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.*;
import java.io.*;
class Main{
  public static void main(String args[]){
    int n=readI();
    int countR=0;
    char ch[]=new char[n];
    for(int i=0;i<n;i++){
      ch[i] = readC();
      if(ch[i]=='R'){countR++;}
    }
    int ans=0;
    for(int i=0;i<countR;i++){
      if(ch[i]=='R'){
        ans++;
      }
    }
    ans = countR-ans;
    pl(ans+"");
  }

  static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  public static void pr(String str){
    System.out.print(str);
  }
  public static void pl(String str){
    System.out.println(str);
  }
  public static String read(){
    try{
      return ctos((char)br.read());
    }catch(IOException e){
      e.printStackTrace();
      return "";
    }
  }
  public static char readC(){
    try{
      return (char)br.read();
    }catch(IOException e){
      e.printStackTrace();
      return (char)-1;
    }
  }
  public static String readL(){
    try{
      return br.readLine();
    }catch(IOException e){
      e.printStackTrace();
      return "";
    }
  }
  public static String readS(){
    StringBuilder sb = new StringBuilder();
    while(true){
      try{
        int k = br.read();
        if(k==-1||(char)k==' '||(char)k=='\n'){break;}
        sb.append((char)k);
      }catch(IOException e){
        e.printStackTrace();
      }
    }
    return sb.toString();
  }
  public static int readI(){
    return stoi(readS());
  }
  public static long readLong(){
    return stol(readS());
  }
  public static long stol(String s){
    return Long.parseLong(s);
  }
  public static String[] readSs(){
    return readL().split(" ");
  }
  public static int[] readIs(){
    return stoi(readSs());
  }
  public static int stoi(String s){
    return Integer.parseInt(s);
  }
  public static int[] stoi(String s[]){
    int a[]=new int[s.length];
    for(int i=0;i<s.length;i++){
      a[i]=stoi(s[i]);
    }
    return a;
  }
  public static String itos(int i){
    return String.valueOf(i);
  }
  public static String[] itos(int[] a){
    String s[]=new String[a.length];
    for(int i=0;i<a.length;i++){
      s[i]=itos(a[i]);
    }
    return s;
  }
  public static String ctos(char c){
    return String.valueOf(c);
  }
  public static String cstos(char[] c){
    return new String(c);
  }
  public static char stoc(String s){
    return s.charAt(0);
  }
  public static char[] stocs(String s){
    return s.toCharArray();
  }
}