Problem:
Score : 400 points

Problem Statement

We have a string S of length N consisting of R, G, and B.

Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:

S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.

j - i \neq k - j.

Constraints

1 \leq N \leq 4000

S is a string of length N consisting of R, G, and B.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of triplets in question.

Sample Input 1

4
RRGB

Sample Output 1

1

Only the triplet (1,~3,~4) satisfies both conditions. The triplet (2,~3,~4) satisfies the first condition but not the second, so it does not count.

Sample Input 2

39
RBRBGRBGGBBRRGBBRRRBGGBRBGBRBGBRBBBGBBB

Sample Output 2

1800

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define ff first
#define ss second
#define endl "\n"
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x.size())
int powmod(int a,int l, int md){a%=md; int res=1;while(l){if(l&1)res=res*a%md;l/=2;a=a*a%md;}return res;}
int binpow(int a,int l){int res=1;while(l){if(l&1)res=res*a;l/=2;a=a*a;}return res;}
int invmod(int a, int md){return powmod(a,md-2,md);}
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
typedef vector<int> vi; typedef pair<int, int> ii; typedef vector< ii > vii;
#define pb push_back
int __set(int b, int i) {return b|(1LL<<i);} //set ith bit
int __unset(int b, int i) {return b&(~(1UL<<i));}
int __check(int b, int i) {return b&(1LL<<i);} //returns 0 if ith bit is 0
int mulmod(int a, int b, int md) {return (((a%md)*(b%md))%md+md)%md;}
int addmod(int a, int b, int md) {return ((a%md+b%md)%md+md)%md;}
int submod(int a, int b, int md) {return (((a%md-b%md)%md)+md)%md;}
int divmod(int a, int b, int md) {return mulmod(a, powmod(b, md-2, md), md);} //if md is prime;
const ll inf = 0xFFFFFFFFFFFFFFFL; //very large number
priority_queue<int, vector<int>, greater<int> > pq; //for min priority_queue
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);


signed main(void)
{
	IOS;
	int n; cin>>n;
	string s; cin>>s;
	vi a(n);
	vector<vi> g(3);
	for(int  i =0 ; i  < n; i++) {
		if(s[i]=='R') {
			a[i]=0;
			g[0].pb(i);
		}
		if(s[i]=='G'){
			a[i]=1;
			g[1].pb(i);
		}
		if(s[i]=='B') {
			a[i]=2;
			g[2].pb(i);
		}
	}
	int ans=0;
	for(int i = 0; i < n; i++) {
		for(int j = i+1; j< n; j++) {
			if(s[i]==s[j]) continue;
			int id=3-a[i]-a[j];
			ans += sz(g[id])-(upper_bound(all(g[id]),j)-g[id].begin());
			auto it=lower_bound(all(g[id]),2*j-i);
			if(it!=g[id].end() && *it==2*j-i)
				ans--;
		}
	}
	cout<<ans<<endl;
	return 0;
}