Given two arrays of integers nums1 and nums2, return the number of triplets formed (type 1 and type 2) under the following rules:
Type 1: Triplet (i, j, k) if nums1[i]2 == nums2[j] * nums2[k] where 0 <= i < nums1.length and 0 <= j < k < nums2.length.
Type 2: Triplet (i, j, k) if nums2[i]2 == nums1[j] * nums1[k] where 0 <= i < nums2.length and 0 <= j < k < nums1.length.
Example 1:
Input: nums1 = [7,4], nums2 = [5,2,8,9]
Output: 1
Explanation: Type 1: (1,1,2), nums1[1]^2 = nums2[1] * nums2[2]. (4^2 = 2 * 8). 
Example 2:
Input: nums1 = [1,1], nums2 = [1,1,1]
Output: 9
Explanation: All Triplets are valid, because 1^2 = 1 * 1.
Type 1: (0,0,1), (0,0,2), (0,1,2), (1,0,1), (1,0,2), (1,1,2).  nums1[i]^2 = nums2[j] * nums2[k].
Type 2: (0,0,1), (1,0,1), (2,0,1). nums2[i]^2 = nums1[j] * nums1[k].
Example 3:
Input: nums1 = [7,7,8,3], nums2 = [1,2,9,7]
Output: 2
Explanation: There are 2 valid triplets.
Type 1: (3,0,2).  nums1[3]^2 = nums2[0] * nums2[2].
Type 2: (3,0,1).  nums2[3]^2 = nums1[0] * nums1[1].
Example 4:
Input: nums1 = [4,7,9,11,23], nums2 = [3,5,1024,12,18]
Output: 0
Explanation: There are no valid triplets.
Constraints:
1 <= nums1.length, nums2.length <= 1000
1 <= nums1[i], nums2[i] <= 10^5
class Solution:
    def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        table1 = {}
        table2 = {}
        res = 0
        
        for n in nums1:
            square = n * n
            table = {}
            for j, m in enumerate(nums2):
                if square % m == 0:
                    remainder = square // m
                    if remainder in table:
                        # print(square, m, remainder, table)
                        res += table[remainder]
                if m not in table:
                    table[m] = 0
                table[m] += 1
                
        for n in nums2:
            square = n * n
            table = {}
            for j, m in enumerate(nums1):
                if square % m == 0:
                    remainder = square // m
                    if remainder in table:
                        # print(square, m, remainder, table)

                        res += table[remainder]
                if m not in table:
                    table[m] = 0
                table[m] += 1  
        return res

class Solution:
    def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        table1 = {}
        table2 = {}
        res = 0
        
        for n in nums1:
            square = n * n
            table = {}
            for j, m in enumerate(nums2):
                if square % m == 0:
                    remainder = square // m
                    if remainder in table:
                        # print(square, m, remainder, table)
                        res += table[remainder]
                if m not in table:
                    table[m] = 0
                table[m] += 1
                
        for n in nums2:
            square = n * n
            table = {}
            for j, m in enumerate(nums1):
                if square % m == 0:
                    remainder = square // m
                    if remainder in table:
                        # print(square, m, remainder, table)

                        res += table[remainder]
                if m not in table:
                    table[m] = 0
                table[m] += 1  
        return res
