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feat: add solutions to lc problems: No.2000+ (doocs#2095)
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‎solution/2000-2099/2000.Reverse Prefix of Word/README_EN.md‎

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## Solutions
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**Solution 1: Simulation**
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First, we find the index $i$ where the character $ch$ first appears. Then, we reverse the characters from index 0ドル$ to index $i$ (including index $i$). Finally, we concatenate the reversed string with the string starting from index $i + 1$.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the string $word$.
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‎solution/2000-2099/2001.Number of Pairs of Interchangeable Rectangles/README.md‎

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为了能够唯一表示矩形,我们需要将矩形的宽高比化简为最简分数。因此,我们可以求出每个矩形的宽高比的最大公约数,然后将宽高比化简为最简分数。接下来,我们使用哈希表统计每个最简分数的矩形数量,然后计算每个最简分数的矩形数量的组合数,即可得到答案。
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时间复杂度 $O(n \log M),ドル空间复杂度 $O(n)$。其中 $n$ 和 $M$ 分别是矩形的数量和矩形的最大边长。
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时间复杂度 $O(n \times \log M),ドル空间复杂度 $O(n)$。其中 $n$ 和 $M$ 分别是矩形的数量和矩形的最大边长。
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‎solution/2000-2099/2001.Number of Pairs of Interchangeable Rectangles/README_EN.md‎

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## Solutions
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**Solution 1: Mathematics + Hash Table**
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In order to uniquely represent a rectangle, we need to simplify the width-to-height ratio of the rectangle to a simplest fraction. Therefore, we can find the greatest common divisor of the width-to-height ratio of each rectangle, and then simplify the width-to-height ratio to the simplest fraction. Next, we use a hash table to count the number of rectangles for each simplest fraction, and then calculate the combination of the number of rectangles for each simplest fraction to get the answer.
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The time complexity is $O(n \times \log M),ドル and the space complexity is $O(n)$. Here, $n$ and $M$ are the number of rectangles and the maximum side length of the rectangles, respectively.
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‎solution/2000-2099/2002.Maximum Product of the Length of Two Palindromic Subsequences/README_EN.md‎

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## Solutions
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**Solution 1: Binary Enumeration**
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We notice that the length of the string $s$ does not exceed 12ドル,ドル so we can use the method of binary enumeration to enumerate all subsequences of $s$. Suppose the length of $s$ is $n,ドル we can use 2ドル^n$ binary numbers of length $n$ to represent all subsequences of $s$. For each binary number, the $i$-th bit being 1ドル$ means the $i$-th character of $s$ is in the subsequence, and 0ドル$ means it is not in the subsequence. For each binary number, we judge whether it is a palindrome subsequence and record it in the array $p$.
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Next, we enumerate each number $i$ in $p$. If $i$ is a palindrome subsequence, then we can enumerate a number $j$ from the complement of $i,ドル $mx = (2^n - 1) \oplus i$. If $j$ is also a palindrome subsequence, then $i$ and $j$ are the two palindrome subsequences we are looking for. Their lengths are the number of 1ドル$s in the binary representation of $i$ and $j,ドル denoted as $a$ and $b,ドル respectively. Then their product is $a \times b$. We take the maximum of all possible $a \times b$.
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The time complexity is $(2^n \times n + 3^n),ドル and the space complexity is $O(2^n)$. Here, $n$ is the length of the string $s$.
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‎solution/2000-2099/2006.Count Number of Pairs With Absolute Difference K/README.md‎

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**方法一:暴力枚举**
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我们注意到,数组 `nums` 的长度不超过 200ドル,ドル因此我们可以枚举所有的数对 $(i, j),ドル其中 $i < j,ドル并判断 $|nums[i] - nums[j]|$ 是否等于 $k,ドル是则答案加一。
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我们注意到,数组 $nums$ 的长度不超过 200ドル,ドル因此我们可以枚举所有的数对 $(i, j),ドル其中 $i < j,ドル并判断 $|nums[i] - nums[j]|$ 是否等于 $k,ドル是则答案加一。
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最后返回答案即可。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(1)$。其中 $n$ 为数组 `nums` 的长度。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(1)$。其中 $n$ 为数组 $nums$ 的长度。
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**方法二:哈希表或数组**
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我们可以使用哈希表或数组记录数组 `nums` 中每个数出现的次数,然后枚举数组 `nums` 中的每个数 $x,ドル判断 $x + k$ 和 $x - k$ 是否在数组 `nums` 中,是则答案加上 $x+k$ 和 $x-k$ 出现的次数之和。
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我们可以使用哈希表或数组记录数组 $nums$ 中每个数出现的次数,然后枚举数组 $nums$ 中的每个数 $x,ドル判断 $x + k$ 和 $x - k$ 是否在数组 $nums$ 中,是则答案加上 $x+k$ 和 $x-k$ 出现的次数之和。
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最后返回答案即可。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 `nums` 的长度。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。
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‎solution/2000-2099/2006.Count Number of Pairs With Absolute Difference K/README_EN.md‎

