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>
#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
typedef long long ll;
typedef long double ld;
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define mod 1000000007
#define inf 1000000000000000000
#define bpc(x) __builtin_popcountll(x)
#define autoit(x,it) for(auto it = x.begin(); it != x.end(); it++)
#define rep(n) for(ll i = 0; i < n; i++)
#define repi(i,n) for(ll i = 0; i < n; i++)

#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 

#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update> 

using namespace std;

int main()
{
  FAST/**/
   
   ll n;
   cin>>n;
   
   string s;
   cin>>s;
   
   ll red = 0;
   rep(n)
        if(s[i] == 'R')
            red++;
   ll ans = min(red,n-red);
   ll r = 0;
   for(ll i=0;i<n-1;i++)
   {
       if(s[i] == 'R')
            r++;
       ll c1 = i+1-r;
       ll c2 = (red-r);
       ll mini = min(c1,c2);
       ans = min(ans, c1 + c2 - mini);
   }
   cout<<ans;
   
   return 0;        
}