=='.') b[n-1-i]=b[n-i]+1;
    else b[n-1-i]=b[n-i];
  }
  
  for(i=0;i<n;i++) ans=min(ans,a[i]+b[i]-1);
  
  printf("%d\n",ans);
  
  return 0;
}