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 set<int> si;
typedef set<ll> sll;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define pvec(v) for(auto e: v) cout << e << " ";
#define IO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define rep(i, a, b) for(int i = (a); i<(b); ++i)
#define pb push_back
#define INF 1000000005
#define MOD 1000000007
#define all(v) v.begin(), v.end()

void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)

ll power(ll x, ll y, ll p = MOD){  
    ll res = 1LL; 
    x = x % p;
    if (x == 0) return 0;
    while (y > 0){
        if (y & 1LL)  
            res = (res*x) % p;
        y = y>>1LL; 
        x = (x*x) % p;  
    }  
    return res;  
}  

int main(){
	IO
	int n;
	string s;
	cin >> n >> s;

	int reds = 0;
	for(char c: s){
		if(c == 'R') ++reds;
	}

	int ans = 0;
	for(int i = 0; i<reds; ++i)
		if(s[i] == 'W') ++ans;
	cout << ans << endl;
	return 0;
}