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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {
    //Fast Reader
    private static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public void resetST() {
            st = null;
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public float nextFloat() {
            return Float.parseFloat(next());
        }

        public String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            st = null;
            return str;
        }

        public int[] nextIntArray(int size) {
            int[] arr = new int[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }

        public long[] nextLongArray(int size) {
            long[] arr = new long[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }

        public double[] nextDoubleArray(int size) {
            double[] arr = new double[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextDouble();
            }
            return arr;
        }
    }

    public static void main(String[] args) {
        FastReader sc = new FastReader();
        PrintWriter pr = new PrintWriter(System.out);
        int n = sc.nextInt();
        String s = sc.nextLine();
        int count = n - 1;
        int ans = 0;
        char[] c = new char[n];
        for (int i = 0; i < n; i++) {
            c[i] = s.charAt(i);
        }
        for (int i = 0; i <= count; i++) {
            if (c[i] == 'W') {
                pr.println(i);
                for (int j = count; j >= i; j--) {
                    pr.println(j+" "+i);
                    if (c[j] == 'R') {
                        c[j] = 'W';
                        c[i] = 'R';
                        ans++;
                        count--;
                        break;
                    }

                }
            }
        }
        pr.println(ans);
        pr.close();
        // write your code here
    }
}