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feat: add solutions to lc problem: No.0713
No.0713.Subarray Product Less Than K
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‎README.md

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- [区间加法](/solution/0300-0399/0370.Range%20Addition/README.md) - 前缀和、差分
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- [用邮票贴满网格图](/solution/2100-2199/2132.Stamping%20the%20Grid/README.md) - 二维前缀和、二维差分
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- [无重复字符的最长子串](/solution/0000-0099/0003.Longest%20Substring%20Without%20Repeating%20Characters/README.md) - 双指针、哈希表
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- [乘积小于 K 的子数组](/solution/0700-0799/0713.Subarray%20Product%20Less%20Than%20K/README.md) - 双指针
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- [位 1 的个数](/solution/0100-0199/0191.Number%20of%201%20Bits/README.md) - 位运算、lowbit
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- [合并区间](/solution/0000-0099/0056.Merge%20Intervals/README.md) - 区间合并
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‎README_EN.md

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@@ -45,6 +45,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
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- [Range Addition](/solution/0300-0399/0370.Range%20Addition/README_EN.md) - Prefix sum, Difference array
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- [ Stamping the Grid](/solution/2100-2199/2132.Stamping%20the%20Grid/README_EN.md) - Prefix sum, Difference array
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- [Longest Substring Without Repeating Characters](/solution/0000-0099/0003.Longest%20Substring%20Without%20Repeating%20Characters/README_EN.md) - Two pointers, Hash table
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- [Subarray Product Less Than K](/solution/0700-0799/0713.Subarray%20Product%20Less%20Than%20K/README_EN.md) - Two pointers
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- [Number of 1 Bits](/solution/0100-0199/0191.Number%20of%201%20Bits/README_EN.md) - Bit manipulation, Lowbit
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- [Merge Intervals](/solution/0000-0099/0056.Merge%20Intervals/README_EN.md) - Merge intervals
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‎solution/0000-0099/0003.Longest Substring Without Repeating Characters/README.md

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时间复杂度 O(n)，其中 n 表示字符串 s 的长度。
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双指针算法模板：
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```java
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for (int i = 0, j = 0; i < n; ++i) {
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while (j < i && check(j, i)) {
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++j;
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}
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// 具体问题的逻辑
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}
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```
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<!-- tabs:start -->
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### **Python3**

‎solution/0700-0799/0713.Subarray Product Less Than K/README.md

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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **C++**
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```cpp
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class Solution {
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public:
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int numSubarrayProductLessThanK(vector<int>& nums, int k) {
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int left = 0, right;
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long mul = 1;
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int count = 0;
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for (right = 0; right < nums.size(); right++) {
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mul *= nums[right];
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while(left <= right && mul >= k) {
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mul /= nums[left++];
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}
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**方法一：双指针**
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count += right >= left? right - left + 1: 0;
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}
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双指针算法模板：
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return count;
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```java
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for (int i = 0, j = 0; i < n; ++i) {
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while (j < i && check(j, i)) {
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++j;
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}
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};
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// 具体问题的逻辑
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}
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```
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时间复杂度 O(n)。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
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ans, s, j = 0, 1, 0
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for i, v in enumerate(nums):
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s *= v
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while j <= i and s >= k:
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s //= nums[j]
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j += 1
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ans += i - j + 1
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int numSubarrayProductLessThanK(int[] nums, int k) {
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int ans = 0;
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for (int i = 0, j = 0, s = 1; i < nums.length; ++i) {
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s *= nums[i];
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while (j <= i && s >= k) {
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s /= nums[j++];
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}
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ans += i - j + 1;
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int numSubarrayProductLessThanK(vector<int>& nums, int k) {
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int ans = 0;
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for (int i = 0, j = 0, s = 1; i < nums.size(); ++i)
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{
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s *= nums[i];
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while (j <= i && s >= k) s /= nums[j++];
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ans += i - j + 1;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func numSubarrayProductLessThanK(nums []int, k int) int {
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ans := 0
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for i, j, s := 0, 0, 1; i < len(nums); i++ {
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s *= nums[i]
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for ; j <= i && s >= k; j++ {
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s /= nums[j]
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}
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ans += i - j + 1
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function numSubarrayProductLessThanK(nums: number[], k: number): number {
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if (k <= 1) {
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return 0;
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}
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const n = nums.length;
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let res = 0;
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for (let i = 0; i < n; i++) {
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let product = 1;
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for (let j = i; j < n; j++) {
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product *= nums[j];
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if (product >= k) {
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break;
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}
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res++;
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let ans = 0;
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for (let i = 0, j = 0, s = 1; i < nums.length; ++i) {
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s *= nums[i];
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while (j <= i && s >= k) {
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s /= nums[j++];
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}
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ans += i - j + 1;
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}
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return res;
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return ans;
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}
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```
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‎solution/0700-0799/0713.Subarray Product Less Than K/README_EN.md

