feat: add Java solutions to lc problems: No.2869~2872

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/)
+
+[English Version](/solution/2800-2899/2869.Minimum%20Operations%20to%20Collect%20Elements/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<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>
+
+<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>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre><b>输入：</b>nums = [3,1,5,4,2], k = 2
+<b>输出：</b>4
+<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 。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre><b>输入：</b>nums = [3,1,5,4,2], k = 5
+<b>输出：</b>5
+<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 。
+</pre>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<pre><b>输入：</b>nums = [3,2,5,3,1], k = 3
+<b>输出：</b>4
+<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 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
+	<li><code>1 &lt;= k &lt;= nums.length</code></li>
+	<li>输入保证你可以收集到元素&nbsp;<code>1, 2, ..., k</code> 。</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int minOperations(List<Integer> nums, int k) {
+        boolean[] isAdded = new boolean[k];
+        int n = nums.size();
+        int count = 0;
+        for (int i = n - 1; i >= 0; i--) {
+            if (nums.get(i) > k || isAdded[nums.get(i) - 1]) {
+                continue;
+            }
+            isAdded[nums.get(i) - 1] = true;
+            count++;
+            if (count == k) {
+                return n - i;
+            }
+        }
+        return n;
+    }
+}
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- 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/)
+
+[中文文档](/solution/2800-2899/2869.Minimum%20Operations%20to%20Collect%20Elements/README.md)
+
+## Description
+
+<p>In one operation, you can remove the last element of the array and add it to your collection.</p>
+
+<p>Return <em>the <strong>minimum number of operations</strong> needed to collect elements</em> <code>1, 2, ..., k</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre><strong>Input:</strong> nums = [3,1,5,4,2], k = 2
+<strong>Output:</strong> 4
+<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.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre><strong>Input:</strong> nums = [3,1,5,4,2], k = 5
+<strong>Output:</strong> 5
+<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.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre><strong>Input:</strong> nums = [3,2,5,3,1], k = 3
+<strong>Output:</strong> 4
+<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.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
+	<li><code>1 &lt;= k &lt;= nums.length</code></li>
+	<li>The input is generated such that you can collect elements <code>1, 2, ..., k</code>.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int minOperations(List<Integer> nums, int k) {
+        boolean[] isAdded = new boolean[k];
+        int n = nums.size();
+        int count = 0;
+        for (int i = n - 1; i >= 0; i--) {
+            if (nums.get(i) > k || isAdded[nums.get(i) - 1]) {
+                continue;
+            }
+            isAdded[nums.get(i) - 1] = true;
+            count++;
+            if (count == k) {
+                return n - i;
+            }
+        }
+        return n;
+    }
+}
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

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

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

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/)
+
+[English Version](/solution/2800-2899/2870.Minimum%20Number%20of%20Operations%20to%20Make%20Array%20Empty/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>你可以对数组执行以下两种操作 <strong>任意次</strong>&nbsp;：</p>
+
+<ul>
+	<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>
+	<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>
+</ul>
+
+<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>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre><strong>输入：</strong>nums = [2,3,3,2,2,4,2,3,4]
+<b>输出：</b>4
+<b>解释：</b>我们可以执行以下操作使数组为空：
+- 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] 。
+- 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] 。
+- 对下标为 0 ，1 和 3 的元素执行第二种操作，得到 nums = [4,4] 。
+- 对下标为 0 和 1 的元素执行第一种操作，得到 nums = [] 。
+至少需要 4 步操作使数组为空。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre><b>输入：</b>nums = [2,1,2,2,3,3]
+<b>输出：</b>-1
+<b>解释：</b>无法使数组为空。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int minOperations(int[] nums) {
+        Map<Integer, Integer> count = new HashMap<>();
+        for (int num : nums) {
+            // count.put(num, count.getOrDefault(num, 0) + 1);
+            count.merge(num, 1, Integer::sum);
+        }
+        int ans = 0;
+        for (Integer c : count.values()) {
+            if (c < 2) {
+                return -1;
+            }
+            int r = c % 3;
+            int d = c / 3;
+            switch (r) {
+                case (0) -> {
+                    ans += d;
+                }
+                default -> {
+                    ans += d + 1;
+                }
+            }
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->