1352. Product of the Last K Numbers - Detailed Explanation

Explanation of the code: In both Python and Java optimized versions, the logic is the same:

• We use a list (prefixProducts or prefix) to store cumulative products. It starts with [1].
• On add(num): if num is zero, we reset the list to [1]. If num is non-zero, we append last_element * num to the list.
• On getProduct(k): if k is at least the length of the prefix list, we return 0 (meaning a zero was encountered in that range). Otherwise, we do the division of the last prefix by the prefix value from k elements ago to get the product of the last k numbers.

Complexity Analysis:

• Brute Force Approach:

  • Time Complexity: add() is O(1) (just appending to a list). getProduct(k) is O(k) because it iterates through k elements to compute the product. In the worst case, k could be as large as the number of elements in the stream. If we have n operations and many of them are getProduct of large segments, the worst-case time could be on the order of O(n * k). For example, in the worst scenario with 40,000 operations, each getProduct(40000) would multiply 40,000 numbers – this would be extremely slow in a language like Python (and even in Java it could be problematic).

  • Space Complexity: O(n) to store all the numbers in the stream (where n is the total numbers added).

• Optimized Prefix-Product Approach:

  • Time Complexity: Both add() and getProduct() are O(1) on average. Each add does a constant amount of work (multiplication and append, or resetting the list). Each getProduct does a constant amount of work (just a couple of array index lookups and one division). This holds true even for large k because we aren’t looping through those k elements, we’re using the stored prefix products. (One subtle detail: if you add a lot of zeros, each zero causes a reset of the prefix list. However, each addition operation is still constant time; we are not carrying out any long loops during resets, just reinitializing a list.)
  • Space Complexity: O(n) in the worst case, where n is the number of elements added. In the worst scenario (no zeros at all), the prefixProducts list will store a prefix product for every element added (plus the initial 1). If zeros occur, the space might be slightly less because we reset, but overall, we still at most store one prefix value per element of the stream. The storage is proportional to the number of elements added (max 40,000). This is efficient for the given input size.

Why O(1) queries matter: The optimized approach ensures that even if we have many queries asking for very large segments, each query is fast. This makes the solution efficient even when k is large or when we have many operations. The brute force approach would not scale well in those cases.

Step-by-Step Walkthrough (Optimized Solution):

Let’s do a detailed walkthrough of the optimized solution using Example 1 ([3, 0, 2, 5, 4, 8] with queries in between) to see how the internal state changes after each operation:

• Start: prefixProducts = [1] (starts with 1 as a neutral product, representing an "empty" product).

• Operation 1: add(3)

  • New number is 3 (non-zero).
  • Compute new prefix = last prefix * 3 = 1 * 3 = 3.
  • Append it: prefixProducts = [1, 3].
  • (This means the product of the last 1 number is 3, which matches the stream [3].)

• Operation 2: add(0)

  • New number is 0.
  • Reset: prefixProducts = [1].
  • (We reset because any product that includes this 0 will be 0. After this reset, the prefixProducts now only accounts for numbers after the zero. The stream so far is [3, 0], but we dropped the product for the prefix that included 3 because it’s irrelevant for any segment after the 0.)

• Operation 3: add(2)

  • New number is 2 (non-zero).
  • New prefix = 1 * 2 = 2.
  • prefixProducts = [1, 2].
  • (Now the stream after the last zero is [2], and the product of [2] is 2.)

• Operation 4: add(5)

  • New number is 5.
  • New prefix = 2 * 5 = 10.
  • prefixProducts = [1, 2, 10].
  • (The stream after the last zero is [2, 5], product is 10.)

• Operation 5: add(4)

  • New number is 4.
  • New prefix = 10 * 4 = 40.
  • prefixProducts = [1, 2, 10, 40].
  • (The stream after the last zero is [2, 5, 4], product is 40.)

• Operation 6: getProduct(2) – product of last 2 numbers.

  • We check k=2 against len(prefixProducts)=4. Since 2 < 4, we have enough data.
  • P_last = prefixProducts[-1] = 40 (product of [2,5,4]).
  • P_before = prefixProducts[-1 - 2] = prefixProducts[-3] = 2 (this is the product of [2], which is the part before the last 2 numbers in the current segment).
  • Result = 40 / 2 = 20.
  • (The last 2 numbers in the stream are [5,4]. Indeed 5×4 = 20.)

• Operation 7: getProduct(3) – product of last 3 numbers.

