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

#[fastout]
fn main() {
    input! {
        n: usize,
        cx: Chars,
    }

    let all_w = cx.iter().filter(|&&c| c == 'W').count();
    let all_r = cx.iter().filter(|&&c| c == 'R').count();

    if all_r == n || all_w == n {
        println!(0);
        return;
    }
    let mut w_fixed = false;
    let mut w_count = 0;
    let mut r_fixed = false;
    let mut r_count = 0;
    for &c in &cx {
        if !w_fixed {
            if c == 'W' {
                w_count += 1;
            } else {
                w_fixed = true;
            }
        }
        if !r_fixed {
            if c == 'R' {
                r_count += 1;
            } else if w_fixed {
                r_fixed = true;
            }
        }
        if w_fixed && r_fixed {
            break;
        }
    }
    let left_ans = w_count.max(all_r - r_count);
    let mut w_fixed = false;
    let mut w_count = 0;
    let mut r_fixed = false;
    let mut r_count = 0;
    for &c in cx.iter().rev() {
        if !r_fixed {
            if c == 'R' {
                r_count += 1;
            } else {
                r_fixed = true;
            }
        }
        if !w_fixed {
            if c == 'W' {
                w_count += 1;
            } else if r_fixed {
                w_fixed = true;
            }
        }
        if r_fixed && w_fixed {
            break;
        }
    }
    let right_ans = r_count.max(all_w - w_count);

    println!("{}", left_ans.min(right_ans));
}