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: Accepted)

#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use itertools::Itertools;
use text_io::*;

const U_INF: usize = 1 << 60;
const I_INF: isize = 1 << 60;

fn main() {
    let n: usize = read!();
    let s: String = read!();

    let s = s.chars().collect_vec();

    let red = s.iter().filter(|c| **c == 'R').count();
    let white = s.iter().filter(|c| **c == 'W').count();

    let mut expected = String::new();

    for _ in 0..red {
        expected.push('R');
    }
    for _ in 0..white {
        expected.push('W');
    }

    let diff = s
        .iter()
        .zip(expected.chars())
        .filter(|(a, b)| **a != *b)
        .count();

    println!("{}", [red, white, diff / 2].iter().min().unwrap());
}