Partition Array into Two Equal Product Subsets

• First, perform a crucial preliminary check: the product of all elements in nums must equal target * target. If not, a valid partition is impossible. Use BigInteger for this calculation to avoid overflow.
• If the check passes, the problem reduces to finding a single non-empty, proper subset with a product equal to target.
• Implement a recursive function, canFindSubset(index, currentProduct), to explore all subsets.
• Base Cases:
  • If currentProduct == target, a subset is found, return true.
  • If currentProduct > target or index is out of bounds, this path is invalid, return false.
• Recursive Step: For each element nums[index], make two recursive calls:
  1. Exclude: canFindSubset(index + 1, currentProduct).
  2. Include: canFindSubset(index + 1, currentProduct * nums[index]). This call is only made if the new product won't immediately exceed target and target is divisible by it.
• If either recursive call returns true, propagate true up the call stack.
• The initial call is canFindSubset(0, 1). If it returns true, a valid partition exists.

The core idea is to systematically generate all 2^N subsets of the input array nums. Before starting the search, we make a vital optimization: we calculate the product of all elements in nums. For a valid partition to exist, this total product must be exactly target * target. If this condition doesn't hold, we can immediately return false. This check requires using BigInteger to handle potentially large numbers without overflow.

If the total product check passes, we then proceed with a recursive search. The function, say canFindSubset(index, currentProduct), tries to build a subset with the desired product. At each step index, it considers nums[index] and branches into two possibilities: one where nums[index] is included in the subset (multiplying it with currentProduct) and one where it's excluded. The recursion terminates when a product equals target (success), exceeds target (failure), or all elements have been considered (failure).

Because the problem guarantees nums contains distinct positive integers and has a length of at least 3, any subset found with a product of target will necessarily be a non-empty, proper subset, thus guaranteeing that the other subset in the partition is also non-empty.

• The overall logic is identical to the brute-force approach, including the initial totalProduct == target * target check.
• The key difference is the use of a memoization table, e.g., Map<Long, Boolean>[] memo, to store the results of subproblems.
• The state for memoization is (index, currentProduct).
• Before computing canFindSubset(index, currentProduct), check if the result is already in memo[index] for the key currentProduct. If yes, return the stored value.
• After computing the result for (index, currentProduct), store it in the memoization table before returning.
• This ensures that each unique subproblem is solved only once.

While the simple backtracking approach can be inefficient due to re-calculating results for the same subproblems, we can optimize it by storing these results. A subproblem can be uniquely identified by the pair (index, currentProduct). We use an array of hash maps, memo, where memo[index] maps a currentProduct to the boolean result of whether a subset with product target can be formed from that state.

When the recursive function canFindSubset(index, currentProduct, memo) is called, it first checks if the result for this state is already in the memo table. If it is, the stored result is returned immediately. Otherwise, the result is computed recursively, as in the brute-force method. Once the result is determined (either true or false), it is stored in memo[index] with currentProduct as the key before being returned. This trade-off of space for time significantly speeds up the computation if there are many overlapping subproblems (i.e., different combinations of numbers leading to the same intermediate product).

• Perform the initial check: if product(nums) != target * target, return false.
• Split the nums array into two halves: leftHalf (from index 0 to N/2 - 1) and rightHalf (from index N/2 to N-1).
• Define a helper function generateProducts to find all possible subset products for a given subarray. This can be done recursively.
• Call generateProducts on leftHalf to get a set of all its subset products, products1.
• Call generateProducts on rightHalf to get a set of its subset products, products2.
• Iterate through each product p1 in products1.
• For each p1, check if target is divisible by p1.
• If it is, calculate the required product p2 = target / p1.
• Check if this p2 exists in the products2 set.
• If a match is found, it means we can form a subset with product target by combining a subset from the left half and a subset from the right half. Return true.
• If the loop completes without finding a match, return false.

This approach begins with the same crucial check of the total product against target * target. The main strategy is to divide and conquer. The nums array is split into two smaller halves.

First, we generate all possible subset products for the first half of the array and store them in a hash set for O(1) average time lookups. The number of subsets for this half is 2^(N/2), which is very manageable for N <= 12.

Next, we process the second half. Instead of storing all its subset products, we generate them one by one. For each subset product p2 from the second half, we calculate the product p1 that we would need from the first half to achieve the target (i.e., p1 = target / p2). We then check if this required p1 exists in our hash set of products from the first half. If we find such a p1, we have successfully found a combination of subsets from both halves that multiply to target, and we can return true.

This method reduces the time complexity from O(2^N) to O(2^(N/2)), which is a significant improvement. For N=12, this changes the number of operations from ~4096 to ~64, making the solution extremely fast.