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yanglbmeyanglbme
authoredMay 14, 2024
feat: update lc problems (#2811)
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‎lcci/17.16.The Masseuse/README.md

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@@ -124,14 +124,14 @@ class Solution {
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func massage(_ nums: [Int]) -> Int {
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var f = 0
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var g = 0
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for x in nums {
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let ff = g + x
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let gg = max(f, g)
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f = ff
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g = gg
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}
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return max(f, g)
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}
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}

‎lcci/17.16.The Masseuse/README_EN.md

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func massage(_ nums: [Int]) -> Int {
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var f = 0
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var g = 0
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for x in nums {
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let ff = g + x
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let gg = max(f, g)
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f = ff
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g = gg
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}
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return max(f, g)
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}
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}

‎lcci/17.18.Shortest Supersequence/README.md

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var need = [Int: Int]()
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var window = [Int: Int]()
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small.forEach { need[$0, default: 0] += 1 }
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var count = needCount
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var minLength = Int.max
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var result = (-1, -1)
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var left = 0
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for right in 0..<big.count {
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let element = big[right]
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count -= 1
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}
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}
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while count == 0 {
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if right - left + 1 < minLength {
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minLength = right - left + 1
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result = (left, right)
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}
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let leftElement = big[left]
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if need[leftElement] != nil {
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window[leftElement]! -= 1
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left += 1
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}
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}
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return result.0 == -1 ? [] : [result.0, result.1]
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}
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}

‎lcci/17.18.Shortest Supersequence/README_EN.md

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var need = [Int: Int]()
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var window = [Int: Int]()
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small.forEach { need[$0, default: 0] += 1 }
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var count = needCount
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var minLength = Int.max
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var result = (-1, -1)
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var left = 0
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for right in 0..<big.count {
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let element = big[right]
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count -= 1
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}
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}
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while count == 0 {
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if right - left + 1 < minLength {
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minLength = right - left + 1
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result = (left, right)
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}
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let leftElement = big[left]
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if need[leftElement] != nil {
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window[leftElement]! -= 1
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left += 1
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}
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}
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return result.0 == -1 ? [] : [result.0, result.1]
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}
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}

‎lcci/17.19.Missing Two/README.md

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func missingTwo(_ nums: [Int]) -> [Int] {
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let n = nums.count + 2
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var xor = 0
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for num in nums {
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xor ^= num
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}
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for i in 1...n {
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xor ^= i
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}
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let diff = xor & (-xor)
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var a = 0
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for num in nums {
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if (num & diff) != 0 {
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a ^= num
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}
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}
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for i in 1...n {
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if (i & diff) != 0 {
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a ^= i
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}
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}
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let b = xor ^ a
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return [a, b]
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}

‎lcci/17.19.Missing Two/README_EN.md

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func missingTwo(_ nums: [Int]) -> [Int] {
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let n = nums.count + 2
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var xor = 0
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for num in nums {
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xor ^= num
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}
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for i in 1...n {
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xor ^= i
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}
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let diff = xor & (-xor)
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var a = 0
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for num in nums {
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if (num & diff) != 0 {
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a ^= num
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}
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}
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for i in 1...n {
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if (i & diff) != 0 {
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a ^= i
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}
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}
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let b = xor ^ a
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return [a, b]
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}

