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feat: add Java solutions to lc problems: No.2869~2872 (#1728)
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‎solution/2800-2899/2869.Minimum Operations to Collect Elements/README.md

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# [2869. Minimum Operations to Collect Elements](https://leetcode.cn/problems/minimum-operations-to-collect-elements/)
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[English Version](/solution/2800-2899/2869.Minimum%20Operations%20to%20Collect%20Elements/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>一次操作中，你可以将数组的最后一个元素删除，将该元素添加到一个集合中。</p>
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<p>请你返回收集元素&nbsp;<code>1, 2, ..., k</code>&nbsp;需要的&nbsp;<strong>最少操作次数</strong>&nbsp;。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<pre><b>输入：</b>nums = [3,1,5,4,2], k = 2
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<b>输出：</b>4
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<b>解释：</b>4 次操作后，集合中的元素依次添加了 2 ，4 ，5 和 1 。此时集合中包含元素 1 和 2 ，所以答案为 4 。
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</pre>
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<p><strong class="example">示例 2：</strong></p>
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<pre><b>输入：</b>nums = [3,1,5,4,2], k = 5
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<b>输出：</b>5
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<b>解释：</b>5 次操作后，集合中的元素依次添加了 2 ，4 ，5 ，1 和 3 。此时集合中包含元素 1 到 5 ，所以答案为 5 。
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</pre>
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<p><strong class="example">示例 3：</strong></p>
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<pre><b>输入：</b>nums = [3,2,5,3,1], k = 3
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<b>输出：</b>4
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<b>解释：</b>4 次操作后，集合中的元素依次添加了 1 ，3 ，5 和 2 。此时集合中包含元素 1 到 3 ，所以答案为 4 。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 50</code></li>
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<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
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<li><code>1 &lt;= k &lt;= nums.length</code></li>
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<li>输入保证你可以收集到元素&nbsp;<code>1, 2, ..., k</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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<!-- 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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```
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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 minOperations(List<Integer> nums, int k) {
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boolean[] isAdded = new boolean[k];
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int n = nums.size();
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int count = 0;
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for (int i = n - 1; i >= 0; i--) {
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if (nums.get(i) > k || isAdded[nums.get(i) - 1]) {
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continue;
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}
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isAdded[nums.get(i) - 1] = true;
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count++;
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if (count == k) {
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return n - i;
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}
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}
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return n;
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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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```
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### **Go**
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```go
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2800-2899/2869.Minimum Operations to Collect Elements/README_EN.md

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# [2869. Minimum Operations to Collect Elements](https://leetcode.com/problems/minimum-operations-to-collect-elements/)
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[中文文档](/solution/2800-2899/2869.Minimum%20Operations%20to%20Collect%20Elements/README.md)
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## Description
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<p>In one operation, you can remove the last element of the array and add it to your collection.</p>
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<p>Return <em>the <strong>minimum number of operations</strong> needed to collect elements</em> <code>1, 2, ..., k</code>.</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><strong>Input:</strong> nums = [3,1,5,4,2], k = 2
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> After 4 operations, we collect elements 2, 4, 5, and 1, in this order. Our collection contains elements 1 and 2. Hence, the answer is 4.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre><strong>Input:</strong> nums = [3,1,5,4,2], k = 5
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<strong>Output:</strong> 5
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<strong>Explanation:</strong> After 5 operations, we collect elements 2, 4, 5, 1, and 3, in this order. Our collection contains elements 1 through 5. Hence, the answer is 5.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre><strong>Input:</strong> nums = [3,2,5,3,1], k = 3
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<strong>Output:</strong> 4
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<strong>Explanation:</strong> After 4 operations, we collect elements 1, 3, 5, and 2, in this order. Our collection contains elements 1 through 3. Hence, the answer is 4.
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</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;= 50</code></li>
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<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
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<li><code>1 &lt;= k &lt;= nums.length</code></li>
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<li>The input is generated such that you can collect elements <code>1, 2, ..., k</code>.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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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 minOperations(List<Integer> nums, int k) {
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boolean[] isAdded = new boolean[k];
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int n = nums.size();
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int count = 0;
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for (int i = n - 1; i >= 0; i--) {
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if (nums.get(i) > k || isAdded[nums.get(i) - 1]) {
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continue;
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}
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isAdded[nums.get(i) - 1] = true;
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count++;
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if (count == k) {
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return n - i;
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}
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}
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return n;
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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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```
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### **Go**
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```go
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2800-2899/2869.Minimum Operations to Collect Elements/Solution.java

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class Solution {
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public int minOperations(List<Integer> nums, int k) {
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boolean[] isAdded = new boolean[k];
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int n = nums.size();
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int count = 0;
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for (int i = n - 1; i >= 0; i--) {
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if (nums.get(i) > k || isAdded[nums.get(i) - 1]) {
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continue;
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}
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isAdded[nums.get(i) - 1] = true;
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count++;
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if (count == k) {
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return n - i;
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}
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}
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return n;
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}
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}

‎solution/2800-2899/2870.Minimum Number of Operations to Make Array Empty/README.md

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# [2870. Minimum Number of Operations to Make Array Empty](https://leetcode.cn/problems/minimum-number-of-operations-to-make-array-empty/)
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[English Version](/solution/2800-2899/2870.Minimum%20Number%20of%20Operations%20to%20Make%20Array%20Empty/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>你可以对数组执行以下两种操作 <strong>任意次</strong>&nbsp;：</p>
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<ul>
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<li>从数组中选择 <strong>两个</strong>&nbsp;值 <strong>相等</strong>&nbsp;的元素，并将它们从数组中 <strong>删除</strong>&nbsp;。</li>
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<li>从数组中选择 <strong>三个</strong>&nbsp;值 <strong>相等</strong>&nbsp;的元素，并将它们从数组中 <strong>删除</strong>&nbsp;。</li>
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</ul>
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<p>请你返回使数组为空的 <strong>最少</strong>&nbsp;操作次数，如果无法达成，请返回 <code>-1</code>&nbsp;。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<pre><strong>输入：</strong>nums = [2,3,3,2,2,4,2,3,4]
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<b>输出：</b>4
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<b>解释：</b>我们可以执行以下操作使数组为空：
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- 对下标为 0 和 3 的元素执行第一种操作，得到 nums = [3,3,2,4,2,3,4] 。
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- 对下标为 2 和 4 的元素执行第一种操作，得到 nums = [3,3,4,3,4] 。
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- 对下标为 0 ，1 和 3 的元素执行第二种操作，得到 nums = [4,4] 。
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- 对下标为 0 和 1 的元素执行第一种操作，得到 nums = [] 。
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至少需要 4 步操作使数组为空。
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</pre>
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<p><strong class="example">示例 2：</strong></p>
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<pre><b>输入：</b>nums = [2,1,2,2,3,3]
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<b>输出：</b>-1
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<b>解释：</b>无法使数组为空。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</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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<!-- 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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```
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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 minOperations(int[] nums) {
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Map<Integer, Integer> count = new HashMap<>();
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for (int num : nums) {
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// count.put(num, count.getOrDefault(num, 0) + 1);
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count.merge(num, 1, Integer::sum);
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}
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int ans = 0;
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for (Integer c : count.values()) {
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if (c < 2) {
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return -1;
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}
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int r = c % 3;
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int d = c / 3;
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switch (r) {
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case (0) -> {
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ans += d;
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}
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default -> {
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ans += d + 1;
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}
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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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```
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### **Go**
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```go
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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