feat: add solutions to lc problems: No.2000+

solution/2000-2099/2000.Reverse Prefix of Word/README_EN.md

@@ -51,6 +51,12 @@ You should not do any reverse operation, the resulting string is "abcd&quot
 
 ## Solutions
 
+**Solution 1: Simulation**
+
+First, we find the index $i$ where the character $ch$ first appears. Then, we reverse the characters from index $0$ to index $i$ (including index $i$). Finally, we concatenate the reversed string with the string starting from index $i + 1$.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $word$.
+
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 ### **Python3**

solution/2000-2099/2001.Number of Pairs of Interchangeable Rectangles/README.md

@@ -55,7 +55,7 @@
 
 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。
 
-时间复杂度 $O(n \log M)$，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。
+时间复杂度 $O(n \times \log M)$，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

@@ -45,6 +45,12 @@
 
 ## Solutions
 
+**Solution 1: Mathematics + Hash Table**
+
+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.
+
+The time complexity is $O(n \times \log M)$, 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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 ### **Python3**

solution/2000-2099/2002.Maximum Product of the Length of Two Palindromic Subsequences/README_EN.md

@@ -48,6 +48,14 @@ The product of their lengths is: 5 * 5 = 25.
 
 ## Solutions
 
+**Solution 1: Binary Enumeration**
+
+We notice that the length of the string $s$ does not exceed $12$, so we can use the method of binary enumeration to enumerate all subsequences of $s$. Suppose the length of $s$ is $n$, we can use $2^n$ binary numbers of length $n$ to represent all subsequences of $s$. For each binary number, the $i$-th bit being $1$ means the $i$-th character of $s$ is in the subsequence, and $0$ 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 $p$.
+
+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 $i$, $mx = (2^n - 1) \oplus i$. 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 $1$s in the binary representation of $i$ and $j$, denoted as $a$ and $b$, respectively. Then their product is $a \times b$. We take the maximum of all possible $a \times b$.
+
+The time complexity is $(2^n \times n + 3^n)$, and the space complexity is $O(2^n)$. Here, $n$ is the length of the string $s$.
+
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 ### **Python3**

solution/2000-2099/2006.Count Number of Pairs With Absolute Difference K/README.md

@@ -61,19 +61,19 @@
 
 **方法一：暴力枚举**
 
-我们注意到，数组 `nums` 的长度不超过 $200$，因此我们可以枚举所有的数对 $(i, j)$，其中 $i < j$，并判断 $|nums[i] - nums[j]|$ 是否等于 $k$，是则答案加一。
+我们注意到，数组 $nums$ 的长度不超过 $200$，因此我们可以枚举所有的数对 $(i, j)$，其中 $i < j$，并判断 $|nums[i] - nums[j]|$ 是否等于 $k$，是则答案加一。
 
 最后返回答案即可。
 
-时间复杂度 $O(n^2)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `nums` 的长度。
+时间复杂度 $O(n^2)$，空间复杂度 $O(1)$。其中 $n$ 为数组 $nums$ 的长度。
 
 **方法二：哈希表或数组**
 
-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 $x$，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。
+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 $x$，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。
 
 最后返回答案即可。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 `nums` 的长度。
+时间复杂度 $O(n)$，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

@@ -56,6 +56,22 @@
 
 ## Solutions
 
+**Solution 1: Brute Force Enumeration**
+
+We notice that the length of the array $nums$ does not exceed $200$, so we can enumerate all pairs $(i, j)$, where $i < j$, and check if $|nums[i] - nums[j]|$ equals $k$. If it does, we increment the answer by one.
+
+Finally, we return the answer.
+
+The time complexity is $O(n^2)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.
+
+**Solution 2: Hash Table or Array**
+
+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 $nums$, 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 $x - k$.
+
+Finally, we return the answer.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
+
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 ### **Python3**

solution/2000-2099/2007.Find Original Array From Doubled Array/README_EN.md

@@ -47,6 +47,18 @@ Other original arrays could be [4,3,1] or [3,1,4].
 