‎solution/3100-3199/3149.Find the Minimum Cost Array Permutation/README.md

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函数 $\text{dfs}(\text{mask}, \text{pre})$ 的计算过程如下：
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- 如果 $\text{mask}$ 的二进制表示中 $1$ 的个数为 $n$，即 $\text{mask} = 2^n - 1$，表示所有数字都已经被选取，此时返回 $\text{abs}(\text{pre} - \text{nums}[0])$；
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- 如果 $\text{mask}$ 的二进制表示中 $1$ 的个数为 $n$，即 $\text{mask} = 2^n - 1$，表示所有数字都已经被选取，此时返回 $|\text{pre} - \text{nums}[0]|$；
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- 否则，我们枚举下一个选取的数字 $\text{cur}$，如果数字 $\text{cur}$ 还未被选取，那么我们可以将数字 $\text{cur}$ 加入到排列中，此时排列的分数为 $|\text{pre} - \text{nums}[\text{cur}]| + \text{dfs}(\text{mask} \, | \, 1 << \text{cur}, \text{cur})$，我们需要取所有 $\text{cur}$ 中分数的最小值。
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最后，我们利用一个函数 $\text{g}(\text{mask}, \text{pre})$ 来构造得到最小分数的排列。我们首先将数字 $\text{pre}$ 加入到排列中，然后枚举下一个选取的数字 $\text{cur}$，如果数字 $\text{cur}$ 还未被选取，且满足 $|\text{pre} - \text{nums}[\text{cur}]| + \text{dfs}(\text{mask} \, | \, 1 << \text{cur}, \text{cur})$ 的值等于 $\text{dfs}(\text{mask}, \text{pre})$，那么我们就可以将数字 $\text{cur}$ 加入到排列中。

‎solution/3100-3199/3149.Find the Minimum Cost Array Permutation/README_EN.md

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@@ -61,7 +61,7 @@ We design a function $\text{dfs}(\text{mask}, \text{pre})$, which represents the
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The calculation process of the function $\text{dfs}(\text{mask}, \text{pre})$ is as follows:
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- If the number of $1$s in the binary representation of $\text{mask}$ is $n$, that is, $\text{mask} = 2^n - 1$, it means that all numbers have been selected, then return $\text{abs}(\text{pre} - \text{nums}[0])$;
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- If the number of $1$s in the binary representation of $\text{mask}$ is $n$, that is, $\text{mask} = 2^n - 1$, it means that all numbers have been selected, then return $|\text{pre} - \text{nums}[0]|$;
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- Otherwise, we enumerate the next selected number $\text{cur}$. If the number $\text{cur}$ has not been selected yet, then we can add the number $\text{cur}$ to the permutation. At this time, the score of the permutation is $|\text{pre} - \text{nums}[\text{cur}]| + \text{dfs}(\text{mask} \, | \, 1 << \text{cur}, \text{cur})$. We need to take the minimum score among all $\text{cur}$.
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Finally, we use a function $\text{g}(\text{mask}, \text{pre})$ to construct the permutation that gets the minimum score. We first add the number $\text{pre}$ to the permutation, and then enumerate the next selected number $\text{cur}$. If the number $\text{cur}$ has not been selected yet, and it satisfies that the value of $|\text{pre} - \text{nums}[\text{cur}]| + \text{dfs}(\text{mask} \, | \, 1 << \text{cur}, \text{cur})$ is equal to $\text{dfs}(\text{mask}, \text{pre})$, then we can add the number $\text{cur}$ to the permutation.

