How do you solve Subarray Product Less Than K on LeetCode?

Subarray Product Less Than K (LeetCode #713) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using a sliding window allows for efficient counting of subarrays without recalculating products from scratch.

What is the brute force solution for Subarray Product Less Than K?

The brute force approach for Subarray Product Less Than K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible subarray and calculates its product to see if it's less than k. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Subarray Product Less Than K?

Subarray Product Less Than K uses the following concepts: Array, Binary Search, Sliding Window, Prefix Sum. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Subarray Product Less Than K?

Always consider edge cases, such as k being 0 or negative. Explain your thought process clearly while coding, especially when using pointers. Practice explaining the sliding window technique, as it's commonly used in array problems.

What are common mistakes when solving Subarray Product Less Than K?

Failing to handle cases where k is less than or equal to 1, which results in no valid subarrays. Not updating the left pointer correctly when the product exceeds k.

#713

Subarray Product Less Than K

Medium

ArrayBinary SearchSliding WindowPrefix SumSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses a sliding window approach to maintain a product of the current subarray. This allows us to efficiently count valid subarrays without recalculating products from scratch.

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Algorithm

4 steps

1Step 1: Initialize two pointers (left and right) and a product variable to 1.
2Step 2: Expand the right pointer to include elements in the product until it reaches or exceeds k.
3Step 3: For each position of the right pointer, count how many subarrays end at right and start from left to right.
4Step 4: If the product exceeds k, move the left pointer to reduce the product.

solution.py11 lines

1def numSubarrayProductLessThanK(nums, k):
2    count = 0
3    left = 0
4    product = 1
5    for right in range(len(nums)):
6        product *= nums[right]
7        while left <= right and product >= k:
8            product //= nums[left]
9            left += 1
10        count += right - left + 1
11    return count

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Complexity note: The time complexity is O(n) because each element is processed at most twice (once by the right pointer and once by the left pointer). The space complexity is O(1) since we only use a few variables for counting and tracking indices.

1Using a sliding window allows for efficient counting of subarrays without recalculating products from scratch.
2The relationship between the left and right pointers helps maintain the product under the constraint.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.