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feat: add solutions to lc problem: No.2892 (#1753)
No.2892.Minimizing Array After Replacing Pairs With Their Product
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‎solution/2800-2899/2892.Minimizing Array After Replacing Pairs With Their Product/README.md

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# [2892. Minimizing Array After Replacing Pairs With Their Product](https://leetcode.cn/problems/minimizing-array-after-replacing-pairs-with-their-product)
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[English Version](/solution/2800-2899/2892.Minimizing%20Array%20After%20Replacing%20Pairs%20With%20Their%20Product/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Given an integer array <code>nums</code> and an integer <code>k</code>, you can perform the following operation on the array any number of times:</p>
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<ul>
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<li>Select two <strong>adjacent</strong> elements of the array like <code>x</code> and <code>y</code>, such that <code>x * y &lt;= k</code>, and replace both of them with a <strong>single element</strong> with value <code>x * y</code> (e.g. in one operation the array <code>[1, 2, 2, 3]</code> with <code>k = 5</code> can become <code>[1, 4, 3]</code> or <code>[2, 2, 3]</code>, but can&#39;t become <code>[1, 2, 6]</code>).</li>
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</ul>
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<p>Return <em>the <strong>minimum</strong> possible length of </em><code>nums</code><em> after any number of operations</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [2,3,3,7,3,5], k = 20
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> We perform these operations:
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1. [<u>2,3</u>,3,7,3,5] -&gt; [<u>6</u>,3,7,3,5]
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2. [<u>6,3</u>,7,3,5] -&gt; [<u>18</u>,7,3,5]
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3. [18,7,<u>3,5</u>] -&gt; [18,7,<u>15</u>]
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It can be shown that 3 is the minimum length possible to achieve with the given operation.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [3,3,3,3], k = 6
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> We can&#39;t perform any operations since the product of every two adjacent elements is greater than 6.
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Hence, the answer is 4.</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
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<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心**
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我们用一个变量 $ans$ 记录当前数组的长度，用一个变量 $y$ 记录当前数组的乘积，初始时 $ans = 1$, $y = nums[0]$。
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我们从数组的第二个元素开始遍历，设当前元素为 $x$：
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- 如果 $x = 0$，那么整个数组的乘积为 $0 \le k$，因此答案数组的最小长度为 $1$，直接返回即可。
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- 如果 $x \times y \le k$，那么我们可以将 $x$ 与 $y$ 合并，即 $y = x \times y$。
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- 如果 $x \times y \gt k$，那么我们无法将 $x$ 与 $y$ 合并，因此我们需要将 $x$ 单独作为一个元素，即 $ans = ans + 1$，并且 $y = x$。
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最终答案即为 $ans$。
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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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<!-- 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 minArrayLength(self, nums: List[int], k: int) -> int:
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ans, y = 1, nums[0]
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for x in nums[1:]:
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if x == 0:
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return 1
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if x * y <= k:
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y *= x
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else:
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y = x
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ans += 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 minArrayLength(int[] nums, int k) {
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int ans = 1;
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long y = nums[0];
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for (int i = 1; i < nums.length; ++i) {
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int x = nums[i];
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if (x == 0) {
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return 1;
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}
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if (x * y <= k) {
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y *= x;
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} else {
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y = x;
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++ans;
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}
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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 minArrayLength(vector<int>& nums, int k) {
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int ans = 1;
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long long y = nums[0];
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for (int i = 1; i < nums.size(); ++i) {
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int x = nums[i];
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if (x == 0) {
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return 1;
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}
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if (x * y <= k) {
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y *= x;
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} else {
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y = x;
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++ans;
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}
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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 minArrayLength(nums []int, k int) int {
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ans, y := 1, nums[0]
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for _, x := range nums[1:] {
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if x == 0 {
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return 1
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}
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if x*y <= k {
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y *= x
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} else {
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y = x
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ans++
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}
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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 minArrayLength(nums: number[], k: number): number {
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let [ans, y] = [1, nums[0]];
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for (const x of nums.slice(1)) {
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if (x === 0) {
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return 1;
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}
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if (x * y <= k) {
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y *= x;
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} else {
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y = x;
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++ans;
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}
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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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```
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```
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<!-- tabs:end -->

‎solution/2800-2899/2892.Minimizing Array After Replacing Pairs With Their Product/README_EN.md

