- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Hash Table, Math, Two Pointers, leetcode-1577, O(m * n), O(m + n)
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]
where0 <= i < nums1.length
and0 <= j < k < nums2.length
. - Type 2: Triplet (i, j, k) if
nums2[i]2 == nums1[j] * nums1[k]
where0 <= i < nums2.length
and0 <= 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