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feat: update lc problems (doocs#3046)
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‎solution/0200-0299/0249.Group Shifted Strings/README_EN.md‎

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<!-- description:start -->
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<p>We can shift a string by shifting each of its letters to its successive letter.</p>
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<p>Perform the following shift operations on a string:</p>
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<ul>
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<li>For example, <code>&quot;abc&quot;</code> can be shifted to be <code>&quot;bcd&quot;</code>.</li>
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<li><strong>Right shift</strong>: Replace every letter with the <strong>successive</strong> letter of the English alphabet, where &#39;z&#39; is replaced by &#39;a&#39;. For example, <code>&quot;abc&quot;</code> can be right-shifted to <code>&quot;bcd&quot; </code>or <code>&quot;xyz&quot;</code> can be right-shifted to <code>&quot;yza&quot;</code>.</li>
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<li><strong>Left shift</strong>: Replace every letter with the <strong>preceding</strong> letter of the English alphabet, where &#39;a&#39; is replaced by &#39;z&#39;. For example, <code>&quot;bcd&quot;</code> can be left-shifted to <code>&quot;abc&quot;<font face="Times New Roman"> or </font></code><code>&quot;yza&quot;</code> can be left-shifted to <code>&quot;xyz&quot;</code>.</li>
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</ul>
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<p>We can keep shifting the string to form a sequence.</p>
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<p>We can keep shifting the string in both directions to form an <strong>endless</strong> <strong>shifting sequence</strong>.</p>
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<ul>
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<li>For example, we can keep shifting <code>&quot;abc&quot;</code> to form the sequence: <code>&quot;abc&quot; -&gt; &quot;bcd&quot; -&gt; ... -&gt; &quot;xyz&quot;</code>.</li>
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<li>For example, shift <code>&quot;abc&quot;</code> to form the sequence: <code>... &lt;-&gt; &quot;abc&quot; &lt;-&gt; &quot;bcd&quot; &lt;-&gt; ... &lt;-&gt; &quot;xyz&quot; &lt;-&gt; &quot;yza&quot; &lt;-&gt; ...</code>.<code> &lt;-&gt; &quot;zab&quot; &lt;-&gt; &quot;abc&quot; &lt;-&gt; ...</code></li>
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</ul>
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<p>Given an array of strings <code>strings</code>, group all <code>strings[i]</code> that belong to the same shifting sequence. You may return the answer in <strong>any order</strong>.</p>
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<p>You are given an array of strings <code>strings</code>, group together all <code>strings[i]</code> that belong to the same shifting sequence. You may return the answer in <strong>any order</strong>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre><strong>Input:</strong> strings = ["abc","bcd","acef","xyz","az","ba","a","z"]
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<strong>Output:</strong> [["acef"],["a","z"],["abc","bcd","xyz"],["az","ba"]]
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</pre><p><strong class="example">Example 2:</strong></p>
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<pre><strong>Input:</strong> strings = ["a"]
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<strong>Output:</strong> [["a"]]
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</pre>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">strings = [&quot;abc&quot;,&quot;bcd&quot;,&quot;acef&quot;,&quot;xyz&quot;,&quot;az&quot;,&quot;ba&quot;,&quot;a&quot;,&quot;z&quot;]</span></p>
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<p><strong>Output:</strong> <span class="example-io">[[&quot;acef&quot;],[&quot;a&quot;,&quot;z&quot;],[&quot;abc&quot;,&quot;bcd&quot;,&quot;xyz&quot;],[&quot;az&quot;,&quot;ba&quot;]]</span></p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">strings = [&quot;a&quot;]</span></p>
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<p><strong>Output:</strong> <span class="example-io">[[&quot;a&quot;]]</span></p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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‎solution/2500-2599/2595.Number of Even and Odd Bits/README_EN.md‎

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<p>You are given a <strong>positive</strong> integer <code>n</code>.</p>
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<p>Let <code>even</code> denote the number of even indices in the binary representation of <code>n</code> (<strong>0-indexed</strong>) with value <code>1</code>.</p>
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<p>Let <code>even</code> denote the number of even indices in the binary representation of <code>n</code> with value 1.</p>
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<p>Let <code>odd</code> denote the number of odd indices in the binary representation of <code>n</code> (<strong>0-indexed</strong>) with value <code>1</code>.</p>
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<p>Let <code>odd</code> denote the number of odd indices in the binary representation of <code>n</code> with value 1.</p>
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<p>Return <em>an integer array </em><code>answer</code><em> where </em><code>answer = [even, odd]</code>.</p>
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<p>Note that bits are indexed from <strong>right to left</strong> in the binary representation of a number.</p>
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<p>Return the array <code>[even, odd]</code>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> n = 17
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<strong>Output:</strong> [2,0]
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<strong>Explanation:</strong> The binary representation of 17 is 10001.
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It contains 1 on the 0<sup>th</sup> and 4<sup>th</sup> indices.
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There are 2 even and 0 odd indices.
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</pre>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">n = 50</span></p>
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<p><strong>Output:</strong> <span class="example-io">[1,2]</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The binary representation of 50 is <code>110010</code>.</p>
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<p>It contains 1 on indices 1, 4, and 5.</p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 2
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<strong>Output:</strong> [0,1]
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<strong>Explanation:</strong> The binary representation of 2 is 10.
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It contains 1 on the 1<sup>st</sup> index.
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There are 0 even and 1 odd indices.
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</pre>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">n = 2</span></p>
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<p><strong>Output:</strong> <span class="example-io">[0,1]</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The binary representation of 2 is <code>10</code>.</p>
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<p>It contains 1 only on index 1.</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>