 ## Solutions
 
+**Solution 1: Sorting + Counting + Traversal**
+
+First, we check if the length $n$ of the array `changed` is odd. If it is, we directly return an empty array.
+
+Then, we sort the array `changed`, and use a hash table or array `cnt` to count the occurrence of each element in `changed`.
+
+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 $1$ each.
+
+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.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array `changed`.
+
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 ### **Python3**

solution/2000-2099/2009.Minimum Number of Operations to Make Array Continuous/README_EN.md

@@ -57,6 +57,26 @@ The resulting array is [1,2,3,4], which is continuous.
 
 ## Solutions
 
+**Solution 1: Sorting + Deduplication + Binary Search**
+
+First, we sort the array and remove duplicates.
+
+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., $ans = \min(ans, n - (j - i))$.
+
+Finally, we return $ans$.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.
+
+**Solution 2: Sorting + Deduplication + Two Pointers**
+
+Similar to Solution 1, we first sort the array and remove duplicates.
+
+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., $ans = \min(ans, n - (j - i))$.
+
+Finally, we return $ans$.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.
+
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 ### **Python3**

solution/2000-2099/2011.Final Value of Variable After Performing Operations/README.md

@@ -73,7 +73,7 @@ X--：X 减 1 ，X = 1 - 1 = 0
 
 遍历数组 `operations`，对于每个操作 $operations[i]$，如果包含 `'+'`，那么答案加 $1$，否则答案减 $1$。
 
-时间复杂度为 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `operations` 的长度。
+时间复杂度为 $O(n)$，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

@@ -63,6 +63,12 @@ X--: X is decremented by 1, X = 1 - 1 = 0.
 
 ## Solutions
 
+**Solution 1: Simulation**
+
+Traverse the array `operations`. For each operation $operations[i]$, if it contains `'+'`, then the answer increases by $1$, otherwise the answer decreases by $1$.
+
+The time complexity is $O(n)$, where $n$ is the length of the array `operations`. The space complexity is $O(1)$.
+
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 ### **Python3**

solution/2000-2099/2012.Sum of Beauty in the Array/README.md

@@ -56,6 +56,16 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：预处理右侧最小值 + 遍历维护左侧最大值**
+
+我们可以预处理出右侧最小值数组 $right$，其中 $right[i]$ 表示 $nums[i..n-1]$ 中的最小值。
+
+然后我们从左到右遍历数组 $nums$，同时维护左侧最大值 $l$。对于每个位置 $i$，我们判断 $l < nums[i] < right[i + 1]$ 是否成立，如果成立则将 $2$ 累加至答案，否则判断 $nums[i - 1] < nums[i] < nums[i + 1]$ 是否成立，如果成立则将 $1$ 累加至答案。
+
+遍历结束后即可得到答案。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。
+
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 ### **Python3**
@@ -66,18 +76,18 @@
 class Solution:
     def sumOfBeauties(self, nums: List[int]) -> int:
         n = len(nums)
-        lmx = [0] * n
-        for i in range(1, n):
-            lmx[i] = max(lmx[i - 1], nums[i - 1])
-        rmi = [100001] * n
+        right = [nums[-1]] * n
         for i in range(n - 2, -1, -1):
-            rmi[i] = min(rmi[i + 1], nums[i + 1])
+            right[i] = min(right[i + 1], nums[i])
         ans = 0
+        l = nums[0]
         for i in range(1, n - 1):
-            if lmx[i] < nums[i] < rmi[i]:
+            r = right[i + 1]
+            if l < nums[i] < r:
                 ans += 2
             elif nums[i - 1] < nums[i] < nums[i + 1]:
                 ans += 1
+            l = max(l, nums[i])
         return ans
 ```
 
@@ -89,70 +99,47 @@ class Solution:
 class Solution {
     public int sumOfBeauties(int[] nums) {
         int n = nums.length;
-        int[] lmx = new int[n];
-        int[] rmi = new int[n];
-        rmi[n - 1] = 100001;
-        for (int i = 1; i < n; ++i) {
-            lmx[i] = Math.max(lmx[i - 1], nums[i - 1]);
-        }
-        for (int i = n - 2; i >= 0; --i) {
-            rmi[i] = Math.min(rmi[i + 1], nums[i + 1]);
+        int[] right = new int[n];
+        right[n - 1] = nums[n - 1];
+        for (int i = n - 2; i > 0; --i) {
+            right[i] = Math.min(right[i + 1], nums[i]);
         }
         int ans = 0;
+        int l = nums[0];
         for (int i = 1; i < n - 1; ++i) {
-            if (lmx[i] < nums[i] && nums[i] < rmi[i]) {
+            int r = right[i + 1];
+            if (l < nums[i] && nums[i] < r) {
                 ans += 2;
             } else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
                 ans += 1;
             }
+            l = Math.max(l, nums[i]);
         }
         return ans;
     }
 }
 ```
 
