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.Scanner;

public class Main {
	static int check(boolean[] c, int index) {
		for (int i = index; i < c.length; i++) {
			if (c[i]) {
				return i;
			}
		}
		
		return -1;
	}
	
	static int check2(boolean[] c, int index) {
		for (int i = c.length - 1; i > index; i--) {
			if (!c[i] && c[i - 1]) {
				return i;
			}
		}
		
		return -1;
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		boolean[] c = new boolean[n];
		String str = sc.next();
		for (int i = 0; i < n; i++) {
			if (str.charAt(i) == 'W') {
				c[i] = true;
			}
		}
		
		int index = 0;
		int index2 = 0;
		
		int count = 0;
		
		while ((index = check2(c, index2)) > (index2 = check(c, index2)) && index2 >= 0) {
			c[index] = true;
			c[index2] = false;
			++count;
		}
		
		for (int i = c.length - 1; i > 0; i--) {
			if (!c[i] && c[i - 1]) {
				++count;
				c[i - 1] = false;
			}
		}
		System.out.println(count);
	}
}