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: Wrong Answer)

#include <bits/stdc++.h>
#define dbg(x) cerr << #x << " = " << x << "\n"
#define fast_io ios_base::sync_with_stdio(0); cin.tie(0); cout.tie()
using namespace std;
using ll = long long;

const int N = int(2e5) + 500;

int n, pf[N], ans;
char c[N];

int main() {
    fast_io;

    cin >> n;
    for (int i = 0; i < n; ++i)
        cin >> c[i];

    for (int i = 0; i < n; ++i)
        pf[i] = (i - 1 < 0 ? 0 : pf[i - 1]) + (c[i] == 'R');

    ans = n - pf[n - 1];
    for (int i = 0; i < n; ++i) {
        int len = i + 1;
        int white_left = len - pf[i];
        int red_right = pf[n - 1] - pf[i];
        int cur = 0;

        int o = min(white_left, red_right);
        white_left -= o;
        red_right -= o;
        cur += o;

        cur += white_left + red_right;

        ans = min(ans, cur);
    }

    cout << ans << "\n";
}