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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <map>
#include <iomanip>
#include <stdio.h>
#include <bitset>
#include <queue>

constexpr long long INFLL = 1e18;

using namespace std;
using ll = long long;
using P = pair<int, int>;

int main()
{

    int n;
    cin >> n;
    vector<int> s(n);
    char temp;
    // int checkW = 0;
    // int checkR = 0;
    for (int i = 0; i < n; i++)
    {
        cin >> temp;
        if (temp == 'R')
        {

            s.at(i) = 1;
        }
        else
        {
            s.at(i) = 0;
        }
    }
    int r = n - 1;
    bool che = false;
    int count = 0;
    int i;

    for (i = 0; i < n; i++)
    {
        if (s.at(i) == 0)
        {
            while (s.at(r) == 0)
            {
                if (i >= r)
                {
                    che = true;
                    break;
                }

                r--;
            }
            if (che)
            {
                break;
            }
            r--;
            count++;
        }
    }
    if (che == false && i == 0 && r == 0)
    {
        count = 0;
    }

    cout << count << endl;
    return 0;
}