What is the time complexity of Subarray Product Less Than K?

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

What data structures are used to solve Subarray Product Less Than K?

This problem uses Array, Binary Search, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Subarray Product Less Than K?

Insight 1: The problem can be efficiently solved using a sliding window technique. Insight 2: The window needs to expand by including right elements and shrink from the left when the product exceeds k.

Which companies ask LeetCode 713. Subarray Product Less Than K in interviews?

This problem is frequently asked at Agoda, Airbnb, Flexport, IBM, Nvidia and 6 more companies.

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Subarray Product Less Than K - LeetCode 713 Solution

Subarray Product Less Than K - Complete Solution Guide

Subarray Product Less Than K is LeetCode problem 713, 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

Given an array of integers nums and an integer k , return the number of contiguous subarrays where the product of all the elements in the subarray is strictly less than k . Example 1: Input: nums = [10,5,2,6], k = 100 Output: 8 Explanation: The 8 subarrays that have product less than 100 are: [10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6] Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k. Example 2: Input: nums = [1,2,3], k = 0 Output: 0 Constraints:

Detailed Explanation

The problem asks us to find the number of contiguous subarrays within a given array of integers (`nums`) whose product is strictly less than a given integer `k`. A contiguous subarray is a sequence of elements that are adjacent in the array. The input consists of an array of integers `nums` and an integer `k`. The output is the number of subarrays that satisfy the given condition (product less than k). The constraints are that the length of `nums` is between 1 and 3 * 10^4, each element in `nums` is between 1 and 1000, and `k` is between 0 and 10^6.

Solution Approach

The provided solution uses the sliding window technique. We maintain a window defined by `left` and `right` pointers. We iterate through the array with the `right` pointer, expanding the window to include the current element. We update the product by multiplying it by the current element. If the product becomes greater than or equal to `k`, we shrink the window from the left until the product is less than `k`. The number of subarrays ending at the current `right` index is then `right - left + 1`. We add this number to the `count`, which accumulates the total number of subarrays that meet the criteria.

Step-by-Step Algorithm

Step 1: Initialize `count` to 0, `product` to 1, and `left` to 0. Handle the edge case where `k <= 1` by returning 0.
Step 2: Iterate through the `nums` array using a `right` pointer.
Step 3: Multiply the current `product` by the element at the `right` index: `product *= nums[right]`.
Step 4: While the `product` is greater than or equal to `k`, divide the `product` by the element at the `left` index and increment the `left` pointer: `product /= nums[left]; left++`.
Step 5: Add the number of subarrays ending at the current `right` index to the `count`: `count += right - left + 1`.
Step 6: After the loop finishes, return the final `count`.

Key Insights

Insight 1: The problem can be efficiently solved using a sliding window technique.
Insight 2: The window needs to expand by including right elements and shrink from the left when the product exceeds k.
Insight 3: We need to handle the edge case where k is less than or equal to 1, as no subarray can have a product less than k in such a scenario.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Sliding Window, Prefix Sum.

Companies

Asked at: Agoda, Airbnb, Flexport, IBM, Nvidia, PayPal, Salesforce, Samsung, ServiceNow, SoFi, Wise.

Subarray Product Less Than K | LeetCode 713

Medium

Array Binary Search Sliding Window Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        if (k <= 1) {
            return 0;
        }
        
        int count = 0;
        int product = 1;
        int left = 0;
        
        for (int right = 0; right < nums.length; right++) {
            product *= nums[right];
            while (product >= k) {
                product /= nums[left];
                left++;
            }
            count += right - left + 1;
        }
        
        return count;
    }
}

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Problem Info

Description

Problem 713: Subarray Product Less Than K - Detailed explanation and solution approaches.

Topics

Array Binary Search Sliding Window Prefix Sum

Companies

AgodaAirbnbFlexportIBMNvidiaPayPalSalesforceSamsung

Stats

Difficulty:Medium

Languages:4

Companies:11

Problem #:713

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

6 steps

1
Step 1: Initialize `count` to 0, `product` to 1, and `left` to 0. Handle the edge case where `k <= 1` by returning 0.
2
Step 2: Iterate through the `nums` array using a `right` pointer.
3
Step 3: Multiply the current `product` by the element at the `right` index: `product *= nums[right]`.
4
Step 4: While the `product` is greater than or equal to `k`, divide the `product` by the element at the `left` index and increment the `left` pointer: `product /= nums[left]; left++`.
5
Step 5: Add the number of subarrays ending at the current `right` index to the `count`: `count += right - left + 1`.
6
Step 6: After the loop finishes, return the final `count`.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because each element of the input array is visited at most twice (once by the right pointer and once by the left pointer). The space complexity is O(1) because we use a constant amount of extra space, irrespective of the input size.

📖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

Brute Force

Generate all possible subarrays and check if their product is less than k. This involves iterating through all possible start and end indices of subarrays.

Time: O(n^2), where n is the length of the input array `nums`.Space: O(1).

Trade-offs: The brute force approach is simple to understand but inefficient for larger arrays. The sliding window approach is much more efficient (O(n)) and is preferred for larger input sizes. Brute force should only be considered for very small input sizes where the simplicity outweighs the performance penalty.

✨

Key Insights

1Insight 1: The problem can be efficiently solved using a sliding window technique.
2Insight 2: The window needs to expand by including right elements and shrink from the left when the product exceeds k.
3Insight 3: We need to handle the edge case where k is less than or equal to 1, as no subarray can have a product less than k in such a scenario.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: `k <= 1`. In this case, no subarray will have a product less than `k`, so the function should return 0. The code explicitly handles this case with the `if k <= 1: return 0` statement.
Edge case 2: `nums` contains large numbers and the product becomes very large, exceeding the maximum integer value. While the problem statement constrains nums[i] to be less than or equal to 1000 and k less than or equal to 10^6, in the C++ solution, using `long long` for the product variable is good practice to prevent potential integer overflow, though the constraints should prevent it. The python and java implementations handle this implicitly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of subarrays. The correct formula is `right - left + 1`, not just `right - left` or some other variation.

✗Common mistake 2: Failing to handle the case when `k <= 1`. This leads to incorrect results or infinite loops in some cases.

✗Common mistake 3: Integer overflow. Multiplying large numbers together can cause the product to exceed the maximum integer value, leading to incorrect results. While problem constraints aim to prevent this, it is good to be mindful of the data types in use.

✅

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. Consider a stream of website visits, where each visit is assigned a 'weight' based on engagement. We can use this algorithm to find the number of time windows (subarrays of visits) where the total 'engagement' (product of weights) is below a certain threshold.

🏦Real-world scenario 2: Financial data analysis. Given a series of stock prices, identify time periods (subarrays) where the product of percentage changes in price is below a risk tolerance level (k). This helps in identifying periods of low volatility.

💡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: November 21, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Subarray Product Less Than K"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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