Leetcode #713: Subarray Product Less Than K

In this guide, we solve Leetcode #713 Subarray Product Less Than K in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Array, Binary Search, Prefix Sum, Sliding Window

Intuition

We are looking for a contiguous region that satisfies a constraint, which is a classic sliding-window signal.

Expanding and shrinking the window lets us maintain validity without restarting the scan.

Approach

Grow the window with a right pointer, and shrink from the left only when the constraint is violated.

Track the best window as you go to keep the solution linear.

Steps:

Expand the right end of the window.
While invalid, move the left end to restore constraints.
Update the best window found.

Example

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.

Python Solution

class Solution:
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
        ans = l = 0
        p = 1
        for r, x in enumerate(nums):
            p *= x
            while l <= r and p >= k:
                p //= nums[l]
                l += 1
            ans += r - l + 1
        return ans

Complexity

The time complexity is $O(n)$ , where $n$ is the length of the array. The space complexity is $O(1)$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

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.

Python Solution

class Solution:
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
        ans = l = 0
        p = 1
        for r, x in enumerate(nums):
            p *= x
            while l <= r and p >= k:
                p //= nums[l]
                l += 1
            ans += r - l + 1
        return ans

Leetcode #713: Subarray Product Less Than K

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #713: Subarray Product Less Than K

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview