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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int k = sc.nextInt();
		sc.nextLine();

		String str = sc.nextLine();
		char[] charr = str.toCharArray();

		ArrayList<Integer> rList = new ArrayList<Integer>();
		ArrayList<Integer> wList = new ArrayList<Integer>();

		// 左側からWがあったら
		// 右側のRと交換

		// Rの番地リストと、Wの番地リストをつくる
		for (int i = 0; i < charr.length; i++) {

			if (charr[i] == (char) 'R') {
				rList.add(i);
			} else {
				wList.add(i);
			}

		}

		// wListを左から、rListを右からたどる
		Collections.reverse(rList);
		int count = 0;

		int listmax;

		if (rList.size() > wList.size()) {
			listmax = wList.size();
		} else {
			listmax = rList.size();
		}

		for (int j = 0; j < listmax; j++) {
			if (rList.get(j) < wList.get(j)) {
				break;
			}
			count++;
		}

		System.out.println(count);
	}

}