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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>

int max(int W,int R)
{
  return (W >= R) ? W : R;
}

int array_min(int array[],int e)
{
  int min = array[0];
  
  for(int i=1; i<e; i++)
    if(min > array[i])
      min = array[i];
  
  return min;
}

int main(void)
{
  int n;
  int left_w = 0,right_r=0;
  
  scanf("%d",&n);
  char c[n];
  int maxs[n+1];
  scanf("%s",c);
  
  for(int i=0; i<n; i++){
    if(c[i] == 'R')
      right_r++;
  }
  
  maxs[0] = max(left_w,right_r);
    
  for(int i=0; i<n; i++){
    if(c[i] == 'W')
      left_w++;
    else
      right_r--;
    
    maxs[i+1] = max(left_w,right_r);
  }
  
  printf("%d\n",array_min(maxs,n+1));
  
  return 0;
}