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feat: update lc problems (doocs#3634)
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‎solution/0200-0299/0297.Serialize and Deserialize Binary Tree/README_EN.md‎

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<p>Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.</p>
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<p><strong>Clarification:</strong> The input/output format is the same as <a href="https://support.leetcode.com/hc/en-us/articles/360011883654-What-does-1-null-2-3-mean-in-binary-tree-representation-" target="_blank">how LeetCode serializes a binary tree</a>. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.</p>
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<p><strong>Clarification:</strong> The input/output format is the same as <a href="https://support.leetcode.com/hc/en-us/articles/32442719377939-How-to-create-test-cases-on-LeetCode#h_01J5EGREAW3NAEJ14XC07GRW1A" target="_blank">how LeetCode serializes a binary tree</a>. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>

‎solution/0500-0599/0593.Valid Square/README.md‎

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<pre>
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<strong>输入:</strong> p1 = [0,0], p2 = [1,1], p3 = [1,0], p4 = [0,1]
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<strong>输出:</strong> True
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<strong>输出:</strong> true
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</pre>
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<p><strong>示例 2:</strong></p>

‎solution/0500-0599/0594.Longest Harmonious Subsequence/README.md‎

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<p>和谐数组是指一个数组里元素的最大值和最小值之间的差别 <strong>正好是 <code>1</code></strong> 。</p>
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<p>给你一个整数数组 <code>nums</code> ,请你在所有可能的子序列中找到最长的和谐子序列的长度。</p>
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<p>给你一个整数数组 <code>nums</code> ,请你在所有可能的 <spandata-keyword="subsequence-array">子序列</span> 中找到最长的和谐子序列的长度。</p>
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<p>数组的 <strong>子序列</strong> 是一个由数组派生出来的序列,它可以通过删除一些元素或不删除元素、且不改变其余元素的顺序而得到。</p>
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‎solution/0500-0599/0594.Longest Harmonious Subsequence/README_EN.md‎

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<p>We define a harmonious array as an array where the difference between its maximum value and its minimum value is <b>exactly</b> <code>1</code>.</p>
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<p>Given an integer array <code>nums</code>, return <em>the length of its longest harmonious subsequence among all its possible subsequences</em>.</p>
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<p>A <strong>subsequence</strong> of array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements.</p>
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<p>Given an integer array <code>nums</code>, return the length of its longest harmonious <span data-keyword="subsequence-array">subsequence</span> among all its possible subsequences.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>

