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;

typedef long long int ll;
typedef vector<int> vi;			
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef pair<int, int> 	pii;
typedef vector<bool> vb;
#define endl '\n'
#define all(x) x.begin(),x.end()
#define sz(x) (ll)x.size()
#define MOD 1000000007
#define F first
#define S second
#define pb push_back
#define pf push_front
#define rb pop_back
#define rf pop_front
#define in insert
#define mp make_pair
#ifdef LOCAL
	#include "../../trace.h"
#else
    #define trace(args...)
#endif

ll div_floor(const ll &a,const ll &b) { return a / b - (((a ^ b) < 0) and a % b);}
ll div_ceil(const ll &a,const ll &b) { return a / b + (((a ^ b) >= 0) and a % b);}

void solve(){
	int n; cin>>n;
	string s; cin>>s;
	int cntr = 0;
	for(auto &c: s) cntr += (c=='R');
	int pres = 0;
	for(int i = 0; i<cntr; i++) pres += (s[i]=='R');
	cout<<cntr-pres<<endl;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	#ifdef LOCAL
		freopen(".inp.txt", "r", stdin);
    	freopen(".out.txt", "w", stdout);
    	freopen(".err.txt", "w", stderr);
	#endif
	int t=1;
	while(t--)
		solve();
}