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;
#define pi acos(-1.0)
#define pb push_back
#define mp make_pair
#define ll long long
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define testcase ll T; cin >> T; for (ll tc = 1; tc <= T; tc++)
#define M 1000000007
#define eps 1e-8
#define eq(x,y) (fabs((x)-(y)) < eps)

int main()
{
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  //testcase {
  int n;
  cin>>n;
  string s;
  cin>>s;
  int c=0,cw=0,cr=0;
  vector<int>suff(n);
  for(int i=n-1;i>=0;i--) {
  	if(s[i]=='W') { cw++; suff[i]=cr; }
  	else { cr++; suff[i]=cr; }
  }
  int p,ans=min(cw,cr);
  for(int i=0;i<n-1;i++) {
  	if(s[i]=='W') { c++; p=max(c,suff[i+1]); ans=min(ans,p); }
  }
  cout<<ans;
  //} 
  return 0;
}