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)

use proconio::input;
use std::cmp::min;

fn main() {
    input! {
        _n: usize,
        cs: String
    }

    let mut cs: Vec<usize> = cs.chars().map(|c| if c == 'R' { 1 } else { 0 }).collect();

    if cs.iter().all(|&c| c == cs[0]) {
        print!("0");
        return;
    }

    let mut total = 0;

    let mut w_idx = 0;
    loop {
        if cs[w_idx] == 0 {
            cs[w_idx] = 1;
            w_idx += 1;
        } else {
            break;
        }
    }
    total += w_idx;

    let mut r_idx = 0;
    let mut w_found = false;
    for i in w_idx..cs.len() {
        if w_idx == 0 {
            break;
        }
        if w_found && cs[i] == 1 {
            cs[i] = 0;
            w_idx -= 1;
        }
        if !w_found && cs[i] == 0 {
            r_idx = i - 1;
            w_found = true;
        }
    }
    let r_sum: usize = cs[r_idx + 1..].iter().sum();
    total += min(r_sum, cs.len() - 1 - r_sum);
    println!("{}", total);
}