In Kadane's Algorithm, a larger earlier value always gives a larger new value. A product does not follow this rule. Multiplying by a negative x reverses the order: the smallest earlier product becomes the largest new product.

So the state has two values for the subarrays that end at the current number. hi is the largest product and lo is the smallest product. lo is often negative, and one more negative x turns it into the next hi.

Example: for nums = [-2, 3, -4], after 3 the pair is hi = 3 and lo = -6. Then x = -4 gives lo * x = 24. The answer is 24. Without lo, the answer would be 3.

A subarray that ends at x is x alone, or an earlier subarray that ends just before x, followed by x.

For the second form, the product is p * x, where p is the product of the earlier subarray. If x > 0, the largest result comes from the largest p, which is hi. If x < 0, it comes from the smallest p, which is lo. If x = 0, every result is 0.

So max(x, hi * x, lo * x) is the largest product that ends at x, and min(x, hi * x, lo * x) is the smallest. The tuple assignment computes both from the old hi and lo.

A zero resets the state. After x = 0, both hi and lo are 0. The next number starts a new run through the candidate x alone.

Trace for nums = [2, 3, -2, 4], as x, hi, lo, best:

• 2, 2, 2, 2 (the start).
• 3, 6, 3, 6.
• -2, -2, -12, 6.
• 4, 4, -48, 6.

The answer is 6, from [2, 3]. Trace for nums = [-2, 0, -1, 3, -4]: best is 12, from the run [-1, 3, -4]. The zero ends the earlier run, and the next run starts at -1.

The time is O(N), with three products per number. The space is O(1).

Products grow fast. With 40 copies of 2, the result is 1,099,511,627,776. Python integers have no limit. In a language with 32-bit or 64-bit integers, check the size limit, which LeetCode states for each problem.

A second method uses prefix products. Scan from the left with a running product, and again from the right. After a zero, restart the running product at 1. The answer is the largest product seen in either scan. The reason is that the best subarray either has an even number of negative numbers, or it drops a negative number at one end. One of the two scans reaches it. This gives the same results as max_product on random arrays with zeros and negative numbers.

The variant Maximum Length of Subarray With Positive Product (LC 1567) keeps lengths and not products. Keep pos, the longest run ending here with a positive product, and neg, the longest with a negative product. A negative x swaps pos and neg, and a zero resets both to 0. This is the same swap idea as hi and lo.