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: Wrong Answer)

#include <stdio.h>

int leftw(char s[200002], int i, int min)
{
    int j;
    for(j=i+1; j<min; j++)
    {
        if(s[j] == 'W')
            return j;
    }
    return j;
}

int rightr(char s[200002], int i, int min)
{
    int j;
    for(j=i-1; j>=0; j--)
    {
        if(s[j] == 'R')
            return j;
    }
    return j;
}

int main(void)
{
  	int n;
  	int i;
  	int j;
  	char s[200002];
  	int wcnt;
  	int rcnt;
  	int cnt;
  	int min;
  	int wtmp;
  	int rtmp;
  
  	scanf("%d", &n);
    scanf("%s", s);

  	wcnt = 0;
  	rcnt = 0;
  	for(i=0; i<n; i++)
    {
      	if(s[i] == 'W')
            wcnt++;
        if(s[i] == 'R')
            rcnt++;
    }
    
    if(wcnt == n)
    {
        printf("0");
        return (0);
    }
    if(rcnt == n)
    {
        printf("0");
        return (0);
    }
    
    if(wcnt < rcnt)
        min = wcnt;
    else
        min = rcnt;
    
    cnt = 0;
    i = 0;
    while(cnt < min)
    {
        if(s[i] == 'W')
            cnt++;
        i++;
    }
    wtmp = cnt;
    
    cnt = 0;
    i = n - 1;
    while(cnt < min)
    {
        if(s[i] == 'R')
            cnt++;
        i--;
    }
    rtmp = cnt;
    
    cnt = 0;
    i = -1;
    j = n;
    while(leftw(s,i,min) <= rightr(s,j,min))
    {
        i = leftw(s, i, min);
        j = rightr(s, j, min);
        cnt++;
    }
    
  	printf("%d", cnt);
  	return (0);
}