What is the brute force approach and complexity for Subarray Product Less Than K?

For each start, multiply to the right until the product reaches k with time complexity O(N²) and space complexity O(1).

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

Keep the longest window ending at right with product < k; add its length with optimal time complexity O(N) and space complexity O(1).

Subarray Product Less Than K

Q: How do you identify the pattern for Subarray Product Less Than K?

Look for key signals in the prompt: "number of contiguous subarrays" indicates Counting problem → add something per right end, not a max.; "product strictly less than k" indicates An upper limit. Shrinking the window always helps, so a sliding window can enforce it.; "positive integers (nums[i] ≥ 1)" indicates The product never goes down as the window grows. That monotonic behaviour is what makes the window safe..

Medium·LeetCode #713 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You are given an array of positive integers nums and an integer k.

Count how many contiguous subarrays have a product of all their elements that is strictly less than k.

Input

An integer array nums (every element ≥ 1) and an integer k.

Output

The number of subarrays whose product is < k.

Example 1

Input: nums = [10, 5, 2, 6], k = 100

Output: 8

The 8 subarrays are [10], [5], [2], [6], [10, 5], [5, 2], [2, 6] and [5, 2, 6]. [10, 5, 2] has product exactly 100, which is not strictly less than 100.

Example 2

Input: nums = [1, 2, 3], k = 0

Output: 0

Every product is at least 1, so no product can be less than 0.

Constraints

• 1 ≤ nums.length ≤ 3·10^4
• 1 ≤ nums[i] ≤ 1000
• 0 ≤ k ≤ 10^6

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“number of contiguous subarrays”

→ Counting problem → add something per right end, not a max.

“product strictly less than k”

→ An upper limit. Shrinking the window always helps, so a sliding window can enforce it.

“positive integers (nums[i] ≥ 1)”

→ The product never goes down as the window grows. That monotonic behaviour is what makes the window safe.

Approach Complexity Ladder

Brute Force — grow from every startO(N²)O(1)

For each start, multiply to the right until the product reaches k

Optimal — sliding window on the productO(N)O(1)

Keep the longest window ending at right with product < k; add its length

Brute Force Intuition

Fix a start. Extend the end one step at a time, keeping a running product. Count every subarray while the product stays below k. Stop as soon as it reaches k, because the product can only grow from there.

Recognition rule for this sub-pattern:

Count subarrays that satisfy a rule. "At most": for each right end, every start from left to right works, so add (right − left + 1). "At least": once the window is good, every longer one is too, so add left (or n − right).

Remember this · Sliding Window

Trigger

Contiguous subarray / substring + longest, shortest, maximum, count, or "of size K".

Think

Can I keep a valid range and move it, instead of checking every subarray from scratch?

Key question

When is my current window valid?

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