‎solution/0900-0999/0975.Odd Even Jump/README.md‎

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<!-- description:start -->
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<p>给定一个整数数组 <code>A</code>,你可以从某一起始索引出发,跳跃一定次数。在你跳跃的过程中,第 1、3、5... 次跳跃称为奇数跳跃,而第 2、4、6... 次跳跃称为偶数跳跃。</p>
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<p>给定一个整数数组 <code>arr</code>,你可以从某一起始索引出发,跳跃一定次数。在你跳跃的过程中,第 1、3、5... 次跳跃称为奇数跳跃,而第 2、4、6... 次跳跃称为偶数跳跃。</p>
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<p>你可以按以下方式从索引 <code>i</code>&nbsp;向后跳转到索引 <code>j</code>(其中 <code>i &lt; j</code>):</p>
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<ul>
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<li>在进行奇数跳跃时(如,第&nbsp;1,3,5... 次跳跃),你将会跳到索引 <code>j</code>,使得 <code>A[i] &lt;=&nbsp;A[j]</code>,<code>A[j]</code> 是可能的最小值。如果存在多个这样的索引 <code>j</code>,你只能跳到满足要求的<strong>最小</strong>索引 <code>j</code> 上。</li>
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<li>在进行偶数跳跃时(如,第&nbsp;2,4,6... 次跳跃),你将会跳到索引&nbsp;<code>j</code>,使得 <code>A[i] &gt;= A[j]</code>,<code>A[j]</code> 是可能的最大值。如果存在多个这样的索引 <code>j</code>,你只能跳到满足要求的<strong>最小</strong>索引 <code>j</code>&nbsp;上。</li>
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<li>在进行奇数跳跃时(如,第&nbsp;1,3,5... 次跳跃),你将会跳到索引 <code>j</code>,使得 <code>arr[i] &lt;=&nbsp;arr[j]</code>,且 <code>arr[j]</code> 的值尽可能小。如果存在多个这样的索引 <code>j</code>,你只能跳到满足要求的<strong>最小</strong>索引 <code>j</code> 上。</li>
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<li>在进行偶数跳跃时(如,第&nbsp;2,4,6... 次跳跃),你将会跳到索引&nbsp;<code>j</code>,使得 <code>arr[i] &gt;= arr[j]</code>,且 <code>arr[j]</code> 的值尽可能大。如果存在多个这样的索引 <code>j</code>,你只能跳到满足要求的<strong>最小</strong>索引 <code>j</code>&nbsp;上。</li>
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<li>(对于某些索引 <code>i</code>,可能无法进行合乎要求的跳跃。)</li>
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</ul>
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<p>如果从某一索引开始跳跃一定次数(可能是 0 次或多次),就可以到达数组的末尾(索引 <code>A.length - 1</code>),那么该索引就会被认为是好的起始索引。</p>
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<p>如果从某一索引开始跳跃一定次数(可能是 0 次或多次),就可以到达数组的末尾(索引 <code>arr.length - 1</code>),那么该索引就会被认为是好的起始索引。</p>
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<p>返回好的起始索引的数量。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre><strong>输入:</strong>[10,13,12,14,15]
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<pre>
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<strong>输入:</strong>[10,13,12,14,15]
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<strong>输出:</strong>2
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<strong>解释: </strong>
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从起始索引 i = 0 出发,我们可以跳到 i = 2,(因为 A[2] 是 A[1],A[2],A[3],A[4] 中大于或等于 A[0] 的最小值),然后我们就无法继续跳下去了。
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从起始索引 i = 0 出发,我们可以跳到 i = 2,(因为 arr[2] 是 arr[1],arr[2],arr[3],arr[4] 中大于或等于 arr[0] 的最小值),然后我们就无法继续跳下去了。
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从起始索引 i = 1 和 i = 2 出发,我们可以跳到 i = 3,然后我们就无法继续跳下去了。
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从起始索引 i = 3 出发,我们可以跳到 i = 4,到达数组末尾。
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从起始索引 i = 4 出发,我们已经到达数组末尾。
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<p><strong>示例&nbsp;2:</strong></p>
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<pre><strong>输入:</strong>[2,3,1,1,4]
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<pre>
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<strong>输入:</strong>[2,3,1,1,4]
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<strong>输出:</strong>3
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<strong>解释:</strong>
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从起始索引 i=0 出发,我们依次可以跳到 i = 1,i = 2,i = 3:
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在我们的第一次跳跃(奇数)中,我们先跳到 i = 1,因为 A[1] 是(A[1],A[2],A[3],A[4])中大于或等于 A[0] 的最小值。
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在我们的第一次跳跃(奇数)中,我们先跳到 i = 1,因为 arr[1] 是(arr[1],arr[2],arr[3],arr[4])中大于或等于 arr[0] 的最小值。
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在我们的第二次跳跃(偶数)中,我们从 i = 1 跳到 i = 2,因为 A[2] 是(A[2],A[3],A[4])中小于或等于 A[1] 的最大值。A[3] 也是最大的值,但 2 是一个较小的索引,所以我们只能跳到 i = 2,而不能跳到 i = 3。
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在我们的第二次跳跃(偶数)中,我们从 i = 1 跳到 i = 2,因为 arr[2] 是(arr[2],arr[3],arr[4])中小于或等于 arr[1] 的最大值。arr[3] 也是最大的值,但 2 是一个较小的索引,所以我们只能跳到 i = 2,而不能跳到 i = 3。
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在我们的第三次跳跃(奇数)中,我们从 i = 2 跳到 i = 3,因为 A[3] 是(A[3],A[4])中大于或等于 A[2] 的最小值。
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在我们的第三次跳跃(奇数)中,我们从 i = 2 跳到 i = 3,因为 arr[3] 是(arr[3],arr[4])中大于或等于 arr[2] 的最小值。
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我们不能从 i = 3 跳到 i = 4,所以起始索引 i = 0 不是好的起始索引。
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<p><strong>示例 3:</strong></p>
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<pre><strong>输入:</strong>[5,1,3,4,2]
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<pre>
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<strong>输入:</strong>[5,1,3,4,2]
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<strong>输出:</strong>3
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<strong>解释: </strong>
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我们可以从起始索引 1,2,4 出发到达数组末尾。
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<p><strong>提示:</strong></p>
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<ol>
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<li><code>1 &lt;= A.length &lt;= 20000</code></li>
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<li><code>0 &lt;= A[i] &lt; 100000</code></li>
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<li><code>1 &lt;= arr.length &lt;= 20000</code></li>
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<li><code>0 &lt;= arr[i] &lt; 100000</code></li>
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</ol>
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<!-- description:end -->

