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feat: add weekly contest 367 (#1808)
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# [2903. 找出满足差值条件的下标 I](https://leetcode.cn/problems/find-indices-with-index-and-value-difference-i)
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[English Version](/solution/2900-2999/2903.Find%20Indices%20With%20Index%20and%20Value%20Difference%20I/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>给你一个下标从 <strong>0</strong> 开始、长度为 <code>n</code> 的整数数组 <code>nums</code> ,以及整数 <code>indexDifference</code> 和整数 <code>valueDifference</code> 。</p>
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<p>你的任务是从范围 <code>[0, n - 1]</code> 内找出&nbsp; <strong>2</strong> 个满足下述所有条件的下标 <code>i</code> 和 <code>j</code> :</p>
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<ul>
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<li><code>abs(i - j) &gt;= indexDifference</code> 且</li>
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<li><code>abs(nums[i] - nums[j]) &gt;= valueDifference</code></li>
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</ul>
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<p>返回整数数组 <code>answer</code>。如果存在满足题目要求的两个下标,则 <code>answer = [i, j]</code> ;否则,<code>answer = [-1, -1]</code> 。如果存在多组可供选择的下标对,只需要返回其中任意一组即可。</p>
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<p><strong>注意:</strong><code>i</code> 和 <code>j</code> 可能 <strong>相等</strong> 。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [5,1,4,1], indexDifference = 2, valueDifference = 4
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<strong>输出:</strong>[0,3]
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<strong>解释:</strong>在示例中,可以选择 i = 0 和 j = 3 。
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abs(0 - 3) &gt;= 2 且 abs(nums[0] - nums[3]) &gt;= 4 。
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因此,[0,3] 是一个符合题目要求的答案。
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[3,0] 也是符合题目要求的答案。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [2,1], indexDifference = 0, valueDifference = 0
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<strong>输出:</strong>[0,0]
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<strong>解释:</strong>
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在示例中,可以选择 i = 0 和 j = 0 。
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abs(0 - 0) &gt;= 0 且 abs(nums[0] - nums[0]) &gt;= 0 。
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因此,[0,0] 是一个符合题目要求的答案。
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[0,1]、[1,0] 和 [1,1] 也是符合题目要求的答案。
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [1,2,3], indexDifference = 2, valueDifference = 4
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<strong>输出:</strong>[-1,-1]
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<strong>解释:</strong>在示例中,可以证明无法找出 2 个满足所有条件的下标。
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因此,返回 [-1,-1] 。</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;= n == nums.length &lt;= 100</code></li>
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<li><code>0 &lt;= nums[i] &lt;= 50</code></li>
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<li><code>0 &lt;= indexDifference &lt;= 100</code></li>
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<li><code>0 &lt;= valueDifference &lt;= 50</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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```
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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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### **...**
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```
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```
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<!-- tabs:end -->
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# [2903. Find Indices With Index and Value Difference I](https://leetcode.com/problems/find-indices-with-index-and-value-difference-i)
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[中文文档](/solution/2900-2999/2903.Find%20Indices%20With%20Index%20and%20Value%20Difference%20I/README.md)
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## Description
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<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> having length <code>n</code>, an integer <code>indexDifference</code>, and an integer <code>valueDifference</code>.</p>
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<p>Your task is to find <strong>two</strong> indices <code>i</code> and <code>j</code>, both in the range <code>[0, n - 1]</code>, that satisfy the following conditions:</p>
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<ul>
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<li><code>abs(i - j) &gt;= indexDifference</code>, and</li>
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<li><code>abs(nums[i] - nums[j]) &gt;= valueDifference</code></li>
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</ul>
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<p>Return <em>an integer array</em> <code>answer</code>, <em>where</em> <code>answer = [i, j]</code> <em>if there are two such indices</em>, <em>and</em> <code>answer = [-1, -1]</code> <em>otherwise</em>. If there are multiple choices for the two indices, return <em>any of them</em>.</p>
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<p><strong>Note:</strong> <code>i</code> and <code>j</code> may be <strong>equal</strong>.</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>
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<strong>Input:</strong> nums = [5,1,4,1], indexDifference = 2, valueDifference = 4
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<strong>Output:</strong> [0,3]
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<strong>Explanation:</strong> In this example, i = 0 and j = 3 can be selected.
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abs(0 - 3) &gt;= 2 and abs(nums[0] - nums[3]) &gt;= 4.
