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 <iostream>
#include <sstream>
#include <vector>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <utility>
#include <map>
#include <list>
#include <climits>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <bitset>
#include <string>

// #define cerr if(false)cerr
#define watch(x) cerr << "> " << #x << ": " << x << "\n";

using namespace std;

template <typename T>
std::ostream &operator <<(std::ostream &out, vector<T> &v) {
	for (typename vector<T>::size_type i = 0; i < v.size(); ++i)
		out << v[i] << " ";
	out << "\n";
    return out;
}
template <typename T, typename N>
std::ostream &operator <<(std::ostream &out, vector<pair<T, N> > &v) {
	for (size_t i = 0; i < v.size(); ++i)
		out << "(" << v[i].first << ", " << v[i].second << ") ";
	out << "\n";
    return out;
}
template <typename T>
std::ostream &operator <<(std::ostream &out, vector<vector<T> > &v) {
	for (size_t i = 0; i < v.size(); ++i) {
		for (size_t j = 0; j < v[i].size(); ++j) {
			out << v[i][j] << " ";
		}
		out << "\n";
	}
   	return out;
}

int main() {

	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n;
	cin >> n;

	string stones;
	cin >> stones;

	int red = 0, white = 0;
	for (int i = 0; i < n; ++i) {
		if (stones[i] == 'R') {
			++red;
		} else {
			++white;
		}
	}

	if (!white || !red) {
		cout << "0\n";
		return 0;
	}

	int red_left = red, white_left = white;
	int ans = min(red, white);
	int i = 0, j = n - 1;
	int swaps = 0;
	while (i < j) {
		if (stones[i] == 'W' && stones[j] == 'R') {
			--red_left;
			--white_left;
			++i;
			while (i < j && stones[i] == 'R') {
				++i;
				--red_left;
			}

			--j;
			while (j > i && stones[j] == 'W') {
				--j;
				--white_left;
			}

			++swaps;
			ans = min(ans, min(swaps + red_left, swaps + white_left));
		} 
		else if (stones[i] == 'R') {
			++i;
			--red_left;
		}
		else {
			--j;
			--white_left;
		}
	}

	cout << ans << "\n";

	return 0;

}