1726. Tuple with Same Product - Detailed Explanation

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Problem Statement

Given an array of distinct positive integers, we need to count the number of tuples (ordered quadruples) (a, b, c, d) such that the product of one pair equals the product of the other pair. In other words, find how many ways we can pick four distinct elements from the array so that a * b = c * d. The order in the tuple matters, meaning (a, b, c, d) is considered different from (c, d, a, b) if the positions are different, even if they contain the same numbers. Our task is to return the count of all such valid tuples.

Example 1:
Input: nums = [2, 3, 4, 6]
Output: 8
Explanation: The pairs (2, 6) and (3, 4) both produce the product 12. These two pairs can form 8 distinct tuples by permuting the order of the four numbers. For instance, (2, 6, 3, 4) and (3, 4, 2, 6) are considered different tuples, among others.
Example 2:
Input: nums = [1, 2, 4, 5, 10]
Output: 16
Explanation: There are two different products that appear more than once: 10 and 20. Product 10 is made by pairs (1, 10) and (2, 5), and product 20 is made by pairs (2, 10) and (4, 5). Each distinct pair combination can form 8 ordered tuples. Thus, product 10 contributes 8 tuples and product 20 contributes another 8, for a total of 16.
Example 3:
Input: nums = [2, 3, 5, 7]
Output: 0
Explanation: No two distinct pairs of numbers have the same product in this array. All pair products are unique (e.g. 2×3=6, 2×5=10, 2×7=14, 3×5=15, 3×7=21, 5×7=35), so there are no tuples that satisfy the condition.

Constraints:

1 <= nums.length <= 1000 (up to 1000 elements in the array)
1 <= nums[i] <= 10^4 (each number is a positive integer up to 10000)
All elements in nums are distinct (no repeats).

The output is a single integer representing the number of valid tuples. We need an efficient solution because a brute-force check of all quadruples would be too slow when nums.length is large (could be up to 1000, which means billions of possible quadruples).

Hints

Hint 1: Instead of checking every combination of four numbers directly, try to break the problem down. Consider how you might identify two pairs of numbers with the same product. What simpler subproblem involves two numbers?
Hint 2: Using a data structure like a hash map (or dictionary) can help store intermediate results. For example, you can use a hash map to count how often each product of two numbers occurs. This way, you can quickly find when a product is seen again and use that information to count tuples.

Approaches

Approach 1: Brute Force (Generate All Quadruples)

Idea: A straightforward approach is to consider every possible set of four distinct indices and check if they form a valid tuple. That means picking two pairs and comparing their products. We ensure the four numbers are distinct by choosing indices in increasing order (i < j < k < l). For each such combination (i, j, k, l), we check if nums[i] * nums[j] == nums[k] * nums[l]. If they are equal, we count it as one tuple.

Step-by-step (conceptual):

Iterate through all possible pairs of indices for the first pair. For example, choose i and j such that i < j.
For each choice of (i, j), iterate through all possible pairs for the second pair. Choose k and l such that k < l and importantly, ensure {i, j} and {k, l} are disjoint sets of indices (one simple way is to start k from j+1 so there is no overlap).
Check if the product of the first pair equals the product of the second pair. If nums[i] * nums[j] == nums[k] * nums[l], then the four values form a valid tuple.
Count all such occurrences.

Example (Brute Force Check):
Consider nums = [2, 3, 4, 6]. The brute force method would check combinations like:

(i=0, j=1, k=2, l=3) → pairs (2,3) and (4,6) → products 6 and 24 (not equal).
(i=0, j=2, k=1, l=3) → pairs (2,4) and (3,6) → products 8 and 18 (not equal).
(i=0, j=3, k=1, l=2) → pairs (2,6) and (3,4) → products 12 and 12 (equal! this forms a valid tuple).
and so on for all combinations. It would find all 8 ordered tuples for that example, but it does a lot of checks. For larger arrays, this becomes impractical.

Python Implementation (Brute Force):

Python3

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Java Implementation (Brute Force):

Java

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Complexity Analysis (Brute Force): Checking all quadruples is O(n^4) in time complexity, since we have four nested loops over n. For n = 1000, this could be on the order of 10^12 operations, which is completely infeasible. Space complexity is O(1) (constant extra space), since we only use a counter. This brute force approach will work only for very small input sizes and is mainly a conceptual stepping stone. It’s too slow for the given constraints, so we need to optimize.

Approach 2: Using a HashMap (Product Count)

Idea: Instead of explicitly checking every quadruple, we can reduce the problem using a hash map to count pair products. The main insight is: if two different pairs have the same product, they can form valid tuples. We can compute the product of every possible pair (i, j) and use a hash map (dictionary) to record how many times each product occurs. Whenever we find a product that has been seen before, it means we found another pair that matches an earlier pair’s product, which leads to new tuples.