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Hence, a valid answer is [0,3].
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[3,0] is also a valid answer.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [2,1], indexDifference = 0, valueDifference = 0
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<strong>Output:</strong> [0,0]
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<strong>Explanation:</strong> In this example, i = 0 and j = 0 can be selected.
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abs(0 - 0) &gt;= 0 and abs(nums[0] - nums[0]) &gt;= 0.
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Hence, a valid answer is [0,0].
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Other valid answers are [0,1], [1,0], and [1,1].
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,3], indexDifference = 2, valueDifference = 4
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<strong>Output:</strong> [-1,-1]
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<strong>Explanation:</strong> In this example, it can be shown that it is impossible to find two indices that satisfy both conditions.
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Hence, [-1,-1] is returned.</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;= n == nums.length &lt;= 100</code></li>
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<li><code>0 &lt;= nums[i] &lt;= 50</code></li>
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<li><code>0 &lt;= indexDifference &lt;= 100</code></li>
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<li><code>0 &lt;= valueDifference &lt;= 50</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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```
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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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### **...**
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```
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```
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<!-- tabs:end -->
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# [2904. 最短且字典序最小的美丽子字符串](https://leetcode.cn/problems/shortest-and-lexicographically-smallest-beautiful-string)
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[English Version](/solution/2900-2999/2904.Shortest%20and%20Lexicographically%20Smallest%20Beautiful%20String/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>给你一个二进制字符串 <code>s</code> 和一个正整数 <code>k</code> 。</p>
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<p>如果 <code>s</code> 的某个子字符串中 <code>1</code> 的个数恰好等于 <code>k</code> ,则称这个子字符串是一个 <strong>美丽子字符串</strong> 。</p>
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<p>令 <code>len</code> 等于 <strong>最短</strong> 美丽子字符串的长度。</p>
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<p>返回长度等于 <code>len</code> 且字典序 <strong>最小</strong> 的美丽子字符串。如果 <code>s</code> 中不含美丽子字符串,则返回一个 <strong>空</strong> 字符串。</p>
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<p>对于相同长度的两个字符串 <code>a</code> 和 <code>b</code> ,如果在 <code>a</code> 和 <code>b</code> 出现不同的第一个位置上,<code>a</code> 中该位置上的字符严格大于 <code>b</code> 中的对应字符,则认为字符串 <code>a</code> 字典序 <strong>大于</strong> 字符串 <code>b</code> 。</p>
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<ul>
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<li>例如,<code>"abcd"</code> 的字典序大于 <code>"abcc"</code> ,因为两个字符串出现不同的第一个位置对应第四个字符,而 <code>d</code> 大于 <code>c</code> 。</li>
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</ul>
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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>
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<strong>输入:</strong>s = "100011001", k = 3
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<strong>输出:</strong>"11001"
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<strong>解释:</strong>示例中共有 7 个美丽子字符串:
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1. 子字符串 "<em><strong>100011</strong></em>001" 。
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2. 子字符串 "<strong><em>1000110</em></strong>01" 。
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3. 子字符串 "<strong><em>100011001</em></strong>" 。
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4. 子字符串 "1<strong><em>00011001</em></strong>" 。
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5. 子字符串 "10<strong><em>0011001</em></strong>" 。
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6. 子字符串 "100<em><strong>011001</strong></em>" 。
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7. 子字符串 "1000<strong><em>11001</em></strong>" 。
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最短美丽子字符串的长度是 5 。
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长度为 5 且字典序最小的美丽子字符串是子字符串 "11001" 。
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</pre>
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<p><strong class="example">示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>s = "1011", k = 2
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<strong>输出:</strong>"11"
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<strong>解释:</strong>示例中共有 3 个美丽子字符串:
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1. 子字符串 "<em><strong>101</strong></em>1" 。
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2. 子字符串 "1<em><strong>011</strong></em>" 。
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3. 子字符串 "10<em><strong>11</strong></em>" 。
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最短美丽子字符串的长度是 2 。
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长度为 2 且字典序最小的美丽子字符串是子字符串 "11" 。
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</pre>
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<p><strong class="example">示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>s = "000", k = 1
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<strong>输出:</strong>""
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<strong>解释:</strong>示例中不存在美丽子字符串。
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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;= s.length &lt;= 100</code></li>
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<li><code>1 &lt;= k &lt;= s.length</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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```
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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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### **...**
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```
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```
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<!-- tabs:end -->

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