feat: add weekly contest 481 & biweekly contest 172

solution/2000-2099/2045.Second Minimum Time to Reach Destination/README_EN.md

@@ -61,8 +61,8 @@ The red path shows the path to get the second minimum time.
 - 3 -> 4: 3 minutes, time elapsed=6
 - Wait at 4 for 4 minutes, time elapsed=10
 - 4 -> 5: 3 minutes, time elapsed=13
-  Hence the second minimum time is 13 minutes.
-  </pre>
+Hence the second minimum time is 13 minutes.
+    </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2045.Second%20Minimum%20Time%20to%20Reach%20Destination/images/eg2.png" style="width: 225px; height: 50px;" />

solution/3700-3799/3774.Absolute Difference Between Maximum and Minimum K Elements/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 简单
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3774.Absolute%20Difference%20Between%20Maximum%20and%20Minimum%20K%20Elements/README.md
+rating: 1206
+source: 第 480 场周赛 Q1
 ---
 
 <!-- problem:start -->

solution/3700-3799/3774.Absolute Difference Between Maximum and Minimum K Elements/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Easy
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3774.Absolute%20Difference%20Between%20Maximum%20and%20Minimum%20K%20Elements/README_EN.md
+rating: 1206
+source: Weekly Contest 480 Q1
 ---
 
 <!-- problem:start -->

solution/3700-3799/3775.Reverse Words With Same Vowel Count/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3775.Reverse%20Words%20With%20Same%20Vowel%20Count/README.md
+rating: 1391
+source: 第 480 场周赛 Q2
 ---
 
 <!-- problem:start -->

solution/3700-3799/3775.Reverse Words With Same Vowel Count/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3775.Reverse%20Words%20With%20Same%20Vowel%20Count/README_EN.md
+rating: 1391
+source: Weekly Contest 480 Q2
 ---
 
 <!-- problem:start -->

solution/3700-3799/3776.Minimum Moves to Balance Circular Array/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3776.Minimum%20Moves%20to%20Balance%20Circular%20Array/README.md
+rating: 1739
+source: 第 480 场周赛 Q3
 ---
 
 <!-- problem:start -->

solution/3700-3799/3776.Minimum Moves to Balance Circular Array/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3776.Minimum%20Moves%20to%20Balance%20Circular%20Array/README_EN.md
+rating: 1739
+source: Weekly Contest 480 Q3
 ---
 
 <!-- problem:start -->

solution/3700-3799/3777.Minimum Deletions to Make Alternating Substring/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3777.Minimum%20Deletions%20to%20Make%20Alternating%20Substring/README.md
+rating: 2201
+source: 第 480 场周赛 Q4
 ---
 
 <!-- problem:start -->

solution/3700-3799/3777.Minimum Deletions to Make Alternating Substring/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3777.Minimum%20Deletions%20to%20Make%20Alternating%20Substring/README_EN.md
+rating: 2201
+source: Weekly Contest 480 Q4
 ---
 
 <!-- problem:start -->

solution/3700-3799/3779.Minimum Number of Operations to Have Distinct Elements/README.md

@@ -0,0 +1,169 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3779.Minimum%20Number%20of%20Operations%20to%20Have%20Distinct%20Elements/README.md
+---
+
+<!-- problem:start -->
+
+# [3779. 得到互不相同元素的最少操作次数](https://leetcode.cn/problems/minimum-number-of-operations-to-have-distinct-elements)
+
+[English Version](/solution/3700-3799/3779.Minimum%20Number%20of%20Operations%20to%20Have%20Distinct%20Elements/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个整数数组 <code>nums</code>。</p>
+
+<p>在一次操作中，你需要移除当前数组的 <strong>前三个元素</strong>。如果剩余元素少于三个，则移除 <strong>所有</strong> 剩余元素。</p>
+
+<p>重复此操作，直到数组为空或不包含任何重复元素为止。</p>
+
+<p>返回一个整数，表示所需的操作次数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1:</strong></p>
+
+<div class="example-block">
+<p><strong>输入:</strong> <span class="example-io">nums = [3,8,3,6,5,8]</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">1</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>在第一次操作中，我们移除前三个元素。剩余的元素 <code>[6, 5, 8]</code> 互不相同，因此停止。仅需要一次操作。</p>
+</div>
+
+<p><strong class="example">示例 2:</strong></p>
+
+<div class="example-block">
+<p><strong>输入:</strong> <span class="example-io">nums = [2,2]</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">1</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>经过一次操作后，数组变为空，满足停止条件。</p>
+</div>
+
+<p><strong class="example">示例 3:</strong></p>
+
+<div class="example-block">
+<p><strong>输入:</strong> <span class="example-io">nums = [4,3,5,1,2]</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">0</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>数组中的所有元素都是互不相同的，因此不需要任何操作。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：哈希表 + 倒序遍历
+
+我们可以倒序遍历数组 $\textit{nums}$，并使用哈希表 $\textit{st}$ 记录已经遍历过的元素。当遍历到元素 $\textit{nums}[i]$ 时，如果 $\textit{nums}[i]$ 已经在哈希表 $\textit{st}$ 中，那么说明我们需要移除 $\textit{nums}[0..i]$ 的所有元素，需要的操作次数为 $\left\lfloor \frac{i}{3} \right\rfloor + 1$。否则，我们将 $\textit{nums}[i]$ 加入哈希表 $\textit{st}$ 中，并继续遍历下一个元素。
+
+遍历结束后，如果没有找到重复的元素，那么数组中的元素已经互不相同，不需要进行任何操作，答案为 $0$。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def minOperations(self, nums: List[int]) -> int:
+        st = set()
+        for i in range(len(nums) - 1, -1, -1):
+            if nums[i] in st:
+                return i // 3 + 1
+            st.add(nums[i])
+        return 0
+```
+
+#### Java
+
+```java
+class Solution {
+    public int minOperations(int[] nums) {
+        Set<Integer> st = new HashSet<>();
+        for (int i = nums.length - 1; i >= 0; --i) {
+            if (!st.add(nums[i])) {
+                return i / 3 + 1;
+            }
+        }
+        return 0;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int minOperations(vector<int>& nums) {
+        unordered_set<int> st;
+        for (int i = nums.size() - 1; ~i; --i) {
+            if (st.contains(nums[i])) {
+                return i / 3 + 1;
+            }
+            st.insert(nums[i]);
+        }
+        return 0;
+    }
+};
+```
+
+#### Go
+
+```go
+func minOperations(nums []int) int {
+	st := make(map[int]struct{})
+	for i := len(nums) - 1; i >= 0; i-- {
+		if _, ok := st[nums[i]]; ok {
+			return i/3 + 1
+		}
+		st[nums[i]] = struct{}{}
+	}
+	return 0
+}
+```
+
+#### TypeScript
+
+```ts
+function minOperations(nums: number[]): number {
+    const st = new Set<number>();
+    for (let i = nums.length - 1; i >= 0; i--) {
+        if (st.has(nums[i])) {
+            return Math.floor(i / 3) + 1;
+        }
+        st.add(nums[i]);
+    }
+    return 0;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->