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: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringJoiner;
import java.util.StringTokenizer;
import java.util.function.IntUnaryOperator;
import java.util.function.LongUnaryOperator;

public class Main {
    static In in = new In();
    static Out out = new Out();
    static final long inf = 0x1fffffffffffffffL;
    static final int iinf = 0x3fffffff;
    static long mod = 1000000007;

    void solve() {
        int n = in.nextInt();
        char[] c = in.nextCharArray();
        int[] a = new int[n];
        int r = 0;
        int w = 0;
        for (int i = 0; i < n; i++) {
            a[i] = c[i] == 'R' ? 0 : 1;
            if (c[i] == 'R') {
                r++;
            } else {
                w++;
            }
        }
        int cr = 0;
        int cw = 0;
        for (int i = 0; i < n; i++) {
            if (c[i] == 'W') {
                if (i < r) {
                    cw++;
                }
            } else {
                if (r <= i) {
                    cr++;
                }
            }
        }
        out.println(Math.max(cw, cr));
    }

    public static void main(String[]$) {
        new Main().solve();
        out.flush();
    }
}

class In {
    private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in), 0x10000);
    private StringTokenizer tokenizer;

    String next() {
        try {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(reader.readLine());
            }
        } catch (IOException ignored) {
        }
        return tokenizer.nextToken();
    }

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

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

    char[] nextCharArray() {
        return next().toCharArray();
    }

    char[][] nextCharGrid(int n, int m) {
        char[][] a = new char[n][m];
        for (int i = 0; i < n; i++) {
            a[i] = next().toCharArray();
        }
        return a;
    }

    int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    int[] nextIntArray(int n, IntUnaryOperator op) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = op.applyAsInt(nextInt());
        }
        return a;
    }

    long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextLong();
        }
        return a;
    }

    long[] nextLongArray(int n, LongUnaryOperator op) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = op.applyAsLong(nextLong());
        }
        return a;
    }
    
    List<List<Integer>> nextEdges(int n, int m) {
        List<List<Integer>> res = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            res.add(new ArrayList<>());
        }
        for (int i = 0; i < m; i++) {
            int u = nextInt() - 1;
            int v = nextInt() - 1;
            res.get(u).add(v);
            res.get(v).add(u);
        }
        return res;
    }
}

class Out {
    private PrintWriter out = new PrintWriter(System.out);
    boolean autoFlush = false;

    void println(Object... a) {
        StringJoiner joiner = new StringJoiner(" ");
        for (Object obj : a) {
            joiner.add(String.valueOf(obj));
        }
        out.println(joiner);
        if (autoFlush) {
            out.flush();
        }
    }

    void println(char[] s) {
        out.println(String.valueOf(s));
        if (autoFlush) {
            out.flush();
        }
    }

    void println(int[] a) {
        StringJoiner joiner = new StringJoiner(" ");
        for (int i : a) {
            joiner.add(Integer.toString(i));
        }
        out.println(joiner);
        if (autoFlush) {
            out.flush();
        }
    }

    void println(long[] a) {
        StringJoiner joiner = new StringJoiner(" ");
        for (long i : a) {
            joiner.add(Long.toString(i));
        }
        out.println(joiner);
        if (autoFlush) {
            out.flush();
        }
    }

    void flush() {
        out.flush();
    }
}