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## Solutions
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**Solution 1: Brute Force Enumeration**
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We notice that the length of the array $nums$ does not exceed 200ドル,ドル so we can enumerate all pairs $(i, j),ドル where $i < j,ドル and check if $|nums[i] - nums[j]|$ equals $k$. If it does, we increment the answer by one.
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Finally, we return the answer.
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The time complexity is $O(n^2),ドル and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.
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**Solution 2: Hash Table or Array**
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We can use a hash table or array to record the occurrence count of each number in the array $nums$. Then, we enumerate each number $x$ in the array $nums,ドル and check if $x + k$ and $x - k$ are in the array $nums$. If they are, we increment the answer by the sum of the occurrence counts of $x + k$ and $x - k$.
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Finally, we return the answer.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
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‎solution/2000-2099/2007.Find Original Array From Doubled Array/README_EN.md‎

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## Solutions
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**Solution 1: Sorting + Counting + Traversal**
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First, we check if the length $n$ of the array `changed` is odd. If it is, we directly return an empty array.
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Then, we sort the array `changed`, and use a hash table or array `cnt` to count the occurrence of each element in `changed`.
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Next, we traverse the array `changed`. For each element $x$ in `changed`, we first check if $x$ exists in the hash table `cnt`. If it does not exist, we directly skip this element. Otherwise, we check if $x \times 2$ exists in `cnt`. If it does not exist, we directly return an empty array. Otherwise, we add $x$ to the answer array `ans`, and decrease the occurrence counts of $x$ and $x \times 2$ in `cnt` by 1ドル$ each.
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After the traversal, we check if the length of the answer array `ans` is $\frac{n}{2}$. If it is, we return `ans`, otherwise we return an empty array.
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The time complexity is $O(n \times \log n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the array `changed`.
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‎solution/2000-2099/2009.Minimum Number of Operations to Make Array Continuous/README_EN.md‎

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## Solutions
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**Solution 1: Sorting + Deduplication + Binary Search**
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First, we sort the array and remove duplicates.
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Then, we traverse the array, enumerating the current element $nums[i]$ as the minimum value of the consecutive array. We use binary search to find the first position $j$ that is greater than $nums[i] + n - 1$. Then, $j-i$ is the length of the consecutive array when the current element is the minimum value. We update the answer, i.e., $ans = \min(ans, n - (j - i))$.
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Finally, we return $ans$.
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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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**Solution 2: Sorting + Deduplication + Two Pointers**
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Similar to Solution 1, we first sort the array and remove duplicates.
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Then, we traverse the array, enumerating the current element $nums[i]$ as the minimum value of the consecutive array. We use two pointers to find the first position $j$ that is greater than $nums[i] + n - 1$. Then, $j-i$ is the length of the consecutive array when the current element is the minimum value. We update the answer, i.e., $ans = \min(ans, n - (j - i))$.
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Finally, we return $ans$.
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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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‎solution/2000-2099/2011.Final Value of Variable After Performing Operations/README.md‎

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遍历数组 `operations`,对于每个操作 $operations[i],ドル如果包含 `'+'`,那么答案加 1ドル,ドル否则答案减 1ドル$。
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时间复杂度为 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为数组 `operations` 的长度。
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时间复杂度为 $O(n),ドル其中 $n$ 为数组 `operations` 的长度。空间复杂度 $O(1)$。
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‎solution/2000-2099/2011.Final Value of Variable After Performing Operations/README_EN.md‎

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## Solutions
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**Solution 1: Simulation**
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Traverse the array `operations`. For each operation $operations[i],ドル if it contains `'+'`, then the answer increases by 1ドル,ドル otherwise the answer decreases by 1ドル$.
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The time complexity is $O(n),ドル where $n$ is the length of the array `operations`. The space complexity is $O(1)$.
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