How it works:

Use a hash map product_count where the key is a product and the value is the number of pairs that produce that product so far.
Loop through all unique pairs (i, j) (with i < j). For each pair, calculate prod = nums[i] * nums[j].
- If prod is not in the map yet, add it with count 1 (meaning this is the first pair with that product).
- If prod is already in the map with some count c, it means we have found c previous pairs with the same product. Each of those existing pairs can form a tuple with the current pair. How many new tuples does that give? 8 for each pair combination (because the tuple (a, b, c, d) can be ordered in 8 ways when we consider all permutations). So if c pairs were seen, the current pair can form c * 8 new tuples. We add that to our answer count. Then, increment the count for prod in the map (now there are c+1 pairs with this product).
In the end, the accumulated count will be the total number of valid tuples.

Another way to think about it: once we have the final counts of all pair products, for each product that appears f times (i.e., f different pairs have that product), the number of unique unordered pair-pair combinations is C(f, 2) (choose any 2 pairs out of f). And since each such combination of pairs can be arranged in 8 different ways as ordered tuples, it contributes 8 * C(f, 2) tuples to the answer.

Example (HashMap approach):
Suppose nums = [1, 2, 4, 5, 10]. We iterate through pairs and use a dictionary to count products:

Start with an empty map.
Take pair (1, 2) → product 2. Not seen before, add product_count[2] = 1. (No tuple yet)
Pair (1, 4) → product 4. Add product_count[4] = 1.
Pair (1, 5) → product 5. Add product_count[5] = 1.
Pair (1, 10) → product 10. Add product_count[10] = 1.
Pair (2, 4) → product 8. Add product_count[8] = 1.
Pair (2, 5) → product 10. This product is already in the map with count 1 (from pair (1,10)). That means we've found a second pair with product 10. We can form 1*8 = 8 new tuples with the pair (2,5) and the previous pair (1,10). Update count: product_count[10] = 2.
Pair (2, 10) → product 20. Add product_count[20] = 1.
Pair (4, 5) → product 20. Product 20 seen before (count 1 from pair (2,10)). We get 8 new tuples from pairing (4,5) with (2,10). Update count: product_count[20] = 2.
Pair (4, 10) → product 40. Add product_count[40] = 1.
Pair (5, 10) → product 50. Add product_count[50] = 1.

At the end, we found 8 tuples from product 10 and 8 from product 20, total 16, which matches the output. This approach found the result by counting pairs, without explicitly enumerating all quadruples.

Python Implementation (HashMap):

Python3

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Java Implementation (HashMap):

Java

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Complexity Analysis (Optimized): Generating all pairs is an O(n^2) operation. The nested loop runs about n*(n-1)/2 times (for n=1000, roughly 500k iterations), which is efficient enough. Each iteration does O(1) work to update the map and result. Thus, time complexity is O(n²). Space complexity is O(n²) in the worst case for the hash map (in the worst scenario where every pair product is distinct, we store ~n(n-1)/2 products). Typically, the space will be less if many products repeat. This is a dramatic improvement over O(n^4). Using a hash map allows us to count tuple combinations as we go, counting in one pass what brute force would take billions of checks to find.

Step-by-Step Walkthrough (Example)

Let's visually walk through the process using a smaller example to see how the counting works. Consider the array nums = [1, 2, 4, 5, 10] (from Example 2):

Step 1: Generate all pair products. List every unique pair of numbers and calculate their product:
- (1, 2) → product 2
- (1, 4) → product 4
- (1, 5) → product 5
- (1, 10) → product 10
- (2, 4) → product 8
- (2, 5) → product 10 (matches a previous product 10!)
- (2, 10) → product 20
- (4, 5) → product 20 (matches a previous product 20!)
- (4, 10) → product 40
- (5, 10) → product 50
Step 2: Group pairs by identical product. We can see which products appear more than once:
- Product 10: formed by pairs (1, 10) and (2, 5) (2 pairs)
- Product 20: formed by pairs (2, 10) and (4, 5) (2 pairs)
  (All other products like 2, 4, 5, 8, 40, 50 are formed by only one pair each, so they don’t contribute to tuples with another pair.)
Step 3: Count tuples from each group of pairs. Whenever a product is produced by multiple distinct pairs, those pairs can combine to form tuples. For each product that has f pairs:
- Calculate the number of ways to choose 2 pairs out of f (this is the combination C(f, 2) = f*(f-1)/2). This gives the number of unique pair-pair combinations.
- Each such pair combination can be arranged in 8 ways as an ordered tuple (because for two pairs (a,b) and (c,d), there are 2! ways to order the pairs and 2! ways to order within each pair: 2 * 2 * 2 = 8).
  For product 10: f = 2, C(2,2) = 1 combination → 1 * 8 = 8 tuples.
  For product 20: f = 2, C(2,2) = 1 combination → 8 tuples.
Step 4: Sum up contributions. Add all tuple counts from all products with duplicates. Here, 8 (from product 10) + 8 (from product 20) = 16 total tuples, which matches the expected result. Each tuple corresponds to an ordering like (1,10,2,5), (1,10,5,2), (2,5,1,10), (2,5,10,1), (10,1,2,5), etc., covering all permutations .

