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https://gitee.com/coder-xiaomo/leetcode-problemset
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# 力扣题库（完整版）
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> 最后更新日期： **2024.02.19**
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> 最后更新日期： **2024.03.01**
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>
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> 使用脚本前请务必仔细完整阅读本 `README.md` 文件
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|
||||
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||||
File diff suppressed because it is too large
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56
leetcode-cn/originData/[no content]binary-tree-nodes.json
Normal file
56
leetcode-cn/originData/[no content]binary-tree-nodes.json
Normal file
@ -0,0 +1,56 @@
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File diff suppressed because one or more lines are too long
File diff suppressed because one or more lines are too long
59
leetcode-cn/originData/[no content]maximize-items.json
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59
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Normal file
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62
leetcode-cn/originData/[no content]snaps-analysis.json
Normal file
62
leetcode-cn/originData/[no content]snaps-analysis.json
Normal file
@ -0,0 +1,62 @@
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63
leetcode-cn/originData/[no content]top-percentile-fraud.json
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63
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File diff suppressed because one or more lines are too long
179
leetcode-cn/originData/earliest-second-to-mark-indices-i.json
Normal file
179
leetcode-cn/originData/earliest-second-to-mark-indices-i.json
Normal file
File diff suppressed because one or more lines are too long
188
leetcode-cn/originData/earliest-second-to-mark-indices-ii.json
Normal file
188
leetcode-cn/originData/earliest-second-to-mark-indices-ii.json
Normal file
File diff suppressed because one or more lines are too long
File diff suppressed because one or more lines are too long
182
leetcode-cn/originData/split-the-array.json
Normal file
182
leetcode-cn/originData/split-the-array.json
Normal file
File diff suppressed because one or more lines are too long
37
leetcode-cn/problem (Chinese)/分割数组 [split-the-array].html
Normal file
37
leetcode-cn/problem (Chinese)/分割数组 [split-the-array].html
Normal file
@ -0,0 +1,37 @@
|
||||
<p>给你一个长度为 <strong>偶数 </strong>的整数数组 <code>nums</code> 。你需要将这个数组分割成 <code>nums1</code> 和 <code>nums2</code> 两部分，要求：</p>
|
||||
|
||||
<ul>
|
||||
<li><code>nums1.length == nums2.length == nums.length / 2</code> 。</li>
|
||||
<li><code>nums1</code> 应包含 <strong>互不相同</strong><strong> </strong>的元素。</li>
|
||||
<li><code>nums2</code>也应包含<strong> 互不相同</strong> 的元素。</li>
|
||||
</ul>
|
||||
|
||||
<p>如果能够分割数组就返回 <code>true</code> ，否则返回 <code>false</code> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong class="example">示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>nums = [1,1,2,2,3,4]
|
||||
<strong>输出：</strong>true
|
||||
<strong>解释：</strong>分割 nums 的可行方案之一是 nums1 = [1,2,3] 和 nums2 = [1,2,4] 。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>输入：</strong>nums = [1,1,1,1]
|
||||
<strong>输出：</strong>false
|
||||
<strong>解释：</strong>分割 nums 的唯一可行方案是 nums1 = [1,1] 和 nums2 = [1,1] 。但 nums1 和 nums2 都不是由互不相同的元素构成。因此，返回 false 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= nums.length <= 100</code></li>
|
||||
<li><code>nums.length % 2 == 0</code></li>
|
||||
<li><code>1 <= nums[i] <= 100</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,71 @@
|
||||
<p>给你两个下标从 <strong>1</strong> 开始的整数数组 <code>nums</code> 和 <code>changeIndices</code> ，数组的长度分别为 <code>n</code> 和 <code>m</code> 。</p>
|
||||
|
||||
<p>一开始，<code>nums</code> 中所有下标都是未标记的，你的任务是标记 <code>nums</code> 中 <strong>所有</strong> 下标。</p>
|
||||
|
||||
<p>从第 <code>1</code> 秒到第 <code>m</code> 秒（<b>包括 </b>第 <code>m</code> 秒），对于每一秒 <code>s</code> ，你可以执行以下操作 <strong>之一</strong> ：</p>
|
||||
|
||||
<ul>
|
||||
<li>选择范围 <code>[1, n]</code> 中的一个下标 <code>i</code> ，并且将 <code>nums[i]</code> <strong>减少</strong> <code>1</code> 。</li>
