# 【LeetCode刷題】【演算法學習筆記】713. Subarray Product Less Than K
* 學習時間：2025/02/0()
* **題目連結（美服）：** [713. Subarray Product Less Than K](https://leetcode.cn/problems/subarray-product-less-than-k/solutions/1320871/jian-dan-yi-dong-xiang-xi-zhu-jie-shuang-jvy3/)
* **題目連結（國服）：**[713. 乘积小于 K 的子数组](https://leetcode.cn/problems/subarray-product-less-than-k/description/)
* **題目標籤：** ==陣列== ==雙指針== ==排序==

### A. 題目
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>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.
Example 2:
Input: nums = [1,2,3], k = 0
Output: 0

### B. 題意（中文）
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>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

### C. 參考程式碼
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:::spoiler Solution

# reference:  [灵茶山艾府](https://leetcode.cn/problems/subarray-product-less-than-k/solutions/1959538/xia-biao-zong-suan-cuo-qing-kan-zhe-by-e-jebq/) 

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:::spoiler Thinking（思路）
reference: [力扣官方题解](https://leetcode.cn/problems/subarray-product-less-than-k/solutions/1463527/cheng-ji-xiao-yu-k-de-zi-shu-zu-by-leetc-92wl/) [灵茶山艾府](https://leetcode.cn/problems/subarray-product-less-than-k/solutions/1959538/xia-biao-zong-suan-cuo-qing-kan-zhe-by-e-jebq/)

#### 官方題解

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#### 靈神題解

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:::spoiler Note（筆記）
**複雜度**
* Time complexity:
* Space complexity:
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