human
Solve in Python:
We have a string S of length N consisting of A, T, C, and G.
Strings T_1 and T_2 of the same length are said to be complementary when, for every i (1 \leq i \leq l), the i-th character of T_1 and the i-th character of T_2 are complementary. Here, A and T are complementary to each other, and so are C and G.
Find the number of non-empty contiguous substrings T of S that satisfies the following condition:
 - There exists a string that is a permutation of T and is complementary to T.
Here, we distinguish strings that originate from different positions in S, even if the contents are the same.

-----Constraints-----
 - 1 \leq N \leq 5000
 - S consists of A, T, C, and G.

-----Input-----
Input is given from Standard Input in the following format:
N S

-----Output-----
Print the number of non-empty contiguous substrings T of S that satisfies the condition.

-----Sample Input-----
4 AGCT

-----Sample Output-----
2

The following two substrings satisfy the condition:
 - GC (the 2-nd through 3-rd characters) is complementary to CG, which is a permutation of GC.
 - AGCT (the 1-st through 4-th characters) is complementary to TCGA, which is a permutation of AGCT.

Can you provide a Python code to find the number of non-empty contiguous substrings T of S that satisfies the following condition:?
assistant
#Here is Python code to find the number of non-empty contiguous substrings T of S that satisfies the following condition:
.

#region Header
#!/usr/bin/env python3
# from typing import *

import sys
import io
import math
import collections
import decimal
import itertools
from queue import PriorityQueue
import bisect

def input():
    return sys.stdin.readline()[:-1]

sys.setrecursionlimit(1000000)
#endregion

# _INPUT = """10 AAATACCGCG
# """
# sys.stdin = io.StringIO(_INPUT)


# def solve(N: int, S: str) -> int:
def solve(N, S):
    # count_a = [0 for _ in range(N+1)]
    # count_g = [0 for _ in range(N+1)]
    # count_c = [0 for _ in range(N+1)]
    # count_t = [0 for _ in range(N+1)]
    # for i in range(1, N + 1):
    #     count_a[i] = count_a[i-1]
    #     count_g[i] = count_g[i-1]
    #     count_c[i] = count_c[i-1]
    #     count_t[i] = count_t[i-1]
    #     if S[i-1] == 'A':
    #         count_a[i] += 1
    #     elif S[i-1] == 'G':
    #         count_g[i] += 1
    #     elif S[i-1] == 'C':
    #         count_c[i] += 1
    #     elif S[i-1] == 'T':
    #         count_t[i] += 1

    # i文字目からj文字目までの（Aの数）=（Tの数）and（Cの数）=（Gの数）となる (i,j)
    n = 0
    for i in range(-1, N-1):
        count_a = 0
        count_g = 0
        count_c = 0
        count_t = 0
        for j in range(i+1, N):
            if S[j] == 'A':
                count_a += 1
            elif S[j] == 'G':
                count_g += 1
            elif S[j] == 'C':
                count_c += 1
            elif S[j] == 'T':
                count_t += 1
            if (count_a == count_t) and (count_c == count_g):
                n += 1

    return n

def main():
    N, S = input().split()
    N = int(N)
    a = solve(N, S)
    print(a)

def __starting_point():
    main()

__starting_point()