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.math.BigDecimal;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		sc.hasNextLine();
		String[] stones = sc.next().split("");
		int redscnt = 0;
		int whitecnt = 0;
		for (int i = 0; i < n; i++) {
			if (stones[i].equals("R")) {
				redscnt++;
			} else {
				whitecnt++;
			}
		}
		
		int min = 0;
		int whiteleft = 0;				// 左辺の白（移動対象）
		int rednow = redscnt;		// 右辺の赤（移動対象）
		for (int i = 0; i <= n; i++) {
			// i = 仕切り位置
			if ( i != 0) {
				if (stones[i-1].equals("W")) {
					// 仕切りの左辺の「白」
					whiteleft++;
				} else {
					// 右辺の「赤」
					rednow--;
				}
			}
			// 多い方の分の移動回数になる
			int max = Math.max(whiteleft, rednow);
			// 最小なら移動回数を更新
			if (min == 0 || max < min) {
				min = max;
			}
		}
		System.out.println(min);
		
	}
}