Problem:
Score : 400 points

Problem Statement

An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; R stands for red and W stands for white.

You can do the following two kinds of operations any number of times in any order:

Choose two stones (not necessarily adjacent) and swap them.

Choose one stone and change its color (from red to white and vice versa).

According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?

Constraints

2 \leq N \leq 200000

c_i is R or W.

Input

Input is given from Standard Input in the following format:

N
c_{1}c_{2}...c_{N}

Output

Print an integer representing the minimum number of operations needed.

Sample Input 1

4
WWRR

Sample Output 1

2

For example, the two operations below will achieve the objective.

Swap the 1-st and 3-rd stones from the left, resulting in RWWR.

Change the color of the 4-th stone from the left, resulting in RWWW.

Sample Input 2

2
RR

Sample Output 2

0

It can be the case that no operation is needed.

Sample Input 3

8
WRWWRWRR

Sample Output 3

3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <cstdio>
#include <vector>
#include <stdio.h>
#include <algorithm>
#include <functional>
#include <queue>
#include <math.h>
#include <cmath>
#include <unordered_map>

#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;

void solve(int n,vector<int> Rlist,vector<int> Wlist){
    int redcount=Rlist.size();
    int Whitecount=Wlist.size();
    int count=0;  
    while(true){
        if(Rlist.empty()||Wlist.empty()){
            break;
        }else{
            int redlast=Rlist[Rlist.size()-1];
            int whitefarst=Wlist[0];
            if(redlast<whitefarst)break;
            count++;
            Rlist.erase(Rlist.end()-1);
            Wlist.erase(Wlist.begin());
        }
        
    }
    cout<<min({redcount,Whitecount,count})<<endl;
}

int main(){
    int n;
    cin >> n;
    vector<int> Rlist(0);
    vector<int> Wlist(0);
    bool isWhite=false;
    char tmp;
    rep(i,n){
        cin>>tmp;
        if(tmp=='R'){
            Rlist.push_back(i);
        }else{
            Wlist.push_back(i);
        }
    }
    solve(n,Rlist,Wlist);


    return 0;
}