|
||||
<li>如果 <code>nums[changeIndices[s]]</code> <strong>等于</strong> <code>0</code> ，<strong>标记</strong> 下标 <code>changeIndices[s]</code> 。</li>
|
||||
<li>什么也不做。</li>
|
||||
</ul>
|
||||
|
||||
<p>请你返回范围 <code>[1, m]</code> 中的一个整数，表示最优操作下，标记 <code>nums</code> 中 <strong>所有</strong> 下标的 <strong>最早秒数</strong> ，如果无法标记所有下标，返回 <code>-1</code> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong class="example">示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>nums = [2,2,0], changeIndices = [2,2,2,2,3,2,2,1]
|
||||
<b>输出：</b>8
|
||||
<b>解释：</b>这个例子中，我们总共有 8 秒。按照以下操作标记所有下标：
|
||||
第 1 秒：选择下标 1 ，将 nums[1] 减少 1 。nums 变为 [1,2,0] 。
|
||||
第 2 秒：选择下标 1 ，将 nums[1] 减少 1 。nums 变为 [0,2,0] 。
|
||||
第 3 秒：选择下标 2 ，将 nums[2] 减少 1 。nums 变为 [0,1,0] 。
|
||||
第 4 秒：选择下标 2 ，将 nums[2] 减少 1 。nums 变为 [0,0,0] 。
|
||||
第 5 秒，标记 changeIndices[5] ，也就是标记下标 3 ，因为 nums[3] 等于 0 。
|
||||
第 6 秒，标记 changeIndices[6] ，也就是标记下标 2 ，因为 nums[2] 等于 0 。
|
||||
第 7 秒，什么也不做。
|
||||
第 8 秒，标记 changeIndices[8] ，也就是标记下标 1 ，因为 nums[1] 等于 0 。
|
||||
现在所有下标已被标记。
|
||||
最早可以在第 8 秒标记所有下标。
|
||||
所以答案是 8 。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>nums = [1,3], changeIndices = [1,1,1,2,1,1,1]
|
||||
<b>输出：</b>6
|
||||
<b>解释：</b>这个例子中，我们总共有 7 秒。按照以下操作标记所有下标：
|
||||
第 1 秒：选择下标 2 ，将 nums[2] 减少 1 。nums 变为 [1,2] 。
|
||||
第 2 秒：选择下标 2 ，将 nums[2] 减少 1 。nums 变为 [1,1] 。
|
||||
第 3 秒：选择下标 2 ，将 nums[2] 减少 1 。nums 变为 [1,0] 。
|
||||
第 4 秒：标记 changeIndices[4] ，也就是标记下标 2 ，因为 nums[2] 等于 0 。
|
||||
第 5 秒：选择下标 1 ，将 nums[1] 减少 1 。nums 变为 [0,0] 。
|
||||
第 6 秒：标记 changeIndices[6] ，也就是标记下标 1 ，因为 nums[1] 等于 0 。
|
||||
现在所有下标已被标记。
|
||||
最早可以在第 6 秒标记所有下标。
|
||||
所以答案是 6 。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [0,1], changeIndices = [2,2,2]
|
||||
<strong>Output:</strong> -1
|
||||
<strong>Explanation:</strong> 这个例子中，无法标记所有下标，因为下标 1 不在 changeIndices 中。
|
||||
所以答案是 -1 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n == nums.length <= 2000</code></li>
|
||||
<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>
|
||||
<li><code>1 <= m == changeIndices.length <= 2000</code></li>
|
||||
<li><code>1 <= changeIndices[i] <= n</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,71 @@
|
||||
<p>给你两个下标从 <strong>1</strong> 开始的整数数组 <code>nums</code> 和 <code>changeIndices</code> ，数组的长度分别为 <code>n</code> 和 <code>m</code> 。</p>
|
||||
|
||||
<p>一开始，<code>nums</code> 中所有下标都是未标记的，你的任务是标记 <code>nums</code> 中 <strong>所有</strong> 下标。</p>
|
||||
|
||||
<p>从第 <code>1</code> 秒到第 <code>m</code> 秒（<b>包括 </b>第 <code>m</code> 秒），对于每一秒 <code>s</code> ，你可以执行以下操作 <strong>之一</strong> ：</p>
|
||||
|
||||
<ul>
|
||||
<li>选择范围 <code>[1, n]</code> 中的一个下标 <code>i</code> ，并且将 <code>nums[i]</code> <strong>减少</strong> <code>1</code> 。</li>
|
||||
<li>将 <code>nums[changeIndices[s]]</code> 设置成任意的 <strong>非负</strong> 整数。</li>
|
||||
<li>选择范围 <code>[1, n]</code> 中的一个下标 <code>i</code> ， 满足 <code>nums[i]</code> <strong>等于</strong> <code>0</code>, 并 <strong>标记</strong> 下标 <code>i</code> 。</li>
|
||||
<li>什么也不做。</li>
|
||||
</ul>
|
||||
|
||||
<p>请你返回范围 <code>[1, m]</code> 中的一个整数，表示最优操作下，标记 <code>nums</code> 中 <strong>所有</strong> 下标的 <strong>最早秒数</strong> ，如果无法标记所有下标，返回 <code>-1</code> 。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>nums = [3,2,3], changeIndices = [1,3,2,2,2,2,3]
|
||||
<b>输出：</b>6
|
||||
<b>解释：</b>这个例子中，我们总共有 7 秒。按照以下操作标记所有下标：
|
||||
第 1 秒：将 nums[changeIndices[1]] 变为 0 。nums 变为 [0,2,3] 。
|
||||
第 2 秒：将 nums[changeIndices[2]] 变为 0 。nums 变为 [0,2,0] 。
|
||||
第 3 秒：将 nums[changeIndices[3]] 变为 0 。nums 变为 [0,0,0] 。
|
||||
第 4 秒：标记下标 1 ，因为 nums[1] 等于 0 。
|
||||
第 5 秒：标记下标 2 ，因为 nums[2] 等于 0 。
|
||||
第 6 秒：标记下标 3 ，因为 nums[3] 等于 0 。
|
||||
现在所有下标已被标记。
|
||||
最早可以在第 6 秒标记所有下标。
|
||||
所以答案是 6 。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 2：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>nums = [0,0,1,2], changeIndices = [1,2,1,2,1,2,1,2]
|
||||
<b>输出：</b>7
|
||||
<b>解释：</b>这个例子中，我们总共有 8 秒。按照以下操作标记所有下标：
|
||||
第 1 秒：标记下标 1 ，因为 nums[1] 等于 0 。
|
||||
第 2 秒：标记下标 2 ，因为 nums[2] 等于 0 。
|
||||
第 3 秒：将 nums[4] 减少 1 。nums 变为 [0,0,1,1] 。
|
||||
第 4 秒：将 nums[4] 减少 1 。nums 变为 [0,0,1,0] 。
|
||||
第 5 秒：将 nums[3] 减少 1 。nums 变为 [0,0,0,0] 。
|
||||
第 6 秒：标记下标 3 ，因为 nums[3] 等于 0 。
|
||||
第 7 秒：标记下标 4 ，因为 nums[4] 等于 0 。
|
||||
