M. Debug it!
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Miamiao writes a program and submits it to the judger. The judger tells her the result of each test case by a string containing P or F. P means to pass the test and F means to fail the test. Now Huihui wants to help her debug the program. He can remove the first or the last letter in
 
 1 minute or remove any other letter in 2 minutes. After removing a letter from the middle the two remaining parts are joined.

Now, Huihui wants to know, how fast can he finish debugging the problem, which means he can obtain a string without any letter F?

Input
The first line of input contains an integer n(
 
1≤n≤106), which means the length of the string.

The second line of input contains a string s(|s|=n,si∈{F,P}), which means Miamiao's result.

Output
Output a number, which means the minimum time needed to debug the problem by Huihui.

Example
inputCopy
4
FPPF
outputCopy
2

#include <bits/stdc++.h>
using
 
 namespace std;
#define ri register int
#define M 1000005

template <class G> void read(G &x)
{
    x=0;char ch=getchar();int f=0;
    while(ch<'0'||ch>'9'){f|=ch=='-';ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    x=f?-x:x;
    return ;
}

long long  a[M],b[M];
int p[M];

int n,m;
int main(){
    
    read(n);
    for(ri i=1;i<=n;i++)
    {
        char c;cin>>c;
        if(c=='P') p[i]=1;
        else p[i]=2;
    }
    
    for(long long  i=1;i<=n;i++)
    {
        if(p[i]==1) a[i]=a[i-1];
        else
        {
            a[i]=min(a[i-1]+2,i);
        }
    }
    for(long long  i=n;i>=1;i--)
    {
        if(p[i]==1) b[i]=b[i+1];
        else
        {
            b[i]=min(b[i+1]+2,n-i+1);
        }
    }
    long long ans=1e18;
    for(ri i=0;i<=n;i++)
    {
        ans=min(ans,a[i]+b[i+1]);
    }
     cout<<ans;
     return 0;
    
}