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<!-- tabs:start -->
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### **C++**
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### **Python3**
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```cpp
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class Solution {
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public:
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int numSubarrayProductLessThanK(vector<int>& nums, int k) {
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int left = 0, right;
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long mul = 1;
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int count = 0;
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```python
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class Solution:
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def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
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ans, s, j = 0, 1, 0
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for i, v in enumerate(nums):
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s *= v
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while j <= i and s >= k:
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s //= nums[j]
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j += 1
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ans += i - j + 1
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return ans
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```
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for (right = 0; right < nums.size(); right++) {
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mul *= nums[right];
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### **Java**
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while(left <= right && mul >= k) {
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mul /= nums[left++];
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```java
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class Solution {
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public int numSubarrayProductLessThanK(int[] nums, int k) {
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int ans = 0;
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for (int i = 0, j = 0, s = 1; i < nums.length; ++i) {
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s *= nums[i];
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while (j <= i && s >= k) {
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s /= nums[j++];
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}
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count += right >= left? right - left + 1: 0;
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ans += i - j + 1;
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}
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return count;
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return ans;
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}
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};
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}
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```
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### **Python3**
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```python
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### **C++**
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```cpp
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class Solution {
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public:
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int numSubarrayProductLessThanK(vector<int>& nums, int k) {
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int ans = 0;
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for (int i = 0, j = 0, s = 1; i < nums.size(); ++i)
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{
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s *= nums[i];
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while (j <= i && s >= k) s /= nums[j++];
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ans += i - j + 1;
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}
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return ans;
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}
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};
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```
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### **Java**
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```java
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### **Go**
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```go
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func numSubarrayProductLessThanK(nums []int, k int) int {
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ans := 0
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for i, j, s := 0, 0, 1; i < len(nums); i++ {
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s *= nums[i]
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for ; j <= i && s >= k; j++ {
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s /= nums[j]
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}
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ans += i - j + 1
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function numSubarrayProductLessThanK(nums: number[], k: number): number {
81-
if (k <= 1) {
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return 0;
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}
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const n = nums.length;
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let res = 0;
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for (let i = 0; i < n; i++) {
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let product = 1;
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for (let j = i; j < n; j++) {
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product *= nums[j];
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if (product >= k) {
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break;
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}
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res++;
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let ans = 0;
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for (let i = 0, j = 0, s = 1; i < nums.length; ++i) {
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s *= nums[i];
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while (j <= i && s >= k) {
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s /= nums[j++];
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}
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ans += i - j + 1;
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}
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return res;
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return ans;
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}
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```
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‎solution/0700-0799/0713.Subarray Product Less Than K/Solution.cpp

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class Solution {
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public:
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int numSubarrayProductLessThanK(vector<int>& nums, int k) {
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int left = 0, right;
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long mul = 1;
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int count = 0;
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for (right = 0; right < nums.size(); right++) {
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mul *= nums[right];
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while(left <= right && mul >= k) {
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mul /= nums[left++];
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}
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count += right >= left? right - left + 1: 0;
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int ans = 0;
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for (int i = 0, j = 0, s = 1; i < nums.size(); ++i)
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{
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s *= nums[i];
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while (j <= i && s >= k) s /= nums[j++];
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ans += i - j + 1;
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}
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return count;
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return ans;
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}
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};

‎solution/0700-0799/0713.Subarray Product Less Than K/Solution.go

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func numSubarrayProductLessThanK(nums []int, k int) int {
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ans := 0
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for i, j, s := 0, 0, 1; i < len(nums); i++ {
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s *= nums[i]
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for ; j <= i && s >= k; j++ {
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s /= nums[j]
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}
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ans += i - j + 1
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}
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return ans
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}

‎solution/0700-0799/0713.Subarray Product Less Than K/Solution.java

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class Solution {
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public int numSubarrayProductLessThanK(int[] nums, int k) {
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int ans = 0;
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for (int i = 0, j = 0, s = 1; i < nums.length; ++i) {
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s *= nums[i];
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while (j <= i && s >= k) {
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s /= nums[j++];
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}
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ans += i - j + 1;
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}
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return ans;
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}
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}

‎solution/0700-0799/0713.Subarray Product Less Than K/Solution.py

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class Solution:
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def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
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ans, s, j = 0, 1, 0
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for i, v in enumerate(nums):
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s *= v
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while j <= i and s >= k:
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s //= nums[j]
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j += 1
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ans += i - j + 1
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return ans

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