What is the best approach for Number of Ways Where Square of Number Is Equal to Product of Two Numbers?

The most efficient approach uses a hash map to count complements while scanning the second array. For each value x, compute target = x^2 and check whether previously seen numbers form a product equal to that target. Hash lookups allow counting valid pairs in constant time, giving an overall time complexity of O(n^2) with O(n) extra space.

Is Number of Ways Where Square of Number Is Equal to Product of Two Numbers asked at Google/Amazon/Meta?

This problem represents a common interview pattern involving pair products and hash-based counting. Variations of the problem appear in interviews at large tech companies where candidates must convert a triple condition into pair lookups using hash maps or frequency tables.

What data structure is used in Number of Ways Where Square of Number Is Equal to Product of Two Numbers?

The optimal solution relies on a hash table (hash map) to store frequencies of numbers while iterating through the array. This allows constant-time lookups for complementary factors that multiply to the target square value.

What is the time complexity of Number of Ways Where Square of Number Is Equal to Product of Two Numbers?

The brute force approach runs in O(n * m^2) time because every squared value is compared with all pairs from the other array. The optimized solution reduces this to O(n^2) by using a hash map to track previously seen numbers and count factor complements in constant time.

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

Q: Is Number of Ways Where Square of Number Is Equal to Product of Two Numbers easy or hard?

The problem is typically rated Medium difficulty. The brute force idea is straightforward, but recognizing that pair products can be converted into factor lookups with a hash map requires algorithmic insight.

Q: Number of Ways Where Square of Number Is Equal to Product of Two Numbers Python/Java solution

Python and Java implementations follow the same pattern: compute the square target, iterate through the second array, and use a hash map to count complements. Each iteration checks if target % num == 0 and adds the frequency of target / num from the map, resulting in O(n^2) time complexity.

Q: How to solve Number of Ways Where Square of Number Is Equal to Product of Two Numbers in O(n^2)?

Iterate through one array and compute the square of each element. While scanning the other array, maintain a frequency map of numbers already seen. If target % current == 0, look up target / current in the map to count matching factors. Repeat the same process with arrays swapped to include both valid triplet types.

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