Problem Description

You are given two arrays of integers nums1 and nums2. Your task is to count the total number of valid triplets that can be formed following two specific rules.

A triplet consists of three indices, and there are two types of valid triplets:

Type 1 Triplet: A triplet (i, j, k) where:

• i is an index from nums1
• j and k are indices from nums2 with j < k
• The relationship nums1[i]² == nums2[j] * nums2[k] must hold

Type 2 Triplet: A triplet (i, j, k) where:

• i is an index from nums2
• j and k are indices from nums1 with j < k
• The relationship nums2[i]² == nums1[j] * nums1[k] must hold

In simpler terms:

• For Type 1: Take one element from nums1, square it, and check if it equals the product of any two elements from nums2 (where the second element comes after the first)
• For Type 2: Take one element from nums2, square it, and check if it equals the product of any two elements from nums1 (where the second element comes after the first)

The function should return the total count of all valid triplets (both Type 1 and Type 2 combined).

For example, if nums1 = [7, 4] and nums2 = [5, 2, 8, 9]:

• A Type 1 triplet could be (1, 0, 2) because nums1[1]² = 16 = nums2[0] * nums2[2] = 2 * 8
• A Type 2 triplet could be (3, 0, 1) because nums2[3]² = 81 = nums1[0] * nums1[1] = 7 * 4 (this is just an example, not necessarily valid)

Intuition

The key insight is to recognize that we're essentially looking for matching relationships between squares and products across two arrays. Instead of checking every possible triplet combination which would be inefficient, we can precompute all possible products and then match them with squares.

Think about it this way: for Type 1 triplets, we need to find cases where nums1[i]² = nums2[j] * nums2[k]. Rather than fixing i and then searching for all valid (j, k) pairs, we can flip our perspective:

1. First, calculate all possible products from pairs in nums2 (where j < k)
2. Then, for each element in nums1, square it and check how many times this squared value appears in our precomputed products

This transforms the problem from a three-way matching problem into a two-step process:

• Step 1: Generate all products from pairs within each array
• Step 2: Match squares from one array with products from the other

The beauty of this approach is that we can use a hash table (Counter) to store the frequency of each product. Why frequency? Because multiple pairs might produce the same product. For instance, 2 * 6 = 12 and 3 * 4 = 12. If we're looking for triplets where some nums1[i]² = 12, we need to count both pairs.

By separating the counting of products from the matching with squares, we avoid redundant calculations. We compute all products once for each array, store them with their frequencies, and then simply look up how many times each squared value appears in the opposite array's product table.

This bidirectional approach (Type 1: squares from nums1 matched with products from nums2, and Type 2: squares from nums2 matched with products from nums1) ensures we capture all valid triplets while maintaining an efficient time complexity.

Learn more about Math and Two Pointers patterns.

Solution Approach

The implementation uses a hash table-based approach with helper functions to modularize the logic:

Step 1: Create a Product Counting Function

The count function takes an array and returns a Counter (hash table) that stores all possible products of pairs (j, k) where j < k:

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

This function iterates through all valid pairs in the array. For each pair, it calculates their product and increments the count in the hash table. The time complexity is O(n²) where n is the length of the array.

Step 2: Create a Calculation Function

The cal function takes an array and a Counter, then calculates how many valid triplets can be formed:

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

For each element x in the array, it squares it (x * x) and looks up how many times this squared value appears in the Counter. The sum of all these counts gives us the total number of valid triplets for this configuration.

Step 3: Apply to Both Types of Triplets

The main solution combines these helper functions:

cnt1 = count(nums1)  # All products from pairs in nums1
cnt2 = count(nums2)  # All products from pairs in nums2
return cal(nums1, cnt2) + cal(nums2, cnt1)

• cal(nums1, cnt2): Counts Type 2 triplets - squares from nums1 matched with products from nums2
• cal(nums2, cnt1): Counts Type 1 triplets - squares from nums2 matched with products from nums1

Note the seemingly reversed logic here: cal(nums1, cnt2) actually counts Type 2 triplets because:

• We're squaring elements from nums1
• We're matching them with products from nums2
• This corresponds to nums1[j] * nums1[k] = nums2[i]² (Type 2)

Time Complexity: O(m² + n²) where m and n are the lengths of nums1 and nums2 respectively, due to generating all pairs.

Space Complexity: O(m² + n²) for storing the products in the hash tables.

Example Walkthrough

Let's walk through a small example with nums1 = [7, 4] and nums2 = [5, 2, 8, 9].

