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)

//clear adj and visited vector declared globally after each test case
//check for long long overflow
//while adding and subs check if mod becomes -ve
//while using an integer directly in a builtin function add ll
//Mod wale question mein last mein if dalo ie. Ans<0 then ans+=mod;
//Dont keep array name as size or any other key word

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define int long long
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);cout.precision(dbl::max_digits10);
#define pb push_back
#define mod 1000000007 //998244353
#define lld long double
#define mii map<int, int>
#define mci map<char, int>
#define msi map<string, int>
#define pii pair<int, int>
#define ff first
#define ss second 
#define all(x) (x).begin(), (x).end()
#define rep(i,x,y) for(int i=x; i<y; i++)
#define fill(a,b) memset(a, b, sizeof(a))
#define vi vector<int>
#define setbits(x) __builtin_popcountll(x)
#define print2d(dp,n,m) for(int i=0;i<=n;i++){for(int j=0;j<=m;j++)cout<<dp[i][j]<<" ";cout<<"\n";}
typedef std::numeric_limits< double > dbl;
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
const long long N=200005, INF=2000000000000000000;
lld pi=3.1415926535897932;
int lcm(int a, int b)
{
    int g=__gcd(a, b);
    return a/g*b;
}
int power(int a, int b, int p)
    {
        if(a==0)
        return 0;
        int res=1;
        a%=p;
        while(b>0)
        {
            if(b&1)
            res=(res*a)%p;
            b>>=1;
            a=(a*a)%p;
        }
        return res;
    }

int32_t main()
{
    IOS;
    int n;
    cin>>n;
    string s;
    cin>>s;
    int a=0,b=0;
    int ar[n+1];
    fill(ar, 0);
    for(int i=0;i<n;i++)
    {
        if(s[i]=='R')
        {
            a++;
            ar[i+1]=1;
        }
        else
        b++;
    }
    for(int i=1;i<=n;i++)
    ar[i]+=ar[i-1];
    int ans=INF;
    for(int i=0;i<=n;i++)
    {
        int cur=abs(a-i);
        cur+=(ar[n]-ar[i]-a+i);
        //cout<<i<<" "<<cur<<"\n";
        ans=min(ans, cur);
    }
    cout<<ans;
}