  • Check k=3 vs len(prefixProducts)=4. 3 < 4, okay.
  • P_last = 40.
  • P_before = prefixProducts[-1 - 3] = prefixProducts[-4] = 1 (this is the initial 1, representing product before the current segment start).
  • Result = 40 / 1 = 40.
  • (Last 3 numbers are [2,5,4]; 2×5×4 = 40.)

• Operation 8: getProduct(4) – product of last 4 numbers.

  • Check k=4 vs len(prefixProducts)=4. Here 4 >= 4, which means we’re asking for a segment length that is not fully covered by our current prefix segment length. In fact, k=4 would include one number prior to the last zero (since our current segment after the last zero has only 3 numbers).
  • Since k >= len(prefixProducts), we return 0 without further calculation.
  • (This correctly matches the fact that the last 4 numbers [0,2,5,4] include a zero, so the product is 0.)

• Operation 9: add(8)

  • New number is 8.
  • Current last prefix was 40, so new prefix = 40 * 8 = 320.
  • prefixProducts = [1, 2, 10, 40, 320].
  • (The segment after the last zero is now [2,5,4,8], product 320.)

• Operation 10: getProduct(2) – product of last 2 numbers.

  • Check k=2 vs len(prefixProducts)=5. 2 < 5, good.
  • P_last = 320.
  • P_before = prefixProducts[-1 - 2] = prefixProducts[-3] = 10.
  • Result = 320 / 10 = 32.
  • (Last 2 numbers are [4,8]; 4×8 = 32.)

Through this walkthrough, you can see how we only ever do a single multiplication when adding a number (or a reset if it’s zero), and a single division when answering a query (plus a check for zero case). This is why the optimized solution is efficient.

Common Mistakes & Edge Cases:

• Forgetting to handle zero properly: The most common pitfall is to use a prefix product approach but not reset on zero. If you kept multiplying even when a zero appears, the prefix product would become 0 and remain 0 for all subsequent elements, which breaks the logic for future queries. It’s crucial to reset the prefix product sequence after a zero, or equivalently, handle segments separated by zeros. If you don’t, you might end up dividing by zero or returning incorrect results. Our solution handles this by resetting the prefixProducts list to [1] whenever a zero is added.

• Using the wrong condition for zero-range check: In getProduct(k), we used if k >= len(prefixProducts): return 0. This condition works because len(prefixProducts) is one more than the number of actual elements in the current segment (due to the initial 1). Another way to write this is if k > (number of elements since last zero): return 0. Off-by-one errors here can cause mistakes. If you mistakenly used > instead of >= or miscomputed the comparison, you might incorrectly handle the case when k exactly equals the current segment length. We chose >= len(prefixProducts) (or in Java, k >= prefix.size()) deliberately to cover the case where k is exactly the length of the prefix list (meaning the range reaches into the previous segment that contained a zero).

• Edge case – all zeros: If the stream is something like [0, 0, 0, ...], every add(0) will reset the prefix list. Any getProduct(k) will find that len(prefixProducts) = 1 (since it resets each time), and if k >= 1 it will return 0. This correctly handles queries on a sequence of zeros. Our approach gracefully handles this scenario: after each zero, the next query will return 0 as long as the query length is >= 1. (Also, note the problem guarantee that "current list has at least k numbers" means we wouldn’t query more elements than exist. If the stream is all zeros and you query for that many zeros, the result is 0.)

• Edge case – querying all elements: If k equals the total number of elements in the stream, the getProduct(k) should return the product of the entire stream. In the optimized approach, this will return 0 if there has been any zero in the stream at all (because the prefix segment length would be smaller than the total stream length). If there have been no zeros, it will correctly return the full product. For example, if the stream is [1,2,3,4] and you call getProduct(4), the prefix list would be [1,1,2,6,24] and the result would be 24/1 = 24. If the stream was [1,2,0,3] and you call getProduct(4), the prefix list after the last zero covers only [3] plus initial 1, which has length 2, so k >= len(prefixProducts) -> returns 0, which is correct (because the entire stream of 4 includes a zero).

• Data type considerations: In languages like Java or C++, be mindful of potential overflow if you were not guaranteed by the problem that the product fits in 32 bits. In this problem, it’s guaranteed, but if it wasn’t, you might need to use a larger data type (e.g., 64-bit) or modulo arithmetic. In Python, integers can grow arbitrarily large so overflow isn’t a concern in implementation (though it could become slow if numbers were huge without the given guarantee). Always heed the problem constraint about product fitting in 32-bit, which simplifies the implementation.

