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_parens)]
#![allow(unused_imports)]
#![allow(non_upper_case_globals)]
#![allow(non_snake_case)]
#![allow(unused_mut)]
#![allow(unused_variables)]
#![allow(dead_code)]

use itertools::Itertools;
use proconio::input;
use proconio::marker::{Chars, Usize1};

#[allow(unused_macros)]
#[cfg(debug_assertions)]
macro_rules! mydbg {
    //($arg:expr) => (dbg!($arg))
    //($arg:expr) => (println!("{:?}",$arg));
      ($($a:expr),*) => {
          eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
      }
}
#[cfg(not(debug_assertions))]
macro_rules! mydbg {
    ($($arg:expr),*) => {};
}

macro_rules! echo {
    ($($a:expr),*) => {
          $(println!("{}",$a))*
      }
}

use std::cmp::*;
use std::collections::*;
use std::ops::{Add, Div, Mul, Sub};

#[allow(dead_code)]
static INF_I64: i64 = 92233720368547758;
#[allow(dead_code)]
static INF_I32: i32 = 21474836;
#[allow(dead_code)]
static INF_USIZE: usize = 18446744073709551;
#[allow(dead_code)]
static M_O_D: usize = 1000000007;
#[allow(dead_code)]
static PAI: f64 = 3.1415926535897932;

trait IteratorExt: Iterator {
    fn toVec(self) -> Vec<Self::Item>;
}

impl<T: Iterator> IteratorExt for T {
    fn toVec(self) -> Vec<Self::Item> {
        self.collect()
    }
}

trait CharExt {
    fn toNum(&self) -> usize;
    fn toAlphabetIndex(&self) -> usize;
    fn toNumIndex(&self) -> usize;
}
impl CharExt for char {
    fn toNum(&self) -> usize {
        return *self as usize;
    }
    fn toAlphabetIndex(&self) -> usize {
        return self.toNum() - 'a' as usize;
    }
    fn toNumIndex(&self) -> usize {
        return self.toNum() - '0' as usize;
    }
}

trait VectorExt {
    fn joinToString(&self, s: &str) -> String;
}
impl<T: ToString> VectorExt for Vec<T> {
    fn joinToString(&self, s: &str) -> String {
        return self
            .iter()
            .map(|x| x.to_string())
            .collect::<Vec<_>>()
            .join(s);
    }
}

trait StringExt {
    fn get_reverse(&self) -> String;
}
impl StringExt for String {
    fn get_reverse(&self) -> String {
        self.chars().rev().collect::<String>()
    }
}

trait UsizeExt {
    fn pow(&self, n: usize) -> usize;
}
impl UsizeExt for usize {
    fn pow(&self, n: usize) -> usize {
        return ((*self as u64).pow(n as u32)) as usize;
    }
}

fn main() {
    use itertools_num::ItertoolsNum as _;
    input! {
       N: usize,
       mut s:Chars,
    }

    let mut ans: usize = 0;
    let mut c = 0_usize;
    let mut f = 0;
    let mut l = 0;
    let mut r = N - 1;
    let mut a = vec![];
    let mut b = vec![];
    for i in 0..N {
        if s[i] == 'W' {
            a.push(i);
        } else {
            b.push(i);
        }
    }
    b.reverse();
    mydbg!(a, b);
    for i in 0..min(a.len(), b.len()) {
        let x = a[i];
        let y = b[i];
        if x > y {
            break;
        }
        s[x] = 'R';
        s[y] = 'W';
        ans += 1;
    }
    mydbg!(s);
    let mut e = 0;
    for i in (0..N) {
        if s[i] == 'R' {
            e = i;
            break;
        }
    }
    for i in 0..e {
        if s[i] == 'W' {
            ans += 1;
        }
    }
    echo!(ans);
}