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>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
// Hey! 僕の提出を見てくれてありがとう.
// ロボ子さん可愛いぞ！！！！！君もV沼にはまろう↓
// https://www.youtube.com/channel/UCDqI2jOz0weumE8s7paEk6g
using namespace std;
namespace mp = boost::multiprecision;
// 任意長整数型
using bint = mp::cpp_int;
// 仮数部長が32の浮動小数点数型
using real32 = mp::number<mp::cpp_dec_float<32>>;
// 仮数部長が1024の浮動小数点数型
using real1024 = mp::number<mp::cpp_dec_float<1024>>;
// ついでに有理数型
using rat = boost::rational<bint>;
using ll = long long;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}//a,bの最大公約数(gcd)を求める
ll lcm(ll a,ll b){return abs(a*b)/gcd(a,b);}//a,bの最小公倍数(lcm)を求める
vector<ll> enum_div(ll n){vector<ll> ret;for(int i=1 ; i*i<=n ; i++){if(n%i == 0){ret.push_back(i);if(i!=1 && i*i!=n)ret.push_back(n/i);}}ret.push_back(n);return ret;}
//↑nの約数を求める
vector<bool> IsPrime; void sieve(size_t max){if(max+1 > IsPrime.size())IsPrime.resize(max+1,true);IsPrime[0] = false;IsPrime[1] = false;for(size_t i=2; i*i<=max; ++i)if(IsPrime[i])for(size_t j=2; i*j<=max; ++j)IsPrime[i*j] = false;}
//↑エラトステネスの篩で素数を求める
#define roundup(divisor,dividend) (divisor + (dividend - 1)) / dividend //切り上げ割り算
#define all(x) (x).begin(),(x).end() //xの初めから終わりまでのポインタ
#define size_t ll //size_tは自動でllに変換される
#define pb(x) push_back(x)
#define pri_queue priority_queue //優先度付きキュー
#define syo(x) fixed << setprecision(x) //iostreamで小数をx桁表示
//sortを降順にする時は greater<型>()


int main(){
    ll N;string c;
    cin >> N >> c;
    ll w=0,r=0;
    for (int i = 0; i < N; i++) {
      if(c[i] == 'R')r++;
      else w++;
    }if(w == N){cout << 0 << endl; return 0;}
    w = 0;
    ll ans = LLONG_MAX;
    for (int i = 0; i < N; i++) {
      if(c[i] == 'W')w++;
      else r--;
      ll x = min(w,r) + abs(r-w);
      ans = min(ans , x);
    }
    cout << ans << endl;
    return 0;
}