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Commit af6edcf

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feat: add solutions to lcci problems (#1773)
* No.10.11.Peaks and Valleys * No.16.01.Swap Numbers * No.16.02.Words Frequency * No.16.05.Factorial Zeros * No.16.07.Maximum * No.16.11.Diving Board * No.16.14.Best Line * No.16.15.Master Mind * No.16.16.Sub Sort * No.16.17.Contiguous Sequence * No.16.18.Pattern Matching * No.16.19.Pond Sizes * No.16.20.T9 * No.16.21.Sum Swap * No.16.22.Langtons Ant * No.16.24.Pairs With Sum
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‎.github/workflows/pretter-lint.yml‎

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name: prettier-linter
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on:
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push: {}
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pull_request: {}
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push: {}
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pull_request: {}
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jobs:
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prettier:

‎lcci/10.11.Peaks and Valleys/README_EN.md‎

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## Solutions
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**Solution 1: Sorting**
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We first sort the array, and then traverse the array and swap the elements at even indices with their next element.
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The time complexity is $O(n \times \log n),ドル and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.
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### **Python3**

‎lcci/16.01.Swap Numbers/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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异或运算。
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**方法一:位运算**
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我们可以使用异或运算 $\oplus$ 来实现两个数的交换。
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异或运算有以下三个性质。
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- 任何数和 0ドル$ 做异或运算,结果仍然是原来的数,即 $a \oplus 0=a$。
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- 任何数和其自身做异或运算,结果是 0ドル,ドル即 $a \oplus a=0$。
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- 异或运算满足交换律和结合律,即 $a \oplus b \oplus a=b \oplus a \oplus a=b \oplus (a \oplus a)=b \oplus 0=b$。
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因此,我们可以对 $numbers$ 中的两个数 $a$ 和 $b$ 进行如下操作:
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- $a=a \oplus b,ドル此时 $a$ 中存储了两个数的异或结果;
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- $b=a \oplus b,ドル此时 $b$ 中存储了原来 $a$ 的值;
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- $a=a \oplus b,ドル此时 $a$ 中存储了原来 $b$ 的值;
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这样,我们就可以实现在不使用临时变量的情况下对两个数进行交换。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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‎lcci/16.01.Swap Numbers/README_EN.md‎

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## Solutions
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**Solution 1: Bitwise Operation**
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We can use the XOR operation $\oplus$ to implement the swap of two numbers.
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The XOR operation has the following three properties:
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- Any number XORed with 0ドル$ remains unchanged, i.e., $a \oplus 0=a$.
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- Any number XORed with itself results in 0ドル,ドル i.e., $a \oplus a=0$.
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- The XOR operation satisfies the commutative and associative laws, i.e., $a \oplus b \oplus a=b \oplus a \oplus a=b \oplus (a \oplus a)=b \oplus 0=b$.
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Therefore, we can perform the following operations on two numbers $a$ and $b$ in the array $numbers$:
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- $a=a \oplus b,ドル now $a$ stores the XOR result of the two numbers;
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- $b=a \oplus b,ドル now $b$ stores the original value of $a$;
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- $a=a \oplus b,ドル now $a$ stores the original value of $b$;
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In this way, we can swap two numbers without using a temporary variable.
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The time complexity is $O(1),ドル and the space complexity is $O(1)$.
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### **Python3**

‎lcci/16.02.Words Frequency/README.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>设计一个方法,找出任意指定单词在一本书中的出现频率。</p>
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<p>你的实现应该支持如下操作:</p>
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<ul>
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**方法一:哈希表**
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我们用哈希表 `cnt` 统计每个单词出现的次数,`get` 函数直接返回 `cnt[word]` 即可。
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我们用哈希表 $cnt$ 统计 $book$ 中每个单词出现的次数。
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调用 `get` 函数时,我们只需要返回 $cnt$ 中对应的单词的出现次数即可。
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初始化哈希表 `cnt` 的时间复杂度为 $O(n),ドル其中 $n$ 为 `book` 的长度。`get` 函数的时间复杂度为 $O(1)$。空间复杂度为 $O(n)$。
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时间复杂度方面,初始化哈希表 $cnt$ 的时间复杂度为 $O(n),ドル其中 $n$ 为 $book$ 的长度。`get` 函数的时间复杂度为 $O(1)$。空间复杂度为 $O(n)$。
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```rust
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use std::collections::HashMap;
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struct WordsFrequency {
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counter: HashMap<String, i32>
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cnt: HashMap<String, i32>
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}
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impl WordsFrequency {
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fn new(book: Vec<String>) -> Self {
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let mut counter = HashMap::new();
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let mut cnt = HashMap::new();
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for word in book.into_iter() {
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*counter.entry(word).or_insert(0) += 1;
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*cnt.entry(word).or_insert(0) += 1;
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}
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Self { counter }
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Self { cnt }
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}
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fn get(&self, word: String) -> i32 {
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*self.counter.get(&word).unwrap_or(&0)
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*self.cnt.get(&word).unwrap_or(&0)
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}
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}
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‎lcci/16.02.Words Frequency/README_EN.md‎

