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solution/0500-0599/0547.Number of Provinces/README.md

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</div>
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</div>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/0500-0599/0547.Number of Provinces/README_EN.md

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<li><code>isConnected[i][j] == isConnected[j][i]</code></li>
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</ul>
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## Solutions
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solution/1500-1599/1576.Replace All 's to Avoid Consecutive Repeating Characters/README.md

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</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1500-1599/1576.Replace All 's to Avoid Consecutive Repeating Characters/README_EN.md

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<li><code>s</code> contains&nbsp;only lower case English letters and <code>&#39;?&#39;</code>.</li>
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</ul>
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## Solutions
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solution/1600-1600/1600.Throne Inheritance/README.md

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@@ -75,7 +75,6 @@ t.getInheritanceOrder(); // 返回 [&quot;king&quot;, &quot;andy&quot;, &quot;ma
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<li>最多调用&nbsp;<code>10</code>&nbsp;次&nbsp;<code>getInheritanceOrder</code>&nbsp;。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1600.Throne Inheritance/README_EN.md

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@@ -73,7 +73,6 @@ t.getInheritanceOrder(); // return [&quot;king&quot;, &quot;andy&quot;, &quot;ma
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<li>At most <code>10</code> calls will be made to <code>getInheritanceOrder</code>.</li>
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</ul>
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## Solutions
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solution/1600-1600/1601.Maximum Number of Achievable Transfer Requests/README.md

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@@ -61,7 +61,6 @@
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<li><code>0 &lt;= from<sub>i</sub>, to<sub>i</sub> &lt; n</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1601.Maximum Number of Achievable Transfer Requests/README_EN.md

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@@ -60,7 +60,6 @@ We can achieve all the requests. </pre>
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<li><code>0 &lt;= from<sub>i</sub>, to<sub>i</sub> &lt; n</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->

solution/1600-1600/1603.Design Parking System/README.md

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@@ -44,7 +44,6 @@ parkingSystem.addCar(1); // 返回 false ,因为没有空的大车位,唯一
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<li>最多会调用 <code>addCar</code> 函数 <code>1000</code> 次</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1603.Design Parking System/README_EN.md

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@@ -40,7 +40,6 @@ parkingSystem.addCar(1); // return false because there is no available slot for
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<li>At most <code>1000</code> calls will be made to <code>addCar</code></li>
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</ul>
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## Solutions
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solution/1600-1600/1604.Alert Using Same Key-Card Three or More Times in a One Hour Period/README.md

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<li><code>keyName[i]</code> 只包含小写英文字母。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1604.Alert Using Same Key-Card Three or More Times in a One Hour Period/README_EN.md

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<li><code>keyName[i] contains only lowercase English letters.</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->

solution/1600-1600/1605.Find Valid Matrix Given Row and Column Sums/README.md

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<li><code>sum(rows) == sum(columns)</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1605.Find Valid Matrix Given Row and Column Sums/README_EN.md

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@@ -68,7 +68,6 @@ Another possible matrix is: [[1,2],
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<li><code>sum(rows) == sum(columns)</code></li>
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</ul>
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## Solutions
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solution/1600-1600/1606.Find Servers That Handled Most Number of Requests/README.md

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<li><code>arrival</code> 保证 <strong>严格递增</strong> 。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1606.Find Servers That Handled Most Number of Requests/README_EN.md

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@@ -77,7 +77,6 @@ Server 0 handled two requests, while servers 1 and 2 handled one request each. H
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<li><code>arrival</code> is <strong>strictly increasing</strong>.</li>
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</ul>
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## Solutions
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solution/1600-1600/1608.Special Array With X Elements Greater Than or Equal X/README.md

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@@ -53,7 +53,6 @@ x 不能取更大的值,因为 nums 中只有两个元素。</pre>
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<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1608.Special Array With X Elements Greater Than or Equal X/README_EN.md

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@@ -54,7 +54,6 @@ x cannot be greater since there are only 2 numbers in nums.
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<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
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</ul>
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## Solutions
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solution/1600-1600/1609.Even Odd Tree/README.md

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<li><code>1 <= Node.val <= 10<sup>6</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1609.Even Odd Tree/README_EN.md

