Count of Subarrays with Product Less Than K

Q: Which edge cases should I test for Count of Subarrays with Product Less Than K?

k <= 1 (no product of positive integers is below 1); Single elements already = k.

Medium Arrays & Hashing Sliding Window · Variable Original on LeetCode

Count of Subarrays with Product Less Than K is a medium arrays & hashing problem solved with the sliding window · variable pattern. The best approach, optimal (variable sliding window), runs in O(n) time and O(1) space. Below are 2 approaches in Java, from brute force up.

Problem

Given an array of positive integers nums and an integer k, count the contiguous subarrays whose product is strictly less than k.

Examples

Example 1

Input: nums = [4, 2, 6, 1], k = 20
Output: 8
Why: Singles [4] [2] [6] [1], pairs [4,2] [2,6] [6,1], and [2,6,1] with product 12. [4,2,6] is 48, too big.

Example 2

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

Constraints

1 <= nums.length <= 3 * 10^4
1 <= nums[i] <= 1000
0 <= k <= 10^6

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Brute force	`O(n²)`	`O(1)`
Optimal (variable sliding window)	`O(n)`	`O(1)`

1Brute force

TimeO(n²)

SpaceO(1)

For each start, extend while the product stays below k, counting each subarray.

For each i: p = 1; for j from i: p *= nums[j]; stop when p >= k, else count++.

Java

class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            long p = 1;
            for (int j = i; j < nums.length; j++) {
                p *= nums[j];
                if (p >= k) break;
                count++;
            }
        }
        return count;
    }
}

2Optimal (variable sliding window)

TimeO(n)

SpaceO(1)

Grow the window on the right and multiply in nums[r]. While the product is k or more, divide out nums[l] and move l. Every subarray that ends at r and starts anywhere in [l, r] is valid, which adds r - l + 1.

If k <= 1 return 0.
For each r: prod *= nums[r]; while prod >= k: prod /= nums[l++].
count += r - l + 1.

Java

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

Edge cases to test

k <= 1 (no product of positive integers is below 1)
Single elements already >= k

Hints

Hint 1

For a window [l, r] whose product is below k, how many valid subarrays end at r?

FAQ

What is the best time complexity for Count of Subarrays with Product Less Than K?

Optimal (variable sliding window) runs in O(n) time and O(1) extra space.

Which pattern does Count of Subarrays with Product Less Than K use?

It is a arrays & hashing problem that uses the sliding window · variable pattern. Other problems with the same pattern: Minimum Size Subarray Sum.

Is there a brute force solution for Count of Subarrays with Product Less Than K?

Yes. Brute force takes O(n²) time and O(1) space. For each start, extend while the product stays below k, counting each subarray.

Which edge cases should I test for Count of Subarrays with Product Less Than K?

k <= 1 (no product of positive integers is below 1); Single elements already = k.