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Commit 808fe99

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feat: update solutions to lc problems: No.2595,2596 (#4084)
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‎solution/2300-2399/2330.Valid Palindrome IV/README.md‎

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<strong>解释:</strong> 能让 s 变成回文,且只用了两步操作的方案如下:
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- 将 s[0] 变成 'b' ,得到 s = "ba" 。
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- 将 s[1] 变成 'b' ,得到s = "bb" 。
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执行两步操作让 s 变成一个回文,所以返回 true 。
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执行两步操作让 s 变成一个回文,所以返回 true 。
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</pre>
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<p><strong>示例 3:</strong></p>
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### 方法一:双指针
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我们可以使用双指针 $i$ 和 $j,ドル分别指向字符串的头尾,然后向中间移动,统计不同字符的个数,如果不同字符的个数大于 2ドル,ドル则返回 `false`,否则返回 `true`。
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我们可以使用双指针 $i$ 和 $j,ドル分别指向字符串的头尾,然后向中间移动,统计不同字符的个数,如果不同字符的个数大于 2ドル,ドル则返回 $\textit{false}$,否则返回 $\textit{true}$。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为字符串 $s$ 的长度。
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‎solution/2300-2399/2330.Valid Palindrome IV/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Two Pointers
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We can use two pointers $i$ and $j,ドル pointing to the beginning and end of the string, respectively, and then move towards the center, counting the number of different characters. If the number of different characters is greater than 2ドル,ドル return $\textit{false}$; otherwise, return $\textit{true}$.
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The time complexity is $O(n),ドル and the space complexity is $O(1)$. Here, $n$ is the length of the string $s$.
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<!-- tabs:start -->
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‎solution/2300-2399/2378.Choose Edges to Maximize Score in a Tree/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Tree DP
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We design a function $dfs(i),ドル which represents the maximum sum of the weights of selected edges in the subtree rooted at node $i,ドル such that no two selected edges are adjacent. This function returns two values $(a, b)$. The first value $a$ represents the sum of the weights of selected edges when the edge between the current node $i$ and its parent node is selected. The second value $b$ represents the sum of the weights of selected edges when the edge between the current node $i$ and its parent node is not selected.
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We can observe the following for the current node $i$:
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- If the edge between $i$ and its parent node is selected, then none of the edges between $i$ and its child nodes can be selected. In this case, the value of $a$ for the current node is the sum of the $b$ values of all its child nodes.
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- If the edge between $i$ and its parent node is not selected, then we can select at most one edge between $i$ and its child nodes. In this case, the value of $b$ for the current node is the sum of the $a$ values of the selected child nodes and the $b$ values of the unselected child nodes, plus the weight of the edge between $i$ and the selected child node.
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We call the function $dfs(0),ドル and the second value returned is the answer, which is the sum of the weights of selected edges when the edge between the root node and its parent node is not selected.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the number of nodes.
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‎solution/2500-2599/2510.Check if There is a Path With Equal Number of 0's And 1's/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Memoization Search
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According to the problem description, we know that the number of 0s and 1s on the path from the top-left corner to the bottom-right corner is equal, and the total number is $m + n - 1,ドル which means the number of 0s and 1s are both $(m + n - 1) / 2$.
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Therefore, we can use memoization search, starting from the top-left corner and moving right or down until reaching the bottom-right corner, to check if the number of 0s and 1s on the path is equal.
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The time complexity is $O(m \times n \times (m + n))$. Here, $m$ and $n$ are the number of rows and columns of the matrix, respectively.
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‎solution/2500-2599/2523.Closest Prime Numbers in Range/README.md‎

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### 方法一:线性筛
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对于给定的范围 $[left, right],ドル我们可以使用线性筛求出所有质数,然后从小到大遍历质数,找到相邻的两个质数,其差值最小的质数对即为答案。
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对于给定的范围 $[\textit{left}, \textit{right}],ドル我们可以使用线性筛求出所有质数,然后从小到大遍历质数,找到相邻的两个质数,其差值最小的质数对即为答案。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n = right$。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n = \textit{right}$。
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<!-- tabs:start -->
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‎solution/2500-2599/2523.Closest Prime Numbers in Range/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Linear Sieve
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For the given range $[\textit{left}, \textit{right}],ドル we can use the linear sieve method to find all prime numbers. Then, we traverse the prime numbers in ascending order to find the pair of adjacent prime numbers with the smallest difference, which will be the answer.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n = \textit{right}$.
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‎solution/2500-2599/2524.Maximum Frequency Score of a Subarray/README.md‎

