feat: add weekly contest 474

Author: yanglbme

solution/3700-3799/3731.Find Missing Elements/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3731.Find%20Missing%20Elements/README.md
+---
+
+<!-- problem:start -->
+
+# [3731. 找出缺失的元素](https://leetcode.cn/problems/find-missing-elements)
+
+[English Version](/solution/3700-3799/3731.Find%20Missing%20Elements/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个整数数组 <code>nums</code> ，数组由若干&nbsp;<b>互不相同</b> 的整数组成。</p>
+
+<p>数组 <code>nums</code> 原本包含了某个范围内的&nbsp;<strong>所有整数&nbsp;</strong>。但现在，其中可能 <strong>缺失</strong> 部分整数。</p>
+
+<p>该范围内的&nbsp;<strong>最小&nbsp;</strong>整数和&nbsp;<strong>最大&nbsp;</strong>整数仍然存在于 <code>nums</code> 中。</p>
+
+<p>返回一个&nbsp;<strong>有序&nbsp;</strong>列表，包含该范围内缺失的所有整数，并&nbsp;<strong>按从小到大排序</strong>。如果没有缺失的整数，返回一个&nbsp;<strong>空&nbsp;</strong>列表。</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 = [1,4,2,5]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">[3]</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>最小整数为 1，最大整数为 5，因此完整的范围应为 <code>[1,2,3,4,5]</code>。其中只有 3 缺失。</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 = [7,8,6,9]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">[]</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>最小整数为 6，最大整数为 9，因此完整的范围为 <code>[6,7,8,9]</code>。所有整数均已存在，因此没有缺失的整数。</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 = [5,1]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">[2,3,4]</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>最小整数为 1，最大整数为 5，因此完整的范围应为 <code>[1,2,3,4,5]</code>。缺失的整数为 2、3 和 4。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：哈希表
+
+我们先找出数组 $\textit{nums}$ 中的最小值和最大值，记为 $\textit{mn}$ 和 $\textit{mx}$。然后我们使用哈希表存储数组 $\textit{nums}$ 中的所有元素。
+
+接下来，我们遍历区间 $[\textit{mn} + 1, \textit{mx} - 1]$，对于每个整数 $x$，如果 $x$ 不在哈希表中，我们就将其加入答案列表中。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$，其中 $n$ 是数组 $\textit{nums}$ 的长度。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def findMissingElements(self, nums: List[int]) -> List[int]:
+        mn, mx = min(nums), max(nums)
+        s = set(nums)
+        return [x for x in range(mn + 1, mx) if x not in s]
+```
+
+#### Java
+
+```java
+class Solution {
+    public List<Integer> findMissingElements(int[] nums) {
+        int mn = 100, mx = 0;
+        Set<Integer> s = new HashSet<>();
+        for (int x : nums) {
+            mn = Math.min(mn, x);
+            mx = Math.max(mx, x);
+            s.add(x);
+        }
+        List<Integer> ans = new ArrayList<>();
+        for (int x = mn + 1; x < mx; ++x) {
+            if (!s.contains(x)) {
+                ans.add(x);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    vector<int> findMissingElements(vector<int>& nums) {
+        int mn = 100, mx = 0;
+        unordered_set<int> s;
+        for (int x : nums) {
+            mn = min(mn, x);
+            mx = max(mx, x);
+            s.insert(x);
+        }
+        vector<int> ans;
+        for (int x = mn + 1; x < mx; ++x) {
+            if (!s.count(x)) {
+                ans.push_back(x);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func findMissingElements(nums []int) (ans []int) {
+	mn, mx := 100, 0
+	s := make(map[int]bool)
+	for _, x := range nums {
+        mn = min(mn, x)
+        mx = max(mx, x)
+		s[x] = true
+	}
+	for x := mn + 1; x < mx; x++ {
+		if !s[x] {
+			ans = append(ans, x)
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function findMissingElements(nums: number[]): number[] {
+    let [mn, mx] = [100, 0];
+    const s = new Set<number>();
+    for (const x of nums) {
+        mn = Math.min(mn, x);
+        mx = Math.max(mx, x);
+        s.add(x);
+    }
+    const ans: number[] = [];
+    for (let x = mn + 1; x < mx; ++x) {
+        if (!s.has(x)) {
+            ans.push(x);
+        }
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3700-3799/3731.Find Missing Elements/README_EN.md