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## Solutions
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**Solution 1: Hash Table**
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We use a hash table $cnt$ to count the number of occurrences of each word in $book$.
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When calling the `get` function, we only need to return the number of occurrences of the corresponding word in $cnt$.
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In terms of time complexity, the time complexity of initializing the hash table $cnt$ is $O(n),ドル where $n$ is the length of $book$. The time complexity of the `get` function is $O(1)$. The space complexity is $O(n)$.
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### **Python3**
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```rust
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use std::collections::HashMap;
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struct WordsFrequency {
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counter: HashMap<String, i32>
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cnt: HashMap<String, i32>
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}
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impl WordsFrequency {
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fn new(book: Vec<String>) -> Self {
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let mut counter = HashMap::new();
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let mut cnt = HashMap::new();
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for word in book.into_iter() {
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*counter.entry(word).or_insert(0) += 1;
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*cnt.entry(word).or_insert(0) += 1;
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}
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Self { counter }
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Self { cnt }
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}
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fn get(&self, word: String) -> i32 {
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*self.counter.get(&word).unwrap_or(&0)
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*self.cnt.get(&word).unwrap_or(&0)
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}
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}
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class WordsFrequency:
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def __init__(self, book: List[str]):
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self.cnt = Counter(book)
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def get(self, word: str) -> int:
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return self.cnt[word]
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# Your WordsFrequency object will be instantiated and called as such:
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# obj = WordsFrequency(book)
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# param_1 = obj.get(word)
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class WordsFrequency:
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def __init__(self, book: List[str]):
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self.cnt = Counter(book)
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def get(self, word: str) -> int:
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return self.cnt[word]
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# Your WordsFrequency object will be instantiated and called as such:
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# obj = WordsFrequency(book)
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# param_1 = obj.get(word)
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use std::collections::HashMap;
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struct WordsFrequency {
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counter: HashMap<String, i32>
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}
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/**
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* `&self` means the method takes an immutable reference.
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* If you need a mutable reference, change it to `&mut self` instead.
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*/
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impl WordsFrequency {
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fn new(book: Vec<String>) -> Self {
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let mut counter = HashMap::new();
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for word in book.into_iter() {
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*counter.entry(word).or_insert(0) += 1;
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}
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Self { counter }
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}
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fn get(&self, word: String) -> i32 {
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*self.counter.get(&word).unwrap_or(&0)
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}
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}
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/**
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* Your WordsFrequency object will be instantiated and called as such:
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* let obj = WordsFrequency::new(book);
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* let ret_1: i32 = obj.get(word);
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use std::collections::HashMap;
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struct WordsFrequency {
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cnt: HashMap<String, i32>
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}
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/**
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* `&self` means the method takes an immutable reference.
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* If you need a mutable reference, change it to `&mut self` instead.
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*/
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impl WordsFrequency {
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fn new(book: Vec<String>) -> Self {
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let mut cnt = HashMap::new();
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for word in book.into_iter() {
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*cnt.entry(word).or_insert(0) += 1;
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}
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Self { cnt }
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}
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fn get(&self, word: String) -> i32 {
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*self.cnt.get(&word).unwrap_or(&0)
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}
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}
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/**
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* Your WordsFrequency object will be instantiated and called as such:
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* let obj = WordsFrequency::new(book);
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* let ret_1: i32 = obj.get(word);
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*/

‎lcci/16.05.Factorial Zeros/README_EN.md‎

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## Solutions
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**Solution 1: Mathematics**
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The problem is actually asking for the number of factors of 5ドル$ in $[1,n]$.
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Let's take 130ドル$ as an example:
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1. Divide 130ドル$ by 5ドル$ for the first time, and get 26ドル,ドル which means there are 26ドル$ numbers containing a factor of 5ドル$.
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2. Divide 26ドル$ by 5ドル$ for the second time, and get 5ドル,ドル which means there are 5ドル$ numbers containing a factor of 5ドル^2$.
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3. Divide 5ドル$ by 5ドル$ for the third time, and get 1ドル,ドル which means there is 1ドル$ number containing a factor of 5ドル^3$.
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4. Add up all the counts to get the total number of factors of 5ドル$ in $[1,n]$.
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The time complexity is $O(\log n),ドル and the space complexity is $O(1)$.
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### **Python3**

‎lcci/16.07.Maximum/README_EN.md‎

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## Solutions
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**Solution 1: Bitwise Operation**
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We can extract the sign bit $k$ of $a-b$. If the sign bit is 1ドル,ドル it means $a \lt b$; if the sign bit is 0ドル,ドル it means $a \ge b$.
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Then the final result is $a \times (k \oplus 1) + b \times k$.
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The time complexity is $O(1),ドル and the space complexity is $O(1)$.
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### **Python3**

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