‎solution/2600-2699/2644.Find the Maximum Divisibility Score/README_EN.md‎

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<p>The <strong>divisibility score</strong> of <code>divisors[i]</code> is the number of indices <code>j</code> such that <code>nums[j]</code> is divisible by <code>divisors[i]</code>.</p>
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<p>Return the integer <code>divisors[i]</code> with the <strong>maximum</strong> divisibility score. If multiple numbers have the maximum score, return the smallest one.</p>
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<p>Return the integer <code>divisors[i]</code> with the <strong>maximum</strong> divisibility score. If multiple integers have the maximum score, return the smallest one.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<p>The divisibility score of <code>divisors[1]</code> is 1 since only <code>nums[0]</code> is divisible by 2.</p>
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<p>The divisibility score of <code>divisors[2]</code> is 3 since <code>nums[2]</code>, <code>nums[3]</code> and <code>nums[4]</code> are divisible by 3.</p>
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<p>The divisibility score of <code>divisors[2]</code> is 3 since <code>nums[2]</code>, <code>nums[3]</code> and <code>nums[4]</code> are divisible by 3.</p>
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<p><strong class="example">Example 3:</strong></p>

‎solution/2800-2899/2862.Maximum Element-Sum of a Complete Subset of Indices/README.md‎

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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<p><strongclass="example">示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [8,7,3,5,7,2,4,9]
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<strong>输出:</strong>16
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<strong>解释:</strong>我们选择了下标 2 和 8 的元素,并且 2 * 8 是一个完全平方数。
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</pre>
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<div class="example-block">
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<p><strong>输入:</strong><span class="example-io">nums = [8,7,3,5,7,2,4,9]</span></p>
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<p><strong>示例 2:</strong></p>
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<p><strong>输出:</strong><spanclass="example-io">16</span></p>
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<pre>
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<strong>输入:</strong>nums = [8,10,3,8,1,13,7,9,4]
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<strong>输出:</strong>20
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<strong>解释:</strong>我们选择了下标 1,4 和 9 的元素。1 * 4,1 * 9,4 * 9 都是完全平方数。
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</pre>
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<p><strong>解释:</strong></p>
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<p>我们选择下标为 2 和 8 的元素,并且&nbsp;<code>1 * 4</code>&nbsp;是一个完全平方数。</p>
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</div>
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<p><strong class="example">示例 2:</strong></p>
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<div class="example-block">
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<p><span class="example-io"><b>输入:</b>nums = [8,10,3,8,1,13,7,9,4]</span></p>
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<p><span class="example-io"><b>输出:</b>20</span></p>
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<p><strong>解释:</strong></p>
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<p>我们选择下标为 1, 4, 9 的元素。<code>1 * 4</code>, <code>1 * 9</code>, <code>4 * 9</code>&nbsp;是完全平方数。</p>
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</div>
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<p>&nbsp;</p>
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‎solution/2800-2899/2862.Maximum Element-Sum of a Complete Subset of Indices/README_EN.md‎

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<p><strong>Explanation:</strong></p>
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<p>We select elements at indices 1 and 4 and <code>1 * 4</code> is a perfect square.</p>
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<p>We select elements at indices 2 and 8 and <code>1 * 4</code> is a perfect square.</p>
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<p><strong class="example">Example 2:</strong></p>

‎solution/3100-3199/3173.Bitwise OR of Adjacent Elements/README.md‎

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我们遍历数组的前 $n - 1$ 个元素,对于每个元素,计算它和它的下一个元素的按位或值,将结果存入答案数组中。
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时间复杂度 $O(n),ドル其中 $n$ 是数组的长度。空间复杂度 $O(1)$。
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时间复杂度 $O(n),ドル其中 $n$ 是数组的长度。忽略答案数组的空间消耗,空间复杂度 $O(1)$。
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‎solution/3100-3199/3173.Bitwise OR of Adjacent Elements/README_EN.md‎

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We iterate through the first $n - 1$ elements of the array. For each element, we calculate the bitwise OR value of it and its next element, and store the result in the answer array.
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The time complexity is $O(n),ドル where $n$ is the length of the array. The space complexity is $O(1)$.
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The time complexity is $O(n),ドル where $n$ is the length of the array. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.
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