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 loop(i,a,n) for(int i=a;i<n;i++)
int main(){
	long long int n;
  	scanf("%lld",&n);
    
    char s[4001];
    scanf("%s",s);
    long long int ans = 0; 
    
    vector<int>r,g,b;
    
    loop(i,0,n){
        if(s[i] == 'R')r.push_back(i);
        else if(s[i] == 'G')g.push_back(i);
        else if(s[i] == 'B')b.push_back(i);
    }
    
    loop(i,0,n){
        loop(j,i+1,n){
            if(s[i] != s[j]){
                int dist = j-i;
                if(s[i]!= 'R' and s[j]!= 'R'){
                    auto it = upper_bound(r.begin(),r.end(),j);
                    if(it != r.end()){
                        ans += (r.end() - it);
                        if(2*j-i < n and s[2*j-i] == 'R')ans --;
                    }
                        
                }
                else if(s[i]!= 'G' and s[j]!= 'G'){
                    auto it = upper_bound(g.begin(),g.end(),j);
                     if(it != g.end()){
                        ans += (g.end() - it);
                        if(2*j-i < n and s[2*j-i] == 'G')ans --;
                    }
                }
                else{
                    auto it = upper_bound(b.begin(),b.end(),j);
                    if(it != b.end()){
                        ans += (b.end() - it);
                        if(2*j-i < n and s[2*j-i] == 'B')ans --;
                    }
                }
            }
        }
    }
    
  	printf("%lld",ans);
  	puts("");
}