现在所有下标已被标记。
|
||||
最早可以在第 7 秒标记所有下标。
|
||||
所以答案是 7 。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 3：</strong></p>
|
||||
|
||||
<pre>
|
||||
<b>输入：</b>nums = [1,2,3], changeIndices = [1,2,3]
|
||||
<b>输出：</b>-1
|
||||
<strong>解释：</strong>这个例子中，无法标记所有下标，因为我们没有足够的秒数。
|
||||
所以答案是 -1 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n == nums.length <= 5000</code></li>
|
||||
<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>
|
||||
<li><code>1 <= m == changeIndices.length <= 5000</code></li>
|
||||
<li><code>1 <= changeIndices[i] <= n</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,50 @@
|
||||
<p>在二维平面上存在 <code>n</code> 个矩形。给你两个下标从 <strong>0</strong> 开始的二维整数数组 <code>bottomLeft</code> 和 <code>topRight</code>，两个数组的大小都是 <code>n x 2</code> ，其中 <code>bottomLeft[i]</code> 和 <code>topRight[i]</code> 分别代表第 <code>i</code> 个矩形的<strong> 左下角 </strong>和 <strong>右上角 </strong>坐标。</p>
|
||||
|
||||
<p>我们定义 <strong>向右 </strong>的方向为 x 轴正半轴（<strong>x 坐标增加</strong>），<strong>向左 </strong>的方向为 x 轴负半轴（<strong>x 坐标减少</strong>）。同样地，定义 <strong>向上 </strong>的方向为 y 轴正半轴（<strong>y 坐标增加</strong>）<strong>，向下 </strong>的方向为 y 轴负半轴（<strong>y 坐标减少</strong>）。</p>
|
||||
|
||||
<p>你可以选择一个区域，该区域由两个矩形的 <strong>交集</strong> 形成。你需要找出能够放入该区域 <strong>内 </strong>的<strong> 最大 </strong>正方形面积，并选择最优解。</p>
|
||||
|
||||
<p>返回能够放入交集区域的正方形的 <strong>最大 </strong>可能面积，如果矩形之间不存在任何交集区域，则返回 <code>0</code>。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong class="example">示例 1：</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/05/example12.png" style="width: 443px; height: 364px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;" />
|
||||
<pre>
|
||||
<strong>输入：</strong>bottomLeft = [[1,1],[2,2],[3,1]], topRight = [[3,3],[4,4],[6,6]]
|
||||
<strong>输出：</strong>1
|
||||
<strong>解释：</strong>边长为 1 的正方形可以放入矩形 0 和矩形 1 的交集区域，或矩形 1 和矩形 2 的交集区域。因此最大面积是边长 * 边长，即 1 * 1 = 1。
|
||||
可以证明，边长更大的正方形无法放入任何交集区域。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 2：</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/04/rectanglesexample2.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 445px; height: 365px;" />
|
||||
<pre>
|
||||
<strong>输入：</strong>bottomLeft = [[1,1],[2,2],[1,2]], topRight = [[3,3],[4,4],[3,4]]
|
||||
<strong>输出：</strong>1
|
||||
<strong>解释：</strong>边长为 1 的正方形可以放入矩形 0 和矩形 1，矩形 1 和矩形 2，或所有三个矩形的交集区域。因此最大面积是边长 * 边长，即 1 * 1 = 1。
|
||||
可以证明，边长更大的正方形无法放入任何交集区域。
|
||||
请注意，区域可以由多于两个矩形的交集构成。
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">示例 3：</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/04/rectanglesexample3.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 444px; height: 364px;" />
|
||||
<pre>
|
||||
<strong>输入：</strong>bottomLeft = [[1,1],[3,3],[3,1]], topRight = [[2,2],[4,4],[4,2]]
|
||||
<strong>输出：</strong>0
|
||||
<strong>解释：</strong>不存在相交的矩形，因此，返回 0 。
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>n == bottomLeft.length == topRight.length</code></li>
|
||||
<li><code>2 <= n <= 10<sup>3</sup></code></li>
|
||||
<li><code>bottomLeft[i].length == topRight[i].length == 2</code></li>
|
||||
<li><code>1 <= bottomLeft[i][0], bottomLeft[i][1] <= 10<sup>7</sup></code></li>
|
||||
<li><code>1 <= topRight[i][0], topRight[i][1] <= 10<sup>7</sup></code></li>
|
||||
<li><code>bottomLeft[i][0] < topRight[i][0]</code></li>
|
||||
<li><code>bottomLeft[i][1] < topRight[i][1]</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,35 @@
|
||||
<p>You are given an integer array <code>nums</code> of <strong>even</strong> length. You have to split the array into two parts <code>nums1</code> and <code>nums2</code> such that:</p>
|
||||
|
||||
<ul>
|
||||
<li><code>nums1.length == nums2.length == nums.length / 2</code>.</li>
|
||||
<li><code>nums1</code> should contain <strong>distinct </strong>elements.</li>
|
||||
<li><code>nums2</code> should also contain <strong>distinct</strong> elements.</li>
|
||||
</ul>
|
||||
|
||||
<p>Return <code>true</code><em> if it is possible to split the array, and </em><code>false</code> <em>otherwise</em><em>.</em></p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,1,2,2,3,4]
|
||||
<strong>Output:</strong> true
|
||||
<strong>Explanation:</strong> One of the possible ways to split nums is nums1 = [1,2,3] and nums2 = [1,2,4].