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# [2892. Minimizing Array After Replacing Pairs With Their Product](https://leetcode.com/problems/minimizing-array-after-replacing-pairs-with-their-product)
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[中文文档](/solution/2800-2899/2892.Minimizing%20Array%20After%20Replacing%20Pairs%20With%20Their%20Product/README.md)
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## Description
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<p>Given an integer array <code>nums</code> and an integer <code>k</code>, you can perform the following operation on the array any number of times:</p>
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<ul>
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<li>Select two <strong>adjacent</strong> elements of the array like <code>x</code> and <code>y</code>, such that <code>x * y &lt;= k</code>, and replace both of them with a <strong>single element</strong> with value <code>x * y</code> (e.g. in one operation the array <code>[1, 2, 2, 3]</code> with <code>k = 5</code> can become <code>[1, 4, 3]</code> or <code>[2, 2, 3]</code>, but can&#39;t become <code>[1, 2, 6]</code>).</li>
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</ul>
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<p>Return <em>the <strong>minimum</strong> possible length of </em><code>nums</code><em> after any number of operations</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [2,3,3,7,3,5], k = 20
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> We perform these operations:
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1. [<u>2,3</u>,3,7,3,5] -&gt; [<u>6</u>,3,7,3,5]
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2. [<u>6,3</u>,7,3,5] -&gt; [<u>18</u>,7,3,5]
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3. [18,7,<u>3,5</u>] -&gt; [18,7,<u>15</u>]
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It can be shown that 3 is the minimum length possible to achieve with the given operation.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [3,3,3,3], k = 6
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> We can&#39;t perform any operations since the product of every two adjacent elements is greater than 6.
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Hence, the answer is 4.</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
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<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
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</ul>
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## Solutions
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**Method 1: Greedy**
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We use a variable $ans$ to record the current length of the array, and a variable $y$ to record the current product of the array. Initially, $ans = 1$ and $y = nums[0]$.
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We start traversing from the second element of the array. Let the current element be $x$:
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- If $x = 0$, then the product of the entire array is $0 \le k$, so the minimum length of the answer array is $1$, and we can return directly.
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- If $x \times y \le k$, then we can merge $x$ and $y$, that is, $y = x \times y$.
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- If $x \times y \gt k$, then we cannot merge $x$ and $y$, so we need to treat $x$ as a separate element, that is, $ans = ans + 1$, and $y = x$.
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The final answer is $ans$.
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The time complexity is $O(n)$, where n is the length of the array. The space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def minArrayLength(self, nums: List[int], k: int) -> int:
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ans, y = 1, nums[0]
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for x in nums[1:]:
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if x == 0:
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return 1
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if x * y <= k:
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y *= x
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else:
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y = x
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ans += 1
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int minArrayLength(int[] nums, int k) {
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int ans = 1;
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long y = nums[0];
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for (int i = 1; i < nums.length; ++i) {
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int x = nums[i];
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if (x == 0) {
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return 1;
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}
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if (x * y <= k) {
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y *= x;
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} else {
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y = x;
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++ans;
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}
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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 minArrayLength(vector<int>& nums, int k) {
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int ans = 1;
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long long y = nums[0];
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for (int i = 1; i < nums.size(); ++i) {
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int x = nums[i];
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if (x == 0) {
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return 1;
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}
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if (x * y <= k) {
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y *= x;
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} else {
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y = x;
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++ans;
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}
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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 minArrayLength(nums []int, k int) int {
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ans, y := 1, nums[0]
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for _, x := range nums[1:] {
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if x == 0 {
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return 1
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}
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if x*y <= k {
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y *= x
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} else {
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y = x
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ans++
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}
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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 minArrayLength(nums: number[], k: number): number {
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let [ans, y] = [1, nums[0]];
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for (const x of nums.slice(1)) {
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if (x === 0) {
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return 1;
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}
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if (x * y <= k) {
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y *= x;
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} else {
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y = x;
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++ans;
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}
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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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```
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```
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<!-- tabs:end -->

‎solution/2800-2899/2892.Minimizing Array After Replacing Pairs With Their Product/Solution.cpp

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class Solution {
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public:
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int minArrayLength(vector<int>& nums, int k) {
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int ans = 1;
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long long y = nums[0];
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for (int i = 1; i < nums.size(); ++i) {
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int x = nums[i];
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if (x == 0) {
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return 1;
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}
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if (x * y <= k) {
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y *= x;
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} else {
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y = x;
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++ans;
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}
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}
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return ans;
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}
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};

‎solution/2800-2899/2892.Minimizing Array After Replacing Pairs With Their Product/Solution.go

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func minArrayLength(nums []int, k int) int {
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ans, y := 1, nums[0]
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for _, x := range nums[1:] {
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if x == 0 {
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return 1
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}
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if x*y <= k {
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y *= x
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} else {
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y = x
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ans++
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}
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}
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return ans
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}

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