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: Wrong Answer)

import java.util.ArrayDeque;
import java.util.Deque;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		String s = sc.next();
		int sum = 0;
		int ans = 0;
		int whiteZantei = 0;
		int redZantei = 0;
		Deque<Integer> white = new ArrayDeque<Integer>();
		Deque<Integer> red = new ArrayDeque<Integer>();
		boolean last = true; //true white  false red
		for(int i = 0 ; i < n ; i++) {
			if(s.charAt(i) == 'W') {
				sum++;
				whiteZantei++;
				last = true;
				if(redZantei != 0) {
					red.push(redZantei);
					redZantei = 0;
				}
			}
			else {
				last = false;
				redZantei++;
				if(whiteZantei != 0) {
					white.push(whiteZantei);
					whiteZantei = 0;
				}
				}
		}
		if(redZantei != 0) {
			red.push(redZantei);
		}
		else
			white.push(whiteZantei);
		int sum2 = 0;

		while(!white.isEmpty() && !red.isEmpty()) {
			if(last) {
				sum -= white.pop();


				last = !last;
			}
			else {
				int i = red.pop();
				sum2 += i;
				if(sum2 >= sum)
					break;
              last = !last;
			}
		}

		sc.close();
		System.out.println(sum);

	}
}