• Using division vs. multiplication: Some might try to maintain a running product of only the last k elements by dividing out the oldest element when a new one comes in (like a sliding window). This becomes tricky when zeros are involved and because k can change for each query (we don’t have a fixed window size – each query might ask for a different k). The prefix product method with reset is a cleaner way to handle it. Trying to manually drop the oldest element’s contribution when the window moves (which is how you might handle a moving window of fixed size) is not directly applicable here, since the "window" size varies per query and the stream is ever-growing rather than sliding with a fixed size.

Alternative Variations:

• Product of last k numbers (fixed k, streaming): If the problem were framed as “always return the product of the last k numbers after each insertion” (with a fixed k), one could maintain a running product in a queue of size k. However, you would still need to handle zeros (which would invalidate the entire product until the zero leaves the window). The given problem is more flexible (k can vary per query), and that’s why the prefix approach is used instead of a fixed-size window approach.

• Sum of the last k numbers: A similar design problem could ask for the sum of the last k numbers. In that case, you could use a prefix sum array or a sliding window sum. In fact, LeetCode has a problem “Moving Average from Data Stream” where you are asked to find the average of the last k elements in a stream. The typical solution uses a queue of size k and a running sum. Sum is easier than product because you don’t have the zero issue (a zero just adds 0 to sum, which is harmless), and you can subtract the element that goes out of the window when the window exceeds size k. The product scenario is trickier because of multiplication and zeros, which is why resetting on zero is crucial.

• Range product queries on an array: The concept of prefix products can be extended. For a static array (not a stream), if you need to answer many queries of the product of elements between indices i and j, you can preprocess prefix products for the array. Then the product of a range [i, j] is prefix[j] / prefix[i-1] (assuming no zeros, or handling zeros by segmenting the array). If updates are happening in the middle of the array, more complex data structures like segment trees or Fenwick trees (binary indexed trees) might be used. But because our problem only appends at the end, the prefix method is very efficient.

• Handling negative numbers or modulo: This problem restricted num to 0 <= num <= 100 (no negatives). If negatives were allowed, the logic still works – a negative number just multiplies into the prefix. The only difference is the product could be negative or positive. If the problem asked for the product modulo some number, you could take the prefix products modulo that number (though then zero handling would mean taking mod after division, which requires modular inverses if not handling zero separately – in short, it complicates things but the idea of prefix products still applies, usually one would avoid division and instead maintain inverses or do segment tree for modulo case). These variations are beyond the scope of the basic problem but are good thought exercises on how the approach adapts.

• Using data structures like Segment Tree/Fenwick Tree: These structures can handle range product queries and point updates more generally. However, for this specific problem, the prefix approach is simpler and perfectly efficient. Segment trees would be useful if, say, you needed to get the product of any arbitrary range (not just suffixes) or if you needed to update arbitrary positions (not just append at end). Since here we always add at the end and query suffixes, the prefix method is ideal.

In summary, the technique used here (prefix products with resets on zero) is a specialized application of the prefix-sum/prefix-product concept for an append-only array with queries on the suffix. This idea can be adapted or appears in various forms in other problems, especially where we want to quickly compute results of contiguous segments of a list.

Related Problems for Further Practice:

To strengthen your understanding, you can practice these related problems and concepts:

• LeetCode 238. Product of Array Except Self: Although not a streaming problem, it uses prefix and suffix products to compute result without directly multiplying every time. It shows how prefix products can be used in array problems.

• LeetCode 303. Range Sum Query - Immutable: This is about prefix sums. You preprocess an array so that any query for the sum of a range can be answered in O(1) time by subtracting prefix sums. (Analogy: prefix sum for addition is like prefix product for multiplication.)

• LeetCode 304. Range Sum Query 2D - Immutable: Extension of the above to 2D grids using 2D prefix sums.

• LeetCode 346. Moving Average from Data Stream: Design a class to get the moving average of the last k values in a stream. It’s similar in spirit but for sums/averages; you can practice maintaining a running sum and a queue of fixed size k.

• LeetCode 1352. Product of the Last K Numbers: (The problem we just solved.) It might be worth reimplementing it in another language or revisiting after some time to ensure the concept sticks.

• LeetCode 1172. Dinner Plate Stacks (or other design problems): Any design question where you have to maintain a data structure and perform operations like add and get queries can help you practice careful consideration of time complexity. (Dinner Plate Stacks is more complex, but it’s an example of designing around constraints.)

• LeetCode 362. Design Hit Counter: Another design problem where you have to return the number of hits in the past 5 minutes. While the specifics are different (and involves time-based sliding windows), it’s practice for designing efficient streaming calculations.