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feat: add Java solutions to lc problems: No.2869~2872 #1728

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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 -->
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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 -->
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Original file line number Diff line number Diff line change
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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;
}
}
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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 -->
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