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 Go (1.14.1) (judge verdict: Accepted)

package main

import "fmt"

type key struct {
	To    int
	Color rune
}

func color3(c1, c2 rune) rune {
	if c1 == 'R' && c2 == 'G' {
		return 'B'
	}
	if c1 == 'R' && c2 == 'B' {
		return 'G'
	}
	if c1 == 'G' && c2 == 'B' {
		return 'R'
	}
	if c1 == 'G' && c2 == 'R' {
		return 'B'
	}
	if c1 == 'B' && c2 == 'R' {
		return 'G'
	}
	if c1 == 'B' && c2 == 'G' {
		return 'R'
	}
	return ' '
}

func main() {
	var n int
	fmt.Scan(&n)
	var s string
	fmt.Scan(&s)

	count := make(map[key]int)
	for i, c := range s {
		if c == 'R' {
			count[key{i, 'R'}] = count[key{i - 1, 'R'}] + 1
			count[key{i, 'G'}] = count[key{i - 1, 'G'}]
			count[key{i, 'B'}] = count[key{i - 1, 'B'}]
		} else if c == 'G' {
			count[key{i, 'R'}] = count[key{i - 1, 'R'}]
			count[key{i, 'G'}] = count[key{i - 1, 'G'}] + 1
			count[key{i, 'B'}] = count[key{i - 1, 'B'}]
		} else {
			count[key{i, 'R'}] = count[key{i - 1, 'R'}]
			count[key{i, 'G'}] = count[key{i - 1, 'G'}]
			count[key{i, 'B'}] = count[key{i - 1, 'B'}] + 1
		}
	}

	sum := 0
	for i := 0; i <= n-3; i++ {
		ci := s[i]
		for j := i; j <= n-2; j++ {
			cj := s[j]
			if ci == cj {
				continue
			}
			ck := color3(rune(ci), rune(cj))
			cnt := count[key{len(s) - 1, ck}] - count[key{j, ck}]
			diff := j - i
			if j+diff <= n-1 && rune(s[j+diff]) == ck {
				cnt--
			}
			sum += cnt
		}
	}
	fmt.Println(sum)
}