‎solution/3100-3199/3150.Invalid Tweets II/README.md

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# [3150. Invalid Tweets II 🔒](https://leetcode.cn/problems/invalid-tweets-ii)
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# [3150. 无效的推文 II 🔒](https://leetcode.cn/problems/invalid-tweets-ii)
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[English Version](/solution/3100-3199/3150.Invalid%20Tweets%20II/README_EN.md)
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<!-- 这里写题目描述 -->
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<p>Table: <code>Tweets</code></p>
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<p>表：<code>Tweets</code></p>
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<pre>
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+----------------+---------+
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| tweet_id | int |
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| content | varchar |
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+----------------+---------+
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tweet_id is the primary key (column with unique values) for this table.
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This table contains all the tweets in a social media app.
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</pre>
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tweet_id 是这个表的主键（有不同值的列）。
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这个表包含某社交媒体 App 中所有的推文。</pre>
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<p>Write a solution to find <strong>invalid tweets</strong>. A tweet is considered invalid if it meets <strong>any</strong> of the following criteria:</p>
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<p>编写一个解决方案来找到 <strong>无效的推文</strong>。如果一条推文满足下面 <strong>任一</strong>&nbsp;条件会被认为无效：</p>
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<ul>
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<li>It exceeds <code>140</code> characters in length.</li>
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<li>It has more than <code>3</code> mentions.</li>
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<li>It includes more than <code><font face="monospace">3</font></code>&nbsp;hashtags.</li>
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<li>长度超过&nbsp;<code>140</code>&nbsp;个字符。</li>
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<li>有超过&nbsp;<code>3</code>&nbsp;次提及。</li>
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<li>有超过&nbsp;<code><font face="monospace">3</font></code>&nbsp;个标签。</li>
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</ul>
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<p>Return <em>the result table ordered by</em> <code>tweet_id</code> <em>in <strong>ascending</strong> order</em>.</p>
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<p>以&nbsp;<code>tweet_id</code> <em><strong>升序</strong>&nbsp;</em>返回结果表。</p>
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<p>The result format is in the following example.</p>
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<p>查询结果格式如下所示：</p>
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<p>&nbsp;</p>
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<p><strong>Example:</strong></p>
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<p><strong>示例：</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong></p>
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<p><b>输入：</b></p>
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<p>Tweets table:</p>
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<p>Tweets 表：</p>
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<pre class="example-io">
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+----------+-----------------------------------------------------------------------------------+
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+----------+-----------------------------------------------------------------------------------+
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</pre>
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<p><strong>Output:</strong></p>
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<p><strong>输出：</strong></p>
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<pre class="example-io">
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</pre>
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<p><strong>Explanation:</strong></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>tweet_id&nbsp;1 contains 4&nbsp;mentions.</li>
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<li>tweet_id 4 contains 4 hashtags.</li>
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<li>tweet_id&nbsp;1 包含 4&nbsp;次提及。</li>
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<li>tweet_id 4 包含 4 个标签。</li>
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</ul>
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Output table is ordered by tweet_id in ascending order.</div>
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输出表以 tweet_id 升序排序。</div>
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## 解法
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‎solution/DATABASE_README.md

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| 3124 | [查找最长的电话](/solution/3100-3199/3124.Find%20Longest%20Calls/README.md) | `数据库` | 中等 | 🔒 |
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| 3126 | [Server Utilization Time](/solution/3100-3199/3126.Server%20Utilization%20Time/README.md) | `数据库` | 中等 | 🔒 |
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| 3140 | [连续空余座位 II](/solution/3100-3199/3140.Consecutive%20Available%20Seats%20II/README.md) | `数据库` | 中等 | 🔒 |
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| 3150 | [Invalid Tweets II](/solution/3100-3199/3150.Invalid%20Tweets%20II/README.md) | | 简单 | 🔒 |
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| 3150 | [无效的推文 II](/solution/3100-3199/3150.Invalid%20Tweets%20II/README.md) | | 简单 | 🔒 |
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## 版权
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‎solution/README.md

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| 3147 | [从魔法师身上吸取的最大能量](/solution/3100-3199/3147.Taking%20Maximum%20Energy%20From%20the%20Mystic%20Dungeon/README.md) | | 中等 | 第 397 场周赛 |
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| 3148 | [矩阵中的最大得分](/solution/3100-3199/3148.Maximum%20Difference%20Score%20in%20a%20Grid/README.md) | | 中等 | 第 397 场周赛 |
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| 3149 | [找出分数最低的排列](/solution/3100-3199/3149.Find%20the%20Minimum%20Cost%20Array%20Permutation/README.md) | | 困难 | 第 397 场周赛 |
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| 3150 | [Invalid Tweets II](/solution/3100-3199/3150.Invalid%20Tweets%20II/README.md) | | 简单 | 🔒 |
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| 3150 | [无效的推文 II](/solution/3100-3199/3150.Invalid%20Tweets%20II/README.md) | | 简单 | 🔒 |
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## 版权
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