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 Rust (1.42.0) (judge verdict: Wrong Answer)

// -*- coding:utf-8-unix -*-

use proconio::input;
use proconio::marker::Chars;

fn main() {
    input! {
        n: u32,
        mut c: Chars,
    }

    // let mut ans = c.iter().filter(|&a| *a == 'W').count();
    // c.reverse();
    let mut ans = 0;
    for i in (0..n).rev() {
        if c[i as usize] == 'R' {
            for j in (0..i).rev() {
                if c[j as usize] == 'W' {
                    ans += 1;
                }
            }
            break;
        }
    }
    let mut num_w_before_r = c.len();
    for i in 0..n {
        if c[i as usize] == 'W' {
            num_w_before_r += 1;
        } else {
            break;
        }
    }
    if ans >= num_w_before_r {
        ans -= 1;
    }

    // // 尻から見る
    // // 寄せられるだけ右に寄せて寄せきれなかったものは置換
    // let mut ans = 0;
    // let mut final_idx = c.len() as u32;
    // c.reverse();
    // // println!("{:?}", c);
    // for i in 0..n - 2 {
    //     if i >= final_idx {
    //         break;
    //     }

    //     if c[i as usize] == 'R' {
    //         for j in (i + 1..n).rev() {
    //             if c[j as usize] == 'W' {
    //                 c[i as usize] = 'W';
    //                 c[j as usize] = 'R';
    //                 ans += 1;
    //                 for k in (i + 1..n).rev() {
    //                     if c[k as usize] == 'R' {
    //                         final_idx = k;
    //                     } else {
    //                         break;
    //                     }
    //                 }
    //                 break;
    //             }
    //         }

    //         if c[i as usize] == 'R' {
    //             c[i as usize] = 'W';
    //             ans += 1;
    //         }

    //         // println!("{:?}", c);
    //     }
    // }
    println!("{}", ans);
}

// reversed (右端の r は無視できる)
// rrwrwwrw
// wrwrwwrr
// wwwrwrrr
// wwwwrrrr

// WRRRRRW
// RRRRRRW