What is the main pattern for solving the Subarray Product Less Than K problem?

The key pattern for this problem is binary search over the valid answer space, combined with a sliding window approach to efficiently track subarray products.

How does the sliding window technique work in this problem?

The sliding window technique expands the window by moving the right pointer, and when the product exceeds k, it shrinks the window from the left to maintain valid subarrays.

Can brute-force solutions work for Subarray Product Less Than K?

Brute-force solutions are inefficient for large arrays, as they would require checking every possible subarray, which is computationally expensive.

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

The time complexity of the optimal solution is O(n * log(n)) due to binary search combined with a sliding window to check subarrays.

How can GhostInterview help with understanding binary search in this problem?

GhostInterview provides practice and explanations to deepen understanding of how binary search can be applied to subarrays and their products efficiently.

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Medium

auto_awesome二分·搜索·答案·空间

LeetCode 题解工作台

乘积小于 K 的子数组

给你一个整数数组 nums 和一个整数 k ，请你返回子数组内所有元素的乘积严格小于 k 的连续子数组的数目。示例 1：输入： nums = [10,5,2,6], k = 100 输出： 8 解释： 8 个乘积小于 100 的子数组分别为：[10]、[5]、[2]、[6]、[10,5]、[5,…

数组二分查找滑动窗口前缀和

题目描述

给你一个整数数组 nums 和一个整数 k ，请你返回子数组内所有元素的乘积严格小于 k 的连续子数组的数目。

示例 1：

输入：nums = [10,5,2,6], k = 100
输出：8
解释：8 个乘积小于 100 的子数组分别为：[10]、[5]、[2]、[6]、[10,5]、[5,2]、[2,6]、[5,2,6]。
需要注意的是 [10,5,2] 并不是乘积小于 100 的子数组。

示例 2：

输入：nums = [1,2,3], k = 0
输出：0

提示:

1 <= nums.length <= 3 * 10⁴
1 <= nums[i] <= 1000
0 <= k <= 10⁶

lightbulb

解题思路

方法一：双指针

我们可以用双指针维护一个滑动窗口，窗口内所有元素的乘积小于 $k$ 。

定义两个指针 $l$ 和 $r$ 分别指向滑动窗口的左右边界，初始时 $l = r = 0$ 。我们用一个变量 $p$ 记录窗口内所有元素的乘积，初始时 $p = 1$ 。

每次，我们将 $r$ 右移一位，将 $r$ 指向的元素 $x$ 加入窗口，更新 $p = p \times x$ 。然后，如果 $p \geq k$ ，我们循环地将 $l$ 右移一位，并更新 $p = p \div \text{nums}[l]$ ，直到 $p < k$ 或者 $l \gt r$ 为止。这样，以 $r$ 结尾的、乘积小于 $k$ 的连续子数组的个数即为 $r - l + 1$ 。然后我们将答案加上这个数量，并继续移动 $r$ ，直到 $r$ 到达数组的末尾。

时间复杂度 $O(n)$ ，其中 $n$ 为数组的长度。空间复杂度 $O(1)$ 。

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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

speed

复杂度分析

指标	值
时间	O(n \cdot \log(n))
空间	O(n)

psychology

面试官常问的追问

外企场景

question_mark
Understand how binary search over the valid space works with respect to subarrays.
question_mark
Candidate uses efficient sliding window or binary search techniques instead of brute-force methods.
question_mark
Look for ability to balance correctness with time complexity optimizations.

warning

常见陷阱

外企场景

error
Failing to adjust the window size properly when the product exceeds k.
error
Using a brute-force approach that evaluates all possible subarrays instead of leveraging efficient algorithms.
error
Misunderstanding the sliding window or binary search approach, resulting in suboptimal performance.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Change the constraint for k to be very large, testing how the solution scales with larger values of k.
arrow_right_alt
Vary the array length significantly to assess performance under different input sizes.
arrow_right_alt
Introduce negative numbers in the array and adjust the problem to handle those.

help

常见问题

外企场景

继续练习

#862 和至少为 K 的最短子数组 #1004 最大连续1的个数 III #209 长度最小的子数组