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1577. Number of Ways Where Square of Number Is Equal to Product of Two Numbers

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]. (42 = 2 * 8).

Example 2:

Input: nums1 = [1,1], nums2 = [1,1,1]
Output: 9
Explanation: All Triplets are valid, because 12 = 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].

Constraints:

  • 1 <= nums1.length, nums2.length <= 1000
  • 1 <= nums1[i], nums2[i] <= 105

Solutions (Python)

1. Solution

class Solution:
    def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        count1 = Counter(nums1)
        count2 = Counter(nums2)
        ret = 0

        for x in nums1:
            x2 = x * x
            for y in nums2:
                if x2 % y == 0 and x2 // y in count2:
                    ret += count2[x2 // y]
                    if y == x:
                        ret -= 1
        for x in nums2:
            x2 = x * x
            for y in nums1:
                if x2 % y == 0 and x2 // y in count1:
                    ret += count1[x2 // y]
                    if y == x:
                        ret -= 1

        return ret // 2