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,1,1,1]
|
||||
<strong>Output:</strong> false
|
||||
<strong>Explanation:</strong> The only possible way to split nums is nums1 = [1,1] and nums2 = [1,1]. Both nums1 and nums2 do not contain distinct elements. Therefore, we return false.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= nums.length <= 100</code></li>
|
||||
<li><code>nums.length % 2 == 0 </code></li>
|
||||
<li><code>1 <= nums[i] <= 100</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,69 @@
|
||||
<p>You are given two <strong>1-indexed</strong> integer arrays, <code>nums</code> and, <code>changeIndices</code>, having lengths <code>n</code> and <code>m</code>, respectively.</p>
|
||||
|
||||
<p>Initially, all indices in <code>nums</code> are unmarked. Your task is to mark <strong>all</strong> indices in <code>nums</code>.</p>
|
||||
|
||||
<p>In each second, <code>s</code>, in order from <code>1</code> to <code>m</code> (<strong>inclusive</strong>), you can perform <strong>one</strong> of the following operations:</p>
|
||||
|
||||
<ul>
|
||||
<li>Choose an index <code>i</code> in the range <code>[1, n]</code> and <strong>decrement</strong> <code>nums[i]</code> by <code>1</code>.</li>
|
||||
<li>If <code>nums[changeIndices[s]]</code> is <strong>equal</strong> to <code>0</code>, <strong>mark</strong> the index <code>changeIndices[s]</code>.</li>
|
||||
<li>Do nothing.</li>
|
||||
</ul>
|
||||
|
||||
<p>Return <em>an integer denoting the <strong>earliest second</strong> in the range </em><code>[1, m]</code><em> when <strong>all</strong> indices in </em><code>nums</code><em> can be marked by choosing operations optimally, or </em><code>-1</code><em> if it is impossible.</em></p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [2,2,0], changeIndices = [2,2,2,2,3,2,2,1]
|
||||
<strong>Output:</strong> 8
|
||||
<strong>Explanation:</strong> In this example, we have 8 seconds. The following operations can be performed to mark all indices:
|
||||
Second 1: Choose index 1 and decrement nums[1] by one. nums becomes [1,2,0].
|
||||
Second 2: Choose index 1 and decrement nums[1] by one. nums becomes [0,2,0].
|
||||
Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [0,1,0].
|
||||
Second 4: Choose index 2 and decrement nums[2] by one. nums becomes [0,0,0].
|
||||
Second 5: Mark the index changeIndices[5], which is marking index 3, since nums[3] is equal to 0.
|
||||
Second 6: Mark the index changeIndices[6], which is marking index 2, since nums[2] is equal to 0.
|
||||
Second 7: Do nothing.
|
||||
Second 8: Mark the index changeIndices[8], which is marking index 1, since nums[1] is equal to 0.
|
||||
Now all indices have been marked.
|
||||
It can be shown that it is not possible to mark all indices earlier than the 8th second.
|
||||
Hence, the answer is 8.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,3], changeIndices = [1,1,1,2,1,1,1]
|
||||
<strong>Output:</strong> 6
|
||||
<strong>Explanation:</strong> In this example, we have 7 seconds. The following operations can be performed to mark all indices:
|
||||
Second 1: Choose index 2 and decrement nums[2] by one. nums becomes [1,2].
|
||||
Second 2: Choose index 2 and decrement nums[2] by one. nums becomes [1,1].
|
||||
Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [1,0].
|
||||
Second 4: Mark the index changeIndices[4], which is marking index 2, since nums[2] is equal to 0.
|
||||
Second 5: Choose index 1 and decrement nums[1] by one. nums becomes [0,0].
|
||||
Second 6: Mark the index changeIndices[6], which is marking index 1, since nums[1] is equal to 0.
|
||||
Now all indices have been marked.
|
||||
It can be shown that it is not possible to mark all indices earlier than the 6th second.
|
||||
Hence, the answer is 6.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [0,1], changeIndices = [2,2,2]
|
||||
<strong>Output:</strong> -1
|
||||
<strong>Explanation:</strong> In this example, it is impossible to mark all indices because index 1 isn't in changeIndices.
