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.*;
import java.io.*;
import java.math.BigInteger;
 
public class Main {
	public static void main(String[] args) {
	Scanner s = new Scanner(System.in);
	int K = s.nextInt();
	String C = s.next();

	int leftWcount = 0;
	int leftRcount = 0;
	int rightRcount = 0;
	int Wcount = 0;
	for(int i = 0; i < C.length(); i++){
		if(C.charAt(i) == 'R') break;
		leftWcount++;
	}
	for(int i = leftWcount; i < C.length(); i++){
		if(C.charAt(i) == 'W') break;
		leftRcount++;
	}
	for(int i = C.length()-1; i >= 0; i--){
		if(C.charAt(i) == 'R'){
			rightRcount++;
		}
		if(C.charAt(i) == 'W'){
			Wcount++;
		}
	}

	if(Wcount < rightRcount){
		System.out.println(Wcount);
	}else{
		System.out.println(Math.max(leftWcount, Math.abs(rightRcount-leftRcount)));
	}

    
  }
}