What is the time complexity of Product of the Last K Numbers?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Product of the Last K Numbers?

This problem uses Array, Math, Design, Data Stream, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Product of the Last K Numbers?

Insight 1: Prefix products allow calculating the product of a range in O(1) time. Insight 2: The presence of zero significantly impacts the calculation since any sub-array containing zero will have a product of zero.

Which companies ask LeetCode 1352. Product of the Last K Numbers in interviews?

This problem is frequently asked at Target, Tekion.

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Product of the Last K Numbers - LeetCode 1352 Solution

Product of the Last K Numbers - Complete Solution Guide

Product of the Last K Numbers is LeetCode problem 1352, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Design an algorithm that accepts a stream of integers and retrieves the product of the last k integers of the stream. Implement the ProductOfNumbers class: ProductOfNumbers() Initializes the object with an empty stream. void add(int num) Appends the integer num to the stream. int getProduct(int k) Returns the product of the last k numbers in the current list. You can assume that always the current list has at least k numbers. The test cases are generated so that, at any time, the product of any

Detailed Explanation

The problem requires designing a data structure (ProductOfNumbers class) that can efficiently handle a stream of integers. It needs to support two operations: `add(num)`, which appends a new integer to the stream, and `getProduct(k)`, which returns the product of the last `k` numbers added to the stream. A crucial constraint is that the product of any contiguous sequence will fit within a 32-bit integer. The follow-up asks to achieve O(1) time complexity for both `add` and `getProduct` methods.

Solution Approach

The solution uses a list (or vector/array) called `products` to store the prefix products. `products[i]` holds the product of all numbers added so far up to index `i`. When a new number `num` is added, if it's 0, the entire `products` list is reset to `[1]`. Otherwise, the product `products[-1] * num` is appended to the `products` list. When `getProduct(k)` is called, the product of the last `k` numbers can be calculated in O(1) time by dividing the last element of the `products` list by the element `k` positions before the end. A special case is handled if `k` is greater than or equal to the current size of `products`, which means that some previous element had a '0', so it returns 0.

Step-by-Step Algorithm

Step 1: Initialize a list `products` with `[1]`. The initial value '1' handles cases where getProduct is called before any numbers are added and simplifies the calculations.
Step 2: In `add(num)`, check if `num` is 0. If it is, clear the `products` list and reset it to `[1]` because any subsequent product including this zero will be zero. If `num` is not zero, append `products[-1] * num` to the `products` list.
Step 3: In `getProduct(k)`, check if `k` is greater than or equal to the size of the `products` list. If it is, return 0, implying that there was a zero within the last `k` elements added.
Step 4: Otherwise, return `products[-1] // products[-(k + 1)]` (Python) or `products.get(products.size() - 1) / products.get(products.size() - (k + 1))` (Java/C++). This calculates the product of the last `k` numbers by dividing the cumulative product by the product of all elements before the last `k` elements.

Key Insights

Insight 1: Prefix products allow calculating the product of a range in O(1) time.
Insight 2: The presence of zero significantly impacts the calculation since any sub-array containing zero will have a product of zero.
Insight 3: When a zero is encountered, the accumulated product series must be reset to handle subsequent `getProduct` requests properly.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(n)

Topics

This problem involves: Array, Math, Design, Data Stream, Prefix Sum.

Companies

Asked at: Target, Tekion.

Product of the Last K Numbers | LeetCode 1352

Medium

Array Math Design Data Stream Prefix Sum

Time: O(1)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class ProductOfNumbers {

    private List<Integer> products;

    public ProductOfNumbers() {
        products = new ArrayList<>();
        products.add(1);
    }
    
    public void add(int num) {
        if (num == 0) {
            products = new ArrayList<>();
            products.add(1);
        } else {
            products.add(products.get(products.size() - 1) * num);
        }
    }
    
    public int getProduct(int k) {
        if (k >= products.size()) {
            return 0;
        } else {
            return products.get(products.size() - 1) / products.get(products.size() - (k + 1));
        }
    }
}

/**
 * Your ProductOfNumbers object will be instantiated and called as such:
 * ProductOfNumbers obj = new ProductOfNumbers();
 * obj.add(num);
 * int param_2 = obj.getProduct(k);
 */

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Problem Info

Description

Problem 1352: Product of the Last K Numbers - Detailed explanation and solution approaches.

Topics

Array Math Design Data Stream Prefix Sum

Companies

TargetTekion

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1352

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

4 steps

1
Step 1: Initialize a list `products` with `[1]`. The initial value '1' handles cases where getProduct is called before any numbers are added and simplifies the calculations.
2
Step 2: In `add(num)`, check if `num` is 0. If it is, clear the `products` list and reset it to `[1]` because any subsequent product including this zero will be zero. If `num` is not zero, append `products[-1] * num` to the `products` list.
3
Step 3: In `getProduct(k)`, check if `k` is greater than or equal to the size of the `products` list. If it is, return 0, implying that there was a zero within the last `k` elements added.
4
Step 4: Otherwise, return `products[-1] // products[-(k + 1)]` (Python) or `products.get(products.size() - 1) / products.get(products.size() - (k + 1))` (Java/C++). This calculates the product of the last `k` numbers by dividing the cumulative product by the product of all elements before the last `k` elements.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The `add` method only performs a constant number of operations, regardless of the input size. Similarly, the `getProduct` method only involves a couple of list lookups and a division, also independent of the input size. The space complexity is O(n) because the `products` list grows linearly with the number of calls to `add`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Store the last K numbers directly

Instead of prefix products, store the last K numbers in a queue or circular buffer. In `getProduct(k)`, calculate the product of the numbers in the buffer.

Time: O(k) for `getProduct` and O(1) for `add`.Space: O(K) if K is a maximum limit on K in getProduct() or O(n) if no limit specified.

Trade-offs: This approach is simpler to implement if space isn't a concern, but it sacrifices the O(1) time complexity of `getProduct` for O(k). It's suitable if `k` is expected to be small. It avoids the need for division, which could be a benefit in some contexts.

✨

Key Insights

1Insight 1: Prefix products allow calculating the product of a range in O(1) time.
2Insight 2: The presence of zero significantly impacts the calculation since any sub-array containing zero will have a product of zero.
3Insight 3: When a zero is encountered, the accumulated product series must be reset to handle subsequent `getProduct` requests properly.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Input `num` is 0. When `num` is 0, the `products` list is reset to `[1]`. This correctly handles the situation where a zero appears in the stream, ensuring future `getProduct` calls correctly return 0 if the last `k` elements include the zero.
Edge case 2: `k` is larger than the number of elements in the stream (or, more precisely, since the last zero). The solution checks if `k >= len(products)`. If it is, it means that there was a '0' element within the last 'k' elements, so it returns 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the zero case correctly. Failing to reset the prefix product when a zero is encountered leads to incorrect results when `getProduct` is called on ranges that include the zero.

✗Common mistake 2: Integer overflow. The problem statement explicitly says the product fits in a 32-bit integer, but it's still a good practice to be mindful of potential overflows when dealing with products, especially if the constraints were different.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing website traffic data. Imagine tracking the number of clicks on a website over time. You might want to quickly calculate the 'product' of clicks over the last 'k' time intervals (e.g., days or weeks). Though a simple sum would be more usual, a product could represent a measure of accelerating popularity (if clicks multiply together).

🏦Real-world scenario 2: Financial data analysis. Analyzing stock prices or trading volumes. While direct multiplication is less common, similar prefix sum/product techniques can be adapted to compute moving averages or cumulative returns over specified periods which can be used to compute more sophisticated trading signals.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Product of the Last K Numbers"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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