|
||||
Hence, the answer is -1.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n == nums.length <= 2000</code></li>
|
||||
<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>
|
||||
<li><code>1 <= m == changeIndices.length <= 2000</code></li>
|
||||
<li><code>1 <= changeIndices[i] <= n</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,69 @@
|
||||
<p>You are given two <strong>1-indexed</strong> integer arrays, <code>nums</code> and, <code>changeIndices</code>, having lengths <code>n</code> and <code>m</code>, respectively.</p>
|
||||
|
||||
<p>Initially, all indices in <code>nums</code> are unmarked. Your task is to mark <strong>all</strong> indices in <code>nums</code>.</p>
|
||||
|
||||
<p>In each second, <code>s</code>, in order from <code>1</code> to <code>m</code> (<strong>inclusive</strong>), you can perform <strong>one</strong> of the following operations:</p>
|
||||
|
||||
<ul>
|
||||
<li>Choose an index <code>i</code> in the range <code>[1, n]</code> and <strong>decrement</strong> <code>nums[i]</code> by <code>1</code>.</li>
|
||||
<li>Set <code>nums[changeIndices[s]]</code> to any <strong>non-negative</strong> value.</li>
|
||||
<li>Choose an index <code>i</code> in the range <code>[1, n]</code>, where <code>nums[i]</code> is <strong>equal</strong> to <code>0</code>, and <strong>mark</strong> index <code>i</code>.</li>
|
||||
<li>Do nothing.</li>
|
||||
</ul>
|
||||
|
||||
<p>Return <em>an integer denoting the <strong>earliest second</strong> in the range </em><code>[1, m]</code><em> when <strong>all</strong> indices in </em><code>nums</code><em> can be marked by choosing operations optimally, or </em><code>-1</code><em> if it is impossible.</em></p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [3,2,3], changeIndices = [1,3,2,2,2,2,3]
|
||||
<strong>Output:</strong> 6
|
||||
<strong>Explanation:</strong> In this example, we have 7 seconds. The following operations can be performed to mark all indices:
|
||||
Second 1: Set nums[changeIndices[1]] to 0. nums becomes [0,2,3].
|
||||
Second 2: Set nums[changeIndices[2]] to 0. nums becomes [0,2,0].
|
||||
Second 3: Set nums[changeIndices[3]] to 0. nums becomes [0,0,0].
|
||||
Second 4: Mark index 1, since nums[1] is equal to 0.
|
||||
Second 5: Mark index 2, since nums[2] is equal to 0.
|
||||
Second 6: Mark index 3, since nums[3] is equal to 0.
|
||||
Now all indices have been marked.
|
||||
It can be shown that it is not possible to mark all indices earlier than the 6th second.
|
||||
Hence, the answer is 6.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [0,0,1,2], changeIndices = [1,2,1,2,1,2,1,2]
|
||||
<strong>Output:</strong> 7
|
||||
<strong>Explanation:</strong> In this example, we have 8 seconds. The following operations can be performed to mark all indices:
|
||||
Second 1: Mark index 1, since nums[1] is equal to 0.
|
||||
Second 2: Mark index 2, since nums[2] is equal to 0.
|
||||
Second 3: Decrement index 4 by one. nums becomes [0,0,1,1].
|
||||
Second 4: Decrement index 4 by one. nums becomes [0,0,1,0].
|
||||
Second 5: Decrement index 3 by one. nums becomes [0,0,0,0].
|
||||
Second 6: Mark index 3, since nums[3] is equal to 0.
|
||||
Second 7: Mark index 4, since nums[4] is equal to 0.
|
||||
Now all indices have been marked.
|
||||
It can be shown that it is not possible to mark all indices earlier than the 7th second.
|
||||
Hence, the answer is 7.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,2,3], changeIndices = [1,2,3]
|
||||
<strong>Output:</strong> -1
|
||||
<strong>Explanation: </strong>In this example, it can be shown that it is impossible to mark all indices, as we don't have enough seconds.
|
||||
Hence, the answer is -1.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n == nums.length <= 5000</code></li>
|
||||
<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>
|
||||
<li><code>1 <= m == changeIndices.length <= 5000</code></li>
|
||||
<li><code>1 <= changeIndices[i] <= n</code></li>
|
||||
</ul>
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||||
@ -0,0 +1,46 @@
|
||||
<p>There exist <code>n</code> rectangles in a 2D plane. You are given two <strong>0-indexed</strong> 2D integer arrays <code>bottomLeft</code> and <code>topRight</code>, both of size <code>n x 2</code>, where <code>bottomLeft[i]</code> and <code>topRight[i]</code> represent the <strong>bottom-left</strong> and <strong>top-right</strong> coordinates of the <code>i<sup>th</sup></code> rectangle respectively.</p>
|
||||
|
||||
<p>You can select a region formed from the <strong>intersection</strong> of two of the given rectangles. You need to find the <strong>largest </strong>area of a <strong>square</strong> that can fit <strong>inside</strong> this region if you select the region optimally.</p>
|
||||
|
||||
<p>Return <em>the <strong>largest </strong>possible area of a square, or </em><code>0</code><em> if there <strong>do not</strong> exist any intersecting regions between the rectangles</em>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/05/example12.png" style="width: 443px; height: 364px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;" />
|
||||
<pre>
|
||||
<strong>Input:</strong> bottomLeft = [[1,1],[2,2],[3,1]], topRight = [[3,3],[4,4],[6,6]]
|
||||
<strong>Output:</strong> 1
|
||||
<strong>Explanation:</strong> A square with side length 1 can fit inside either the intersecting region of rectangle 0 and rectangle 1, or the intersecting region of rectangle 1 and rectangle 2. Hence the largest area is side * side which is 1 * 1 == 1.
|
||||
It can be shown that a square with a greater side length can not fit inside any intersecting region.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/04/rectanglesexample2.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 445px; height: 365px;" />
|
||||
<pre>
|
||||
<strong>Input:</strong> bottomLeft = [[1,1],[2,2],[1,2]], topRight = [[3,3],[4,4],[3,4]]
|
||||
<strong>Output:</strong> 1
|
||||
<strong>Explanation:</strong> A square with side length 1 can fit inside either the intersecting region of rectangle 0 and rectangle 1, the intersecting region of rectangle 1 and rectangle 2, or the intersection region of all 3 rectangles. Hence the largest area is side * side which is 1 * 1 == 1.