Step 1: Count all products from pairs

For nums1 = [7, 4]:

• Only one pair: (0, 1) → 7 * 4 = 28
• cnt1 = {28: 1}

For nums2 = [5, 2, 8, 9]:

• Pair (0, 1): 5 * 2 = 10
• Pair (0, 2): 5 * 8 = 40
• Pair (0, 3): 5 * 9 = 45
• Pair (1, 2): 2 * 8 = 16
• Pair (1, 3): 2 * 9 = 18
• Pair (2, 3): 8 * 9 = 72
• cnt2 = {10: 1, 40: 1, 45: 1, 16: 1, 18: 1, 72: 1}

Step 2: Match squares with products

For Type 1 triplets (square from nums1, products from nums2):

• nums1[0]² = 7² = 49: Check if 49 exists in cnt2 → No match
• nums1[1]² = 4² = 16: Check if 16 exists in cnt2 → Yes! Count = 1
• This gives us triplet (1, 1, 2) where nums1[1]² = 16 = nums2[1] * nums2[2] = 2 * 8

For Type 2 triplets (square from nums2, products from nums1):

• nums2[0]² = 5² = 25: Check if 25 exists in cnt1 → No match
• nums2[1]² = 2² = 4: Check if 4 exists in cnt1 → No match
• nums2[2]² = 8² = 64: Check if 64 exists in cnt1 → No match
• nums2[3]² = 9² = 81: Check if 81 exists in cnt1 → No match

Step 3: Sum up all matches

Total valid triplets = 1 (from Type 1) + 0 (from Type 2) = 1

The algorithm efficiently found the single valid triplet (1, 1, 2) by precomputing products and matching them with squares, avoiding the need to check all possible three-element combinations.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m^2 + n^2 + m + n)

The time complexity breaks down as follows:

• The count function is called twice - once for nums1 and once for nums2
  • For nums1: The nested loops iterate through all pairs (j, k) where j < k, resulting in O(m^2) operations
  • For nums2: Similarly, this takes O(n^2) operations
• The cal function is called twice
  • First call iterates through nums1 and looks up values in cnt2, taking O(m) time
  • Second call iterates through nums2 and looks up values in cnt1, taking O(n) time
• Total time complexity: O(m^2) + O(n^2) + O(m) + O(n) = O(m^2 + n^2 + m + n)

Space Complexity: O(m^2 + n^2)

The space complexity is determined by:

• cnt1 stores all possible products of pairs from nums1. In the worst case, all pairs produce unique products, resulting in O(m^2) space
• cnt2 stores all possible products of pairs from nums2. Similarly, this requires O(n^2) space in the worst case
• Total space complexity: O(m^2) + O(n^2) = O(m^2 + n^2)

Where m is the length of nums1 and n is the length of nums2.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Confusion About Type 1 vs Type 2 Mapping

The Pitfall: The most common mistake is getting confused about which calculation corresponds to which triplet type. In the code, calculate_triplets(nums2, products_nums1) actually counts Type 1 triplets, not Type 2. This seems counterintuitive because:

• Type 1 is defined as: nums1[i]² == nums2[j] * nums2[k]
• But we're calling calculate_triplets(nums2, products_nums1)

Many developers mistakenly think they should square elements from nums1 for Type 1 triplets.

Why This Happens: The confusion arises from misreading the problem statement. For Type 1:

• We need ONE element from nums1 (at index i)
• We need TWO elements from nums2 (at indices j and k)
• So we square the single element and match it against products of pairs

The Solution: Remember the pattern:

• When you need to match one squared element from array A with products of pairs from array B
• You compute products from B and square elements from A
• The array providing the single element gets squared; the array providing pairs gets multiplied

2. Integer Overflow Issues

The Pitfall: When dealing with large integers, squaring them or multiplying pairs can cause overflow issues. For example:

• If nums1 = [100000] and nums2 = [100000, 1]
• Computing 100000² = 10,000,000,000 which exceeds 32-bit integer limits

The Solution: In Python, this is handled automatically as integers can be arbitrarily large. However, in other languages like Java or C++, you should:

# Use long/int64 for calculations
square = int(num) * int(num)  # Explicit casting if needed
product = int(nums[j]) * int(nums[k])

3. Counting Duplicate Products Incorrectly

The Pitfall: Not properly handling cases where the same product can be formed by different pairs. For example:

• nums = [2, 3, 4, 6]
• Product 12 can be formed by: (2, 6) and (3, 4)
• Both should be counted separately

The Solution: The Counter approach handles this correctly by incrementing the count for each occurrence:

product_counter[product] += 1  # Correctly accumulates all occurrences

4. Including Invalid Pairs (j >= k)

The Pitfall: Accidentally counting pairs where j >= k or counting the same pair twice (once as (j,k) and once as (k,j)).

Incorrect approach:

# Wrong: This counts each pair twice
for j in range(len(nums)):
    for k in range(len(nums)):
        if j != k:  # Still wrong!
            product_counter[nums[j] * nums[k]] += 1

The Solution: Always ensure j < k in your loop structure:

for j in range(len(nums)):
    for k in range(j + 1, len(nums)):  # Ensures j < k
        product_counter[nums[j] * nums[k]] += 1

5. Edge Cases with Empty or Single-Element Arrays

The Pitfall: Not handling cases where:

• One or both arrays are empty
• An array has only one element (can't form pairs)

The Solution: The current implementation handles these gracefully:

• Empty array → No pairs → Empty Counter → Zero triplets
• Single element → No valid pairs where j < k → Zero products → Zero triplets

However, you could add explicit checks for clarity:

if len(nums1) < 2 and len(nums2) < 2:
    return 0  # No valid triplets possible