Number of Ways Where Square of Number Is Equal to Product of Two Numbers — LeetCode 1577 Python Solution

The problem

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

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).

Python solution

class Solution:
    def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        def count(nums: List[int]) -> Counter:
            cnt = Counter()
            for j in range(len(nums)):
                for k in range(j + 1, len(nums)):
                    cnt[nums[j] * nums[k]] += 1
            return cnt

        def cal(nums: List[int], cnt: Counter) -> int:
            return sum(cnt[x * x] for x in nums)

        cnt1 = count(nums1)
        cnt2 = count(nums2)
        return cal(nums1, cnt2) + cal(nums2, cnt1)

Complexity

Pattern: Two Pointers

Use the order already in the input to discard half the search space at every step. LeetCode 1577. Number of Ways Where Square of Number Is Equal to Product of Two Numbers is filed here because LeetCode tags it Two Pointers, which is the vocabulary this hub collects.

The two pointers guide has the Python template for the pattern and the 201 LeetCode problems that use it.

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