|
||||
It can be shown that a square with a greater side length can not fit inside any intersecting region.
|
||||
Note that the region can be formed by the intersection of more than 2 rectangles.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/04/rectanglesexample3.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 444px; height: 364px;" />
|
||||
<pre>
|
||||
<strong>Input:</strong> bottomLeft = [[1,1],[3,3],[3,1]], topRight = [[2,2],[4,4],[4,2]]
|
||||
<strong>Output:</strong> 0
|
||||
<strong>Explanation:</strong> No pair of rectangles intersect, hence, we return 0.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>n == bottomLeft.length == topRight.length</code></li>
|
||||
<li><code>2 <= n <= 10<sup>3</sup></code></li>
|
||||
<li><code>bottomLeft[i].length == topRight[i].length == 2</code></li>
|
||||
<li><code>1 <= bottomLeft[i][0], bottomLeft[i][1] <= 10<sup>7</sup></code></li>
|
||||
<li><code>1 <= topRight[i][0], topRight[i][1] <= 10<sup>7</sup></code></li>
|
||||
<li><code>bottomLeft[i][0] < topRight[i][0]</code></li>
|
||||
<li><code>bottomLeft[i][1] < topRight[i][1]</code></li>
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||||
</ul>
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||||
File diff suppressed because it is too large
Load Diff
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@ -0,0 +1,56 @@
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||||
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||||
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"question": {
|
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"questionId": "3357",
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"questionFrontendId": "3057",
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||||
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||||
"titleSlug": "employees-project-allocation",
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@ -0,0 +1,58 @@
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||||
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||||
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||||
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@ -0,0 +1,54 @@
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"metaData": "{\"mysql\": [\"Create Table if Not Exists Friends( user_id1 int, user_id2 int)\"], \"mssql\": [\"Create Table Friends( user_id1 int, user_id2 int)\"], \"oraclesql\": [\"Create Table Friends( user_id1 int, user_id2 int)\"], \"database\": true, \"name\": \"friends_with_no_mutual_friends\", \"pythondata\": [\"Friends = pd.DataFrame([], columns=['user_id1', 'user_id2']).astype({'user_id1':'Int64', 'user_id2':'Int64'})\\n\"], \"postgresql\": [\"CREATE TABLE Friends (\\n user_id1 int,\\n user_id2 int\\n);\"], \"database_schema\": {\"Friends\": {\"user_id1\": \"INT\", \"user_id2\": \"INT\"}}}",
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File diff suppressed because one or more lines are too long
45
leetcode/originData/[no content]linked-list-frequency.json
Normal file
45
leetcode/originData/[no content]linked-list-frequency.json
Normal file
File diff suppressed because one or more lines are too long
56
leetcode/originData/[no content]maximize-items.json
Normal file
56
leetcode/originData/[no content]maximize-items.json
Normal file
@ -0,0 +1,56 @@
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||||
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"dislikes": 1,
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"exampleTestcases": "{\"headers\":{\"Inventory\":[\"item_id\",\"item_type\",\"item_category\",\"square_footage\"]},\"rows\":{\"Inventory\":[[1374,\"prime_eligible\",\"Watches\",68.00],[4245,\"not_prime\",\"Art\",26.40],[5743,\"prime_eligible\",\"Software\",325.00],[8543,\"not_prime\",\"Clothing\",64.50],[2556,\"not_prime\",\"Shoes\",15.00],[2452,\"prime_eligible\",\"Scientific\",85.00],[3255,\"not_prime\",\"Furniture\",22.60],[1672,\"prime_eligible\",\"Beauty\",8.50],[4256,\"prime_eligible\",\"Furniture\",55.50],[6325,\"prime_eligible\",\"Food\",13.20]]}}",
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||||
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||||
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||||
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59
leetcode/originData/[no content]snaps-analysis.json
Normal file
59
leetcode/originData/[no content]snaps-analysis.json
Normal file
@ -0,0 +1,59 @@
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{
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||||
60
leetcode/originData/[no content]top-percentile-fraud.json
Normal file
60
leetcode/originData/[no content]top-percentile-fraud.json
Normal file
@ -0,0 +1,60 @@
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||||
{
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File diff suppressed because one or more lines are too long
176
leetcode/originData/earliest-second-to-mark-indices-i.json
Normal file
176
leetcode/originData/earliest-second-to-mark-indices-i.json
Normal file
File diff suppressed because one or more lines are too long
185
leetcode/originData/earliest-second-to-mark-indices-ii.json
Normal file
185
leetcode/originData/earliest-second-to-mark-indices-ii.json
Normal file
File diff suppressed because one or more lines are too long
File diff suppressed because one or more lines are too long
179
leetcode/originData/split-the-array.json
Normal file
179
leetcode/originData/split-the-array.json
Normal file
File diff suppressed because one or more lines are too long
69
leetcode/problem/earliest-second-to-mark-indices-i.html
Normal file
69
leetcode/problem/earliest-second-to-mark-indices-i.html
Normal file
@ -0,0 +1,69 @@
|
||||
<p>You are given two <strong>1-indexed</strong> integer arrays, <code>nums</code> and, <code>changeIndices</code>, having lengths <code>n</code> and <code>m</code>, respectively.</p>
|
||||
|
||||
<p>Initially, all indices in <code>nums</code> are unmarked. Your task is to mark <strong>all</strong> indices in <code>nums</code>.</p>
|
||||
|
||||
<p>In each second, <code>s</code>, in order from <code>1</code> to <code>m</code> (<strong>inclusive</strong>), you can perform <strong>one</strong> of the following operations:</p>
|
||||
|
||||
<ul>
|
||||
<li>Choose an index <code>i</code> in the range <code>[1, n]</code> and <strong>decrement</strong> <code>nums[i]</code> by <code>1</code>.</li>
|
||||
<li>If <code>nums[changeIndices[s]]</code> is <strong>equal</strong> to <code>0</code>, <strong>mark</strong> the index <code>changeIndices[s]</code>.</li>
|
||||
<li>Do nothing.</li>
|
||||
</ul>
|
||||
|
||||
<p>Return <em>an integer denoting the <strong>earliest second</strong> in the range </em><code>[1, m]</code><em> when <strong>all</strong> indices in </em><code>nums</code><em> can be marked by choosing operations optimally, or </em><code>-1</code><em> if it is impossible.</em></p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [2,2,0], changeIndices = [2,2,2,2,3,2,2,1]
|
||||
<strong>Output:</strong> 8
|
||||
<strong>Explanation:</strong> In this example, we have 8 seconds. The following operations can be performed to mark all indices:
|
||||
Second 1: Choose index 1 and decrement nums[1] by one. nums becomes [1,2,0].