-### \*\*TypeScript
-
-```ts
-function sumOfBeauties(nums: number[]): number {
-    let n = nums.length;
-    let prefix = new Array(n),
-        postfix = new Array(n);
-    prefix[0] = nums[0];
-    postfix[n - 1] = nums[n - 1];
-    for (let i = 1, j = n - 2; i < n; ++i, --j) {
-        prefix[i] = Math.max(nums[i], prefix[i - 1]);
-        postfix[j] = Math.min(nums[j], postfix[j + 1]);
-    }
-    let ans = 0;
-    for (let i = 1; i < n - 1; ++i) {
-        if (prefix[i - 1] < nums[i] && nums[i] < postfix[i + 1]) {
-            ans += 2;
-        } else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
-            ans += 1;
-        }
-    }
-    return ans;
-}
-```
-
 ### **C++**
 
 ```cpp
 class Solution {
 public:
     int sumOfBeauties(vector<int>& nums) {
         int n = nums.size();
-        vector<int> lmx(n);
-        vector<int> rmi(n, 100001);
-        for (int i = 1; i < n; ++i) lmx[i] = max(lmx[i - 1], nums[i - 1]);
-        for (int i = n - 2; i >= 0; --i) rmi[i] = min(rmi[i + 1], nums[i + 1]);
+        vector<int> right(n, nums[n - 1]);
+        for (int i = n - 2; i; --i) {
+            right[i] = min(right[i + 1], nums[i]);
+        }
         int ans = 0;
-        for (int i = 1; i < n - 1; ++i) {
-            if (lmx[i] < nums[i] && nums[i] < rmi[i])
+        for (int i = 1, l = nums[0]; i < n - 1; ++i) {
+            int r = right[i + 1];
+            if (l < nums[i] && nums[i] < r) {
                 ans += 2;
-            else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1])
+            } else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
                 ans += 1;
+            }
+            l = max(l, nums[i]);
         }
         return ans;
     }
@@ -162,26 +149,46 @@ public:
 ### **Go**
 
 ```go
-func sumOfBeauties(nums []int) int {
+func sumOfBeauties(nums []int) (ans int) {
 	n := len(nums)
-	lmx := make([]int, n)
-	rmi := make([]int, n)
-	rmi[n-1] = 100001
-	for i := 1; i < n; i++ {
-		lmx[i] = max(lmx[i-1], nums[i-1])
-	}
-	for i := n - 2; i >= 0; i-- {
-		rmi[i] = min(rmi[i+1], nums[i+1])
+	right := make([]int, n)
+	right[n-1] = nums[n-1]
+	for i := n - 2; i > 0; i-- {
+		right[i] = min(right[i+1], nums[i])
 	}
-	ans := 0
-	for i := 1; i < n-1; i++ {
-		if lmx[i] < nums[i] && nums[i] < rmi[i] {
+	for i, l := 1, nums[0]; i < n-1; i++ {
+		r := right[i+1]
+		if l < nums[i] && nums[i] < r {
 			ans += 2
 		} else if nums[i-1] < nums[i] && nums[i] < nums[i+1] {
-			ans += 1
+			ans++
 		}
+		l = max(l, nums[i])
 	}
-	return ans
+	return
+}
+```
+
+### **TypeScript**
+
+```ts
+function sumOfBeauties(nums: number[]): number {
+    const n = nums.length;
+    const right: number[] = Array(n).fill(nums[n - 1]);
+    for (let i = n - 2; i; --i) {
+        right[i] = Math.min(right[i + 1], nums[i]);
+    }
+    let ans = 0;
+    for (let i = 1, l = nums[0]; i < n - 1; ++i) {
+        const r = right[i + 1];
+        if (l < nums[i] && nums[i] < r) {
+            ans += 2;
+        } else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
+            ans += 1;
+        }
+        l = Math.max(l, nums[i]);
+    }
+    return ans;
 }
 ```