Dictionary in Python

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 As a big fan of Hash table in general, the usage of Hash tables in Python, or Dictionary, is also pretty straightforward. This problem illustrates it well: https://leetcode.com/problems/number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers/

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

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
Accepted
6,704
Submissions
18,737

Since the limits are small, you can solve it in N^2. Push the squares to a dictionary in Python, and go thru the N^2 looking for the keys. Code is below, cheers, ACC.


def numTriplets(self, nums1, nums2):
    """
    :type nums1: List[int]
    :type nums2: List[int]
    :rtype: int
    """
    cache1 = dict()
    for i in range(len(nums1)):
        key = nums1[i] * nums1[i]
        if cache1.get(key) == None:
            cache1[key] = 1
        else:
            cache1[key] = cache1[key] + 1

    cache2 = dict()
    for i in range(len(nums2)):
        key = nums2[i] * nums2[i]
        if cache2.get(key) == None:
            cache2[key] = 1
        else:
            cache2[key] = cache2[key] + 1

    retVal = 0
    for i in range(len(nums1)):
        for j in range(i+1, len(nums1)):
            key = nums1[i] * nums1[j]
            if cache2.get(key) != None:
                retVal += cache2[key]
    for i in range(len(nums2)):
        for j in range(i+1, len(nums2)):
            key = nums2[i] * nums2[j]
            if cache1.get(key) != None:
                retVal += cache1[key]

    return retVal

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