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+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3731.Find%20Missing%20Elements/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3731. Find Missing Elements](https://leetcode.com/problems/find-missing-elements)
+
+[中文文档](/solution/3700-3799/3731.Find%20Missing%20Elements/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an integer array <code>nums</code> consisting of <strong>unique</strong> integers.</p>
+
+<p>Originally, <code>nums</code> contained <strong>every integer</strong> within a certain range. However, some integers might have gone <strong>missing</strong> from the array.</p>
+
+<p>The <strong>smallest</strong> and <strong>largest</strong> integers of the original range are still present in <code>nums</code>.</p>
+
+<p>Return a <strong>sorted</strong> list of all the missing integers in this range. If no integers are missing, return an <strong>empty</strong> list.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,4,2,5]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[3]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The smallest integer is 1 and the largest is 5, so the full range should be <code>[1,2,3,4,5]</code>. Among these, only 3 is missing.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [7,8,6,9]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The smallest integer is 6 and the largest is 9, so the full range is <code>[6,7,8,9]</code>. All integers are already present, so no integer is missing.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [5,1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[2,3,4]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The smallest integer is 1 and the largest is 5, so the full range should be <code>[1,2,3,4,5]</code>. The missing integers are 2, 3, and 4.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Hash Table
+
+We first find the minimum and maximum values in the array $\textit{nums}$, denoted as $\textit{mn}$ and $\textit{mx}$. Then we use a hash table to store all elements in the array $\textit{nums}$.
+
+Next, we iterate through the interval $[\textit{mn} + 1, \textit{mx} - 1]$. For each integer $x$, if $x$ is not in the hash table, we add it to the answer list.
+
+The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def findMissingElements(self, nums: List[int]) -> List[int]:
+        mn, mx = min(nums), max(nums)
+        s = set(nums)
+        return [x for x in range(mn + 1, mx) if x not in s]
+```
+
+#### Java
+
+```java
+class Solution {
+    public List<Integer> findMissingElements(int[] nums) {
+        int mn = 100, mx = 0;
+        Set<Integer> s = new HashSet<>();
+        for (int x : nums) {
+            mn = Math.min(mn, x);
+            mx = Math.max(mx, x);
+            s.add(x);
+        }
+        List<Integer> ans = new ArrayList<>();
+        for (int x = mn + 1; x < mx; ++x) {
+            if (!s.contains(x)) {
+                ans.add(x);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    vector<int> findMissingElements(vector<int>& nums) {
+        int mn = 100, mx = 0;
+        unordered_set<int> s;
+        for (int x : nums) {
+            mn = min(mn, x);
+            mx = max(mx, x);
+            s.insert(x);
+        }
+        vector<int> ans;
+        for (int x = mn + 1; x < mx; ++x) {
+            if (!s.count(x)) {
+                ans.push_back(x);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func findMissingElements(nums []int) (ans []int) {
+	mn, mx := 100, 0
+	s := make(map[int]bool)
+	for _, x := range nums {
+        mn = min(mn, x)
+        mx = max(mx, x)
+		s[x] = true
+	}
+	for x := mn + 1; x < mx; x++ {
+		if !s[x] {
+			ans = append(ans, x)
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function findMissingElements(nums: number[]): number[] {
+    let [mn, mx] = [100, 0];
+    const s = new Set<number>();
+    for (const x of nums) {
+        mn = Math.min(mn, x);
+        mx = Math.max(mx, x);
+        s.add(x);
+    }
+    const ans: number[] = [];
+    for (let x = mn + 1; x < mx; ++x) {
+        if (!s.has(x)) {
+            ans.push(x);
+        }
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3700-3799/3731.Find Missing Elements/Solution.cpp

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+class Solution {
+public:
+    vector<int> findMissingElements(vector<int>& nums) {
+        int mn = 100, mx = 0;
+        unordered_set<int> s;
+        for (int x : nums) {
+            mn = min(mn, x);
+            mx = max(mx, x);
+            s.insert(x);
+        }
+        vector<int> ans;
+        for (int x = mn + 1; x < mx; ++x) {
+            if (!s.count(x)) {
+                ans.push_back(x);
+            }
+        }
+        return ans;
+    }
+};