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)

/***
 ** "Everything before the word 'but' is horseshit."
 **     -- Jon snow
***/

/**
  *  Author: dhruv_gheewala
  *  Problem: __________
  *  Date:   02.08.2020
  *  Time:   17:58:33
**/

#if __has_include("debug.h")
    #include "debug.h"
#else
    #include<bits/stdc++.h>
    using namespace std;
    
    #define fundri 108
    #define debug(...) 1729
#endif

#define int int64_t
typedef pair<int, int> pii;
typedef vector<int> vi;

const bool testcases = false;
#define ff first
#define ss second
#define endl '\n'
#define all(s) s.begin(), s.end()

// if vector isn't soted then it will remove consecutive same elements
#define make_unique(vec) vec.erase(unique(all(vec)), vec.end())
#define test int TC = 1;if(testcases)cin >> TC;for(int tc = 1 ; tc <= TC ; tc++)solve(), fundri
#define sort_by(cont,expr) sort(all(cont),[&](const auto &lhs,const auto &rhs){return expr;})

int seed; mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int rnd(int l = 0, int r = INT_MAX) {return uniform_int_distribution<int>(l, r)(rng);}

inline bool in_range(int a, int l, int r) {return (a >= l && a <= r);}
template<typename H, typename ...T>void inp(H &head) {cin >> head;}
template<typename H, typename ...T>void inp(H &head, T &...tail) {cin >> head;inp(tail...);}
template<typename T>bool isPrime(T n) {for(T i = 2; i * i <= n; i++)if(n % i == 0)return false; return n >= 2;}
template<typename T>inline istream &operator >>(istream &in, vector<T> &a) {for(T &x : a)in >> x; return in;}
template<typename T, typename U>inline istream &operator >>(istream &in, pair<T, U> &a) {in >> a.ff >> a.ss; return in;}

void solve();
void fast_io();

int32_t main(int32_t argc, char **argv)
{
    fast_io();
    test;
    return 0;
}

void solve()
{
    int n;
    cin >> n;
    vector<char> a(n);
    cin >> a;

    int b[2] = {0, 0};
    for(int i = 0 ; i < n ; i++) {
        if(a[i] == 'R')
            b[0]++;
        else
            b[1]++;
    }

    int ans = min(b[0], b[1]);
    int m = (n - 1) / 2;
    int cnt = 0;
    for(int i = 0 ; i <= m ; i++)
        cnt += (a[i] == 'W');
    ans = min(ans, cnt);
    cnt = 0;
    for(int i = m ; i < n ; i++)
        cnt += (a[i] == 'R');
    ans = min(ans, cnt);

    cout << ans << endl;
}

void fast_io()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

#ifdef DHRUV_GHEEWALA
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    freopen("debug.txt", "w", stderr);
#endif
}