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### 方法一:哈希表 + 滑动窗口 + 快速幂
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我们用哈希表 `cnt` 维护窗口大小为 $k$ 的元素及其出现的次数。
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我们用哈希表 $\textit{cnt}$ 维护窗口大小为 $k$ 的元素及其出现的次数。
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先算出初始窗口为 $k$ 的所有元素的分数。然后利用滑动窗口,每次加入一个元素,并移除最左边的元素,同时利用快速幂更新分数。
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时间复杂度 $O(n \times \log n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 `nums` 的长度。
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时间复杂度 $O(n \times \log n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{nums}$ 的长度。
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‎solution/2500-2599/2524.Maximum Frequency Score of a Subarray/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Hash Table + Sliding Window + Fast Power
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We use a hash table $\textit{cnt}$ to maintain the elements of the window of size $k$ and their frequencies.
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First, calculate the score of all elements in the initial window of size $k$. Then, use a sliding window to add one element at a time and remove the leftmost element, while updating the score using fast power.
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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 $\textit{nums}$.
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‎solution/2500-2599/2595.Number of Even and Odd Bits/README.md‎

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<pre><strong>输入:</strong>n = 17
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<strong>输出:</strong>[2,0]
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<strong>解释:</strong>17 的二进制形式是 10001 。
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下标 0 和 下标 4 对应的值为 1 。
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<strong>解释:</strong>17 的二进制形式是 10001 。
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下标 0 和 下标 4 对应的值为 1 。
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共有 2 个偶数下标,0 个奇数下标。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre><strong>输入:</strong>n = 2
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<strong>输出:</strong>[0,1]
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<strong>解释:</strong>2 的二进制形式是 10 。
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下标 1 对应的值为 1 。
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<strong>解释:</strong>2 的二进制形式是 10 。
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下标 1 对应的值为 1 。
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共有 0 个偶数下标,1 个奇数下标。
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</pre>
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```ts
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function evenOddBit(n: number): number[] {
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const ans = newArray(2).fill(0);
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const ans = Array(2).fill(0);
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for (let i = 0; n > 0; n >>= 1, i ^= 1) {
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ans[i] += n & 1;
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<!-- solution:start -->
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### 方法二
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### 方法二:位运算
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我们可以定义一个掩码 $\textit{mask} = \text{0x5555},ドル它的二进制表示为 $\text{0101 0101 0101 0101}_2$。那么 $n$ 与 $\textit{mask}$ 进行按位与运算,就可以得到 $n$ 的二进制表示中偶数下标的位,而 $n$ 与 $\textit{mask}$ 取反后再进行按位与运算,就可以得到 $n$ 的二进制表示中奇数下标的位。统计这两个结果中 1ドル$ 的个数即可。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。其中 $n$ 为给定的整数。
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‎solution/2500-2599/2595.Number of Even and Odd Bits/README_EN.md‎

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```ts
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function evenOddBit(n: number): number[] {
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const ans = Array(2).fill(0);
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for (let i = 0; n > 0; n >>= 1, i ^= 1) {
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<!-- solution:start -->
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### Solution 2
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### Solution 2: Bit Manipulation
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We can define a mask $\textit{mask} = \text{0x5555},ドル which is represented in binary as $\text{0101 0101 0101 0101}_2$. Then, performing a bitwise AND operation between $n$ and $\textit{mask}$ will give us the bits at even indices in the binary representation of $n$. Performing a bitwise AND operation between $n$ and the complement of $\textit{mask}$ will give us the bits at odd indices in the binary representation of $n$. We can count the number of 1s in these two results.
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The time complexity is $O(1),ドル and the space complexity is $O(1)$. Here, $n$ is the given integer.
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