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 <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <chrono>
#include <set>

using namespace std;
using namespace chrono;
 
typedef long long ll;
typedef long double ld;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;

//#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define loop(x,n) for(int x = 0; x < n; ++x)
#define FORd(i,a,b) for (int i = (b)-1; i >= (a); i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto& a : x)

#define mp make_pair
#define pb push_back
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound

#define sz(x) (int)x.size()
#define beg(x) x.begin()
#define en(x) x.end()
#define all(x) beg(x), en(x)
#define resz resize

const int MOD = 1000000007; // 998244353
const ll INF = 1e18;
const int MX = 100001;

string get_str() {
    string temp;
    cin >> temp;
    return temp;
}
int get_int() {
    int temp;
    cin >> temp;
    return temp;
}

void solve() {

    int n; string s;
    cin >> n >> s;

    int r = 0;

    loop(i, n) {
        if(s[i] == 'R') {
            r++;
        }
    }   

    if(r == 0 || r == n) {
        cout << 0 << endl;
        return;
    }

    int p1 = 0, p2 = n-1, ans = 0;

    while(p1 < p2) {
        while(p1< p2 && s[p1] == 'R') {
            p1++;
        }
        while(p1 < p2 && s[p2] == 'W') {            
            p2--;
        }

        if(p1<p2) {
            ans++;
            p1++;
            p2--;
        }
    }
    cout << min(min(r, n-r), ans) << endl;

}

int main() {
    
    solve();

}