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>
#define pb push_back
#define SZ(x) ((int)(x.size()))
#define FOR(i,s,n) for (ll i = (s); (i) < (n); ++i)
#define FORD(i,s,l) for (ll i = (s); (i) >= l; --i)
#define F first
#define S second
#define TC int __tc; cin >> __tc; FOR(case_num,1,__tc+1)
#define TEST(x,i) ((x)&(1ll<<(i)))
#define SET(x,i) ((x)|(1ll<<(i)))
#define FLIP(x,i) ((x)^(1ll<<(i)))
#define CLEAR(x,i) ((x)&~(1ll<<(i)))

const double pi = 4 * atan(1);
using namespace std;
typedef long long ll;

const ll MOD = 1000000007;
const int MAXN = 200005;

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	int n;
	string s;
	cin >> n >> s;
	
	map<char,int> cnt;
	for(auto v : s) {
		cnt[v]++;
	}
	
	if (!cnt['R'] || !cnt['W']) {
		cout << "0\n";
		return 0;
	}
	
	map<char,int> l;
	for (int i = 0; i < cnt['R']; i++){
		l[s[i]]++;
	}
	
	map<char,int> r;
	r['R'] = cnt['R'] - l['R'];
	r['W'] = cnt['W'] - l['W'];
	
	int ans = r['R']+l['L'] - min(r['R'],l['L'] );
	cout << ans << "\n";

}