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.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int rk = 0;
    int wk = 0;
    String s = sc.next();
    for(int i = 0; i < n; i++) {
      if(s.charAt(i) == 'W') {
        wk++;
      } else {
        rk++;
      }
    }
    int[] rui = new int[n];
    if(s.charAt(0) == 'W') rui[0] = 1;
    for(int i = 1; i < n; i++) {
      if(s.charAt(i) == 'W') {
        rui[i] = rui[i - 1] + 1;
      } else {
        rui[i] = rui[i - 1];
      }
    }
    int ans = n;
    for(int k = 0; k <= n; k++) {
      int t = Math.abs(rk - k);
      if(k > 0) {
        if(rk <= k) {
          t += (rui[k - 1] - Math.abs(rk - k));
        } else {
          t += rui[k - 1];
        }
      }
      ans = Math.min(ans, t);
    }
    System.out.println(ans);
  }
}