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

#define ll long long 
#define FI first 
#define SE second
#define pb push_back 
#define eb emplace_back
#define mod 1000000007
#define all(c) (c).begin(),(c).end()
#define LB  lower_bound
#define UB  upper_bound
#define max3(a,b,c) max(c,max(a,b))
#define min3(a,b,c) min(c,min(a,b))
#define mems(s, n) memset(s, n, sizeof(s))
#define NINF -1e18
#define INF 1e18
#define int ll int
#define double long double
#define OOK order_of_key  //no of elements less than
#define FBO find_by_order //iterator pointing kth element;indexing starts from 0
#define CK3(x,y,z) cout<<(x)<<" "<<(y)<<" "<<(z)<<endl
#define CK4(a,b,c,d) cout<<(a)<<" "<<(b)<<" "<<(c)<<" "<<(d)<<endl

typedef pair<int,int> PII;
typedef pair<pair<int,int>,int> PPII;
typedef pair<int,pair<int,int>> PIPI;
typedef map<int,int>MII;
typedef vector<int> VI;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

/*int power(int a, int b)
{int x=1,y=a;while(b>0){if(b%2){x =(x*y)%mod;}y =(y*y)%mod;b/=2;}return x%mod;}
*/


signed main()
{
ios::sync_with_stdio(false); cin.tie(0);
int T=1,T1=0;//cin>>T;
while(T1++<T)
{//cout<<"Case #"<<T1<<": ";
    int n;cin>>n;
    string s;
    cin>>s;
    map<char,int>mr;
    mr['W']=0;mr['R']=1;
    int ans=0,x=0,y=0,x1=0,y1=0;
    for(int i=0;i<n;i++)
    {
        if(s[i]=='R')x++;
        else y++;
    }
    ans=min(x,y);
    for(int i=0;i<n;i++)
    {
        if(s[i]=='R'){x--;x1++;}
        else {y--;y1++;}
        ans=min(ans,max(y1,x));
    }
    cout<<ans<<endl;
}


return 0;
}