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<li><code>1 &lt;= Node.val &lt;= 10<sup>6</sup></code></li>
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</ul>
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## Solutions
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solution/1600-1600/1610.Maximum Number of Visible Points/README.md

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<li><code>0 <= pos<sub>x</sub>, pos<sub>y</sub>, x<sub>i</sub>, y<sub>i</sub> <= 100</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1610.Maximum Number of Visible Points/README_EN.md

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<li><code>0 &lt;= pos<sub>x</sub>, pos<sub>y</sub>, x<sub>i</sub>, y<sub>i</sub> &lt;= 100</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->

solution/1600-1600/1611.Minimum One Bit Operations to Make Integers Zero/README.md

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<li><code>0 <= n <= 10<sup>9</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1611.Minimum One Bit Operations to Make Integers Zero/README_EN.md

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<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->

solution/1600-1600/1614.Maximum Nesting Depth of the Parentheses/README.md

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<li>题目数据保证括号表达式 <code>s</code> 是 <strong>有效的括号表达式</strong></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1614.Maximum Nesting Depth of the Parentheses/README_EN.md

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<li>It is guaranteed that parentheses expression <code>s</code> is a <strong>VPS</strong>.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->

solution/1600-1600/1615.Maximal Network Rank/README.md

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<li>每对城市之间 <strong>最多只有一条</strong> 道路相连</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1615.Maximal Network Rank/README_EN.md

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<li>Each&nbsp;pair of cities has <strong>at most one</strong> road connecting them.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->

solution/1600-1600/1616.Split Two Strings to Make Palindrome/README.md

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@@ -61,7 +61,6 @@ b<sub>prefix</sub> = "jiz", b<sub>suffix</sub> = "alu"
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<li><code>a</code> 和 <code>b</code> 都只包含小写英文字母</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1616.Split Two Strings to Make Palindrome/README_EN.md

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<li><code>a</code> and <code>b</code> consist of lowercase English letters</li>
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</ul>
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## Solutions
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<!-- tabs:start -->

solution/1600-1600/1617.Count Subtrees With Max Distance Between Cities/README_EN.md

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<p>There are <code>n</code> cities numbered from <code>1</code> to <code>n</code>. You are given an array <code>edges</code> of size <code>n-1</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> represents a bidirectional edge between cities <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code>. There exists a unique path between each pair of cities. In other words, the cities form a <strong>tree</strong>.</p>
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<p>A <strong>subtree</strong> is a subset of cities where every city is reachable from every other city in the subset, where the path between each pair passes through only the cities from the subset. Two subtrees are different if there is a city in one subtree that is not present in the other.</p>
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<p>For each <code>d</code> from <code>1</code> to <code>n-1</code>, find the number of subtrees in which the <strong>maximum distance</strong> between any two cities in the subtree is equal to <code>d</code>.</p>
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<p>Return <em>an array of size</em> <code>n-1</code> <em>where the </em><code>d<sup>th</sup></code><em> </em><em>element <strong>(1-indexed)</strong> is the number of subtrees in which the <strong>maximum distance</strong> between any two cities is equal to </em><code>d</code>.</p>
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<p><strong>Notice</strong>&nbsp;that&nbsp;the <strong>distance</strong> between the two cities is the number of edges in the path between them.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> n = 2, edges = [[1,2]]
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> n = 3, edges = [[1,2],[2,3]]
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>2 &lt;= n &lt;= 15</code></li>
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<li><code>edges.length == n-1</code></li>

solution/1600-1600/1619.Mean of Array After Removing Some Elements/README.md

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<li><code>0 <= arr[i] <= 10<sup>5</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1619.Mean of Array After Removing Some Elements/README_EN.md

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<li><code><font face="monospace">0 &lt;= arr[i] &lt;= 10<sup>5</sup></font></code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->

solution/1600-1600/1620.Coordinate With Maximum Network Quality/README.md

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<li><code>1 <= radius <= 50</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

solution/1600-1600/1620.Coordinate With Maximum Network Quality/README_EN.md

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<li><code>1 &lt;= radius &lt;= 50</code></li>
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## Solutions
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solution/1600-1600/1621.Number of Sets of K Non-Overlapping Line Segments/README.md

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<li><code>1 <= k <= n-1</code></li>
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</ul>
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## 解法
6665

6766
<!-- 这里可写通用的实现逻辑 -->

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