This walkthrough shows how we systematically find and count the tuples without checking every quadruple individually.

Common Mistakes

Not accounting for order (permutations): A common mistake is to count each combination of two pairs only once. Remember that the tuple is ordered – for each pair-pair combination, there are 8 possible ordered tuples. Forgetting to multiply by 8 (or misunderstanding that (a,b,c,d) is different from (c,d,a,b)) will underestimate the count.
Double counting the same pair twice: Ensure that the two pairs are made of four distinct numbers. If you iterate pairs incorrectly, you might accidentally consider cases where the pairs share an element (which isn’t allowed). Using indices i < j < k < l or ensuring the pairs come from different index ranges (as we did) prevents this.
Inefficient iteration: Trying to brute-force every quadruple (four loops) for n up to 1000 will not finish in reasonable time. It’s a mistake to implement the brute force approach for large input without optimizing. Using the hashing method or at least precomputing pair products is necessary.
Duplicate pair generation: When generating pairs, make sure each unique pair (i, j) is only considered once. For example, if you loop i and j over the whole array without constraining j > i, you might count (a,b) and (b,a) as two separate pairs even though they are the same combination. Always use an order (like i < j) to avoid such duplication.

Edge Cases

Not enough elements: If the array has fewer than 4 elements, it’s impossible to form two pairs, so the result is 0. The code should handle this gracefully (for example, the loops simply won’t run if n < 4).
No valid tuples: It’s possible that no two distinct pairs have the same product. For example, in an array of primes like [2,3,5,7], every product will be unique, yielding 0 tuples. The solution should return 0 in such cases.
Large outputs: In cases where there are several pairs with the same product, the number of tuples can be large. For example, an array that has multiple pairs sharing products will lead to a big count (as in the case of output 40 in one of the examples). Make sure to use a data type that can handle the potential size of the result. In Java, using an int might overflow if the count is very large, so using a long for the accumulation is safer (we cast back to int at the end because the final answer will fit within int range for given constraints).
Distinctness guarantee: Since the input guarantees all numbers are distinct, we don’t have to worry about the scenario of the same number appearing more than once. If duplicates were allowed, the problem would need additional handling to ensure distinct indices, but here it’s straightforward.

This problem is related to the general theme of finding combinations that satisfy a condition. Once you understand the pairing technique, you can try similar problems that use hashing or combination counting:

Two Sum (LeetCode 1): Find two numbers that add up to a target value. This classic uses a hash map for pairs (in this case, sums) and is a simpler scenario of using a hash map to detect a matching pair.
4Sum / 4Sum II: In 4Sum (LeetCode 18), you find unique quadruplets that sum to a target (order doesn’t matter in that problem, unlike here). 4Sum II (LeetCode 454) involves four numbers across two arrays that sum to zero, and it uses a hash map to count pair sums (very similar in spirit to counting pair products). These teach combination counting with maps.
Interchangeable Rectangles (LeetCode 2001): A problem where you count pairs of rectangles with the same width-to-height ratio. The approach is analogous – use a hash map to count equal ratios, then for each ratio with frequency f, the number of interchangeable pairs is C(f,2). It’s a good practice for using combination counting on grouped values.
Combination Sum, 3Sum, etc.: These are problems where you find combinations of numbers that meet certain criteria (like summing to a target). While they don’t use exactly the same hashing technique, they also require careful consideration of combinations and sometimes leveraging counting or two-pointer strategies. They can help improve understanding of enumerating combinations efficiently.

1726. Tuple with Same Product - Detailed Explanation

Problem Statement

Hints

Approaches

Approach 1: Brute Force (Generate All Quadruples)

Approach 2: Using a HashMap (Product Count)

Step-by-Step Walkthrough (Example)

Common Mistakes

Edge Cases

Alternative Variations and Related Problems