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 min(int a,int b){
	if(a<b)
	return a;
	else
	return b;
}
int max(int a,int b){
	if(a>b)
	return a;
	else
	return b;
}
int main(){
	int n;
	scanf("%d",&n);
	char ar[n];
	for(int i=0;i<n;i++){
		scanf("%c",&ar[i]);
	}
	int cr1=0,cr2=0,cw1=0,cw2=0,min1,min2,ans;
	if(n%2==0){
		int mid=n/2;
		for(int i=0;i<mid;i++){
			if(ar[i]=='R'){
				cr1++;
			}
			else{
				cw1++;
			}
		}
		for(int i=mid;i<n;i++){
			if(ar[i]=='R'){
				cr2++;
			}
			else{
				cw2++;
			}
		}

		int min1=min(cw1,cr2);
		int max1=max(cw1,cr2);
		//printf("%d%d%d%d",cr1,cr2,cw1,cw2);
		ans=min1+(max1-min1);
	}
	if(n%2!=0){
		int mid=n/2;
		for(int i=0;i<mid;i++){
			if(ar[i]=='R'){
				cr1+=1;
			}
			else{
				cw1+=1;
			}
		}
		for(int i=mid+1;i<n;i++){
			if(ar[i]=='R'){
				cr2+=1;
			}
			else{
				cw2+=1;
			}
		}
		int min1=min(cw1,cr2);
		int max1=max(cw1,cr2);
		ans=min1+(max1-min1);
	}
	printf("%d",ans);
}