|
||||
Second 2: Choose index 1 and decrement nums[1] by one. nums becomes [0,2,0].
|
||||
Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [0,1,0].
|
||||
Second 4: Choose index 2 and decrement nums[2] by one. nums becomes [0,0,0].
|
||||
Second 5: Mark the index changeIndices[5], which is marking index 3, since nums[3] is equal to 0.
|
||||
Second 6: Mark the index changeIndices[6], which is marking index 2, since nums[2] is equal to 0.
|
||||
Second 7: Do nothing.
|
||||
Second 8: Mark the index changeIndices[8], which is marking index 1, since nums[1] is equal to 0.
|
||||
Now all indices have been marked.
|
||||
It can be shown that it is not possible to mark all indices earlier than the 8th second.
|
||||
Hence, the answer is 8.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,3], changeIndices = [1,1,1,2,1,1,1]
|
||||
<strong>Output:</strong> 6
|
||||
<strong>Explanation:</strong> In this example, we have 7 seconds. The following operations can be performed to mark all indices:
|
||||
Second 1: Choose index 2 and decrement nums[2] by one. nums becomes [1,2].
|
||||
Second 2: Choose index 2 and decrement nums[2] by one. nums becomes [1,1].
|
||||
Second 3: Choose index 2 and decrement nums[2] by one. nums becomes [1,0].
|
||||
Second 4: Mark the index changeIndices[4], which is marking index 2, since nums[2] is equal to 0.
|
||||
Second 5: Choose index 1 and decrement nums[1] by one. nums becomes [0,0].
|
||||
Second 6: Mark the index changeIndices[6], which is marking index 1, since nums[1] is equal to 0.
|
||||
Now all indices have been marked.
|
||||
It can be shown that it is not possible to mark all indices earlier than the 6th second.
|
||||
Hence, the answer is 6.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [0,1], changeIndices = [2,2,2]
|
||||
<strong>Output:</strong> -1
|
||||
<strong>Explanation:</strong> In this example, it is impossible to mark all indices because index 1 isn't in changeIndices.
|
||||
Hence, the answer is -1.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n == nums.length <= 2000</code></li>
|
||||
<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>
|
||||
<li><code>1 <= m == changeIndices.length <= 2000</code></li>
|
||||
<li><code>1 <= changeIndices[i] <= n</code></li>
|
||||
</ul>
|
||||
69
leetcode/problem/earliest-second-to-mark-indices-ii.html
Normal file
69
leetcode/problem/earliest-second-to-mark-indices-ii.html
Normal file
@ -0,0 +1,69 @@
|
||||
<p>You are given two <strong>1-indexed</strong> integer arrays, <code>nums</code> and, <code>changeIndices</code>, having lengths <code>n</code> and <code>m</code>, respectively.</p>
|
||||
|
||||
<p>Initially, all indices in <code>nums</code> are unmarked. Your task is to mark <strong>all</strong> indices in <code>nums</code>.</p>
|
||||
|
||||
<p>In each second, <code>s</code>, in order from <code>1</code> to <code>m</code> (<strong>inclusive</strong>), you can perform <strong>one</strong> of the following operations:</p>
|
||||
|
||||
<ul>
|
||||
<li>Choose an index <code>i</code> in the range <code>[1, n]</code> and <strong>decrement</strong> <code>nums[i]</code> by <code>1</code>.</li>
|
||||
<li>Set <code>nums[changeIndices[s]]</code> to any <strong>non-negative</strong> value.</li>
|
||||
<li>Choose an index <code>i</code> in the range <code>[1, n]</code>, where <code>nums[i]</code> is <strong>equal</strong> to <code>0</code>, and <strong>mark</strong> index <code>i</code>.</li>
|
||||
<li>Do nothing.</li>
|
||||
</ul>
|
||||
|
||||
<p>Return <em>an integer denoting the <strong>earliest second</strong> in the range </em><code>[1, m]</code><em> when <strong>all</strong> indices in </em><code>nums</code><em> can be marked by choosing operations optimally, or </em><code>-1</code><em> if it is impossible.</em></p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [3,2,3], changeIndices = [1,3,2,2,2,2,3]
|
||||
<strong>Output:</strong> 6
|
||||
<strong>Explanation:</strong> In this example, we have 7 seconds. The following operations can be performed to mark all indices:
|
||||
Second 1: Set nums[changeIndices[1]] to 0. nums becomes [0,2,3].