‎solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README.md‎

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<p><strong>示例 1:</strong></p>
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<pre><strong>输入:</strong>n = 234
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<strong>输出:</strong>15
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<strong>输出:</strong>15
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<strong>解释:</strong>
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各位数之积 = 2 * 3 * 4 = 24
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各位数之和 = 2 + 3 + 4 = 9
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各位数之积 = 2 * 3 * 4 = 24
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各位数之和 = 2 + 3 + 4 = 9
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结果 = 24 - 9 = 15
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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 = 4421
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<strong>输出:</strong>21
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<strong>解释:
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</strong>各位数之积 = 4 * 4 * 2 * 1 = 32
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各位数之和 = 4 +たす 4 +たす 2 +たす 1 =わ 11
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<strong>解释:
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</strong>各位数之积 = 4 * 4 * 2 * 1 = 32
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各位数之和 = 4 +たす 4 +たす 2 +たす 1 =わ 11
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结果 = 32 - 11 = 21
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</pre>
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‎solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README_EN.md‎

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<pre>
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<strong>Input:</strong> n = 234
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<strong>Output:</strong> 15
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<b>Explanation:</b>
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Product of digits = 2 * 3 * 4 = 24
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Sum of digits = 2 + 3 + 4 = 9
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<strong>Output:</strong> 15
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<b>Explanation:</b>
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Product of digits = 2 * 3 * 4 = 24
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Sum of digits = 2 + 3 + 4 = 9
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Result = 24 - 9 = 15
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</pre>
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<pre>
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<strong>Input:</strong> n = 4421
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<strong>Output:</strong> 21
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<b>Explanation:
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</b>Product of digits = 4 * 4 * 2 * 1 = 32
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Sum of digits = 4 +たす 4 +たす 2 +たす 1 =わ 11
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<b>Explanation:
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</b>Product of digits = 4 * 4 * 2 * 1 = 32
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Sum of digits = 4 +たす 4 +たす 2 +たす 1 =わ 11
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Result = 32 - 11 = 21
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</pre>
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‎solution/1200-1299/1295.Find Numbers with Even Number of Digits/README.md‎

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<pre>
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<strong>输入:</strong>nums = [555,901,482,1771]
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<strong>输出:</strong>1
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<strong>输出:</strong>1
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<strong>解释: </strong>
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只有 1771 是位数为偶数的数字。
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</pre>

‎solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md‎

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<pre>
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<strong>Input:</strong> nums = [12,345,2,6,7896]
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<strong>Output:</strong> 2
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<strong>Explanation:
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<strong>Explanation:
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</strong>12 contains 2 digits (even number of digits).&nbsp;
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345 contains 3 digits (odd number of digits).&nbsp;
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2 contains 1 digit (odd number of digits).&nbsp;
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<pre>
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<strong>Input:</strong> nums = [555,901,482,1771]
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<strong>Output:</strong> 1
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<strong>Output:</strong> 1
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<strong>Explanation: </strong>
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Only 1771 contains an even number of digits.
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</pre>

‎solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README_EN.md‎

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<pre>
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<strong>Input:</strong> arr = [17,18,5,4,6,1]
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<strong>Output:</strong> [18,6,6,6,1,-1]
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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- index 0 --&gt; the greatest element to the right of index 0 is index 1 (18).
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- index 1 --&gt; the greatest element to the right of index 1 is index 4 (6).
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- index 2 --&gt; the greatest element to the right of index 2 is index 4 (6).

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