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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef vector<int> VI;

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()





int main() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    vector<int> ctr_r(n), ctr_w(n);
    if (s[0] == 'W') {
        ctr_w[0] = 1;
    } else {
        ctr_w[0] = 0;
    }
    rep (i, 1, n) {
        if (s[i] == 'W') {
            ctr_w[i] = ctr_w[i-1] + 1;
        } else {
            ctr_w[i] = ctr_w[i-1];
        }
    }
    if (s[n-1] == 'R') {
        ctr_r[n-1] = 1;
    } else {
        ctr_r[n-1] = 0;
    }
    rep (i, 1, n) {
        if (s[n-1-i] == 'R') {
            ctr_r[n-1-i] = ctr_r[n-i] + 1;
        } else {
            ctr_r[n-1-i] = ctr_r[n-i];
        }
    }
    int ans = 1e9+7;
    rep (i, n) {
        if(ctr_w[i] == ctr_r[i]) {
            chmin(ans, ctr_w[i]);
        }
    }
    if (ans == 1e9+7) {
        cout << 0 << endl;
    } else {
        cout << ans << endl;
    }
}