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 <bits/stdc++.h>
using namespace std;
#define MOD 1000000007
//#define N 1000005
typedef long long int ll;
//const int mx=1e5+7;
//int n,m;
//vector<int> a(mx),b(mx);
// vector<int> g[mx];
// int deg[mx];
//vector<int> adj[mx];
//vector<int> ans;
//---dfs--
// bool visited[mx];
// vector<ll> val(mx,0);

// void dfs(int s,int e,int c1,int c2){
// 	if(a[s]^c1!=b[s]) {ans.push_back(s), c1^=1;}
// 	for(auto u:adj[s]){
// 		if(u!=e){
// 			dfs(u,s,c2,c1);
// 		}
// 	}
// }
//---dfs---

//---primes---
// const int N = 1e7 + 10;

// bool prime[N];
 
// int spf[N];
 
// void calc(){
// 	for(int i = 2; i<1e4; i++){
// 		if(prime[i])
// 			continue;
// 		spf[i] = i;
// 		for(int j = i * i; j < N; j += i){
// 			prime[j] = 1;
// 			spf[j] = i;
// 		}
// 	}
// 	for(int i = 2; i < N; i++)
// 		if(!spf[i])
// 			spf[i] = i;
// }

// vector<pair<int,int>> getp(int x){
// 	int curr=x;
// 	vector<pair<int,int>> ans;
// 	while(curr>1){
// 		int a=spf[curr];
// 		int count=0;
// 		while(curr%a==0){ curr/=a; count++;}
// 		ans.push_back(make_pair(a,count));
// 	}
// 	return ans;
// }

// bool ispr(int n) {
// 	if(n==1) return false;
//     for (int i=2;i*i<=n;++i){
//         if(n%i==0){
//             return false;
//         }
//     }
//     return true;
// }
//---primes---

//---segment_tree---
// int upd(int t,int val,int l,int r,int y, int z){
// 	if(y<l||y>r) return tr[val];
// 	if(l==r) return tr[val]=z;
// 	int s=(l+r)/2;
// 	upd(t^1,2*val+1,l,s,y,z);
// 	upd(t^1,2*val+2,s+1,r,y,z);
// 	if(t) tr[val]=tr[2*val+1]|tr[2*val+2];
// 	else tr[val]=tr[2*val+1]^tr[2*val+2];
// 	return tr[val];
// }
//int query(){}
//---segment_tree---

int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);cout<<setprecision(25);
	int t=1;
	//cin>>t;
	while(t--){
		int n;
		cin>>n;
		string s;
		cin>>s;
		int cnt=0;
		for(int i=0;i<n;i++){
			if(s[i]=='W') cnt++;
		}
		int ans=0;
		if(cnt>0){
			for(int i=0;i<n-cnt;i++){
				if(s[i]=='W') ans++;
			}
		}
		cout<<ans<<"\n";
 	}
}