|
||||
Second 2: Set nums[changeIndices[2]] to 0. nums becomes [0,2,0].
|
||||
Second 3: Set nums[changeIndices[3]] to 0. nums becomes [0,0,0].
|
||||
Second 4: Mark index 1, since nums[1] is equal to 0.
|
||||
Second 5: Mark index 2, since nums[2] is equal to 0.
|
||||
Second 6: Mark index 3, since nums[3] is equal to 0.
|
||||
Now all indices have been marked.
|
||||
It can be shown that it is not possible to mark all indices earlier than the 6th second.
|
||||
Hence, the answer is 6.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [0,0,1,2], changeIndices = [1,2,1,2,1,2,1,2]
|
||||
<strong>Output:</strong> 7
|
||||
<strong>Explanation:</strong> In this example, we have 8 seconds. The following operations can be performed to mark all indices:
|
||||
Second 1: Mark index 1, since nums[1] is equal to 0.
|
||||
Second 2: Mark index 2, since nums[2] is equal to 0.
|
||||
Second 3: Decrement index 4 by one. nums becomes [0,0,1,1].
|
||||
Second 4: Decrement index 4 by one. nums becomes [0,0,1,0].
|
||||
Second 5: Decrement index 3 by one. nums becomes [0,0,0,0].
|
||||
Second 6: Mark index 3, since nums[3] is equal to 0.
|
||||
Second 7: Mark index 4, since nums[4] is equal to 0.
|
||||
Now all indices have been marked.
|
||||
It can be shown that it is not possible to mark all indices earlier than the 7th second.
|
||||
Hence, the answer is 7.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,2,3], changeIndices = [1,2,3]
|
||||
<strong>Output:</strong> -1
|
||||
<strong>Explanation: </strong>In this example, it can be shown that it is impossible to mark all indices, as we don't have enough seconds.
|
||||
Hence, the answer is -1.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= n == nums.length <= 5000</code></li>
|
||||
<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>
|
||||
<li><code>1 <= m == changeIndices.length <= 5000</code></li>
|
||||
<li><code>1 <= changeIndices[i] <= n</code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,46 @@
|
||||
<p>There exist <code>n</code> rectangles in a 2D plane. You are given two <strong>0-indexed</strong> 2D integer arrays <code>bottomLeft</code> and <code>topRight</code>, both of size <code>n x 2</code>, where <code>bottomLeft[i]</code> and <code>topRight[i]</code> represent the <strong>bottom-left</strong> and <strong>top-right</strong> coordinates of the <code>i<sup>th</sup></code> rectangle respectively.</p>
|
||||
|
||||
<p>You can select a region formed from the <strong>intersection</strong> of two of the given rectangles. You need to find the <strong>largest </strong>area of a <strong>square</strong> that can fit <strong>inside</strong> this region if you select the region optimally.</p>
|
||||
|
||||
<p>Return <em>the <strong>largest </strong>possible area of a square, or </em><code>0</code><em> if there <strong>do not</strong> exist any intersecting regions between the rectangles</em>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/05/example12.png" style="width: 443px; height: 364px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;" />
|
||||
<pre>
|
||||
<strong>Input:</strong> bottomLeft = [[1,1],[2,2],[3,1]], topRight = [[3,3],[4,4],[6,6]]
|
||||
<strong>Output:</strong> 1
|
||||
<strong>Explanation:</strong> A square with side length 1 can fit inside either the intersecting region of rectangle 0 and rectangle 1, or the intersecting region of rectangle 1 and rectangle 2. Hence the largest area is side * side which is 1 * 1 == 1.
|
||||
It can be shown that a square with a greater side length can not fit inside any intersecting region.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/04/rectanglesexample2.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 445px; height: 365px;" />
|
||||
<pre>
|
||||
<strong>Input:</strong> bottomLeft = [[1,1],[2,2],[1,2]], topRight = [[3,3],[4,4],[3,4]]
|
||||
<strong>Output:</strong> 1
|
||||
<strong>Explanation:</strong> A square with side length 1 can fit inside either the intersecting region of rectangle 0 and rectangle 1, the intersecting region of rectangle 1 and rectangle 2, or the intersection region of all 3 rectangles. Hence the largest area is side * side which is 1 * 1 == 1.
|
||||
It can be shown that a square with a greater side length can not fit inside any intersecting region.
|
||||
Note that the region can be formed by the intersection of more than 2 rectangles.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 3:</strong></p>
|
||||
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/04/rectanglesexample3.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 444px; height: 364px;" />
|
||||
<pre>
|
||||
<strong>Input:</strong> bottomLeft = [[1,1],[3,3],[3,1]], topRight = [[2,2],[4,4],[4,2]]
|
||||
<strong>Output:</strong> 0
|
||||
<strong>Explanation:</strong> No pair of rectangles intersect, hence, we return 0.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>n == bottomLeft.length == topRight.length</code></li>
|
||||
<li><code>2 <= n <= 10<sup>3</sup></code></li>
|
||||
<li><code>bottomLeft[i].length == topRight[i].length == 2</code></li>
|
||||
<li><code>1 <= bottomLeft[i][0], bottomLeft[i][1] <= 10<sup>7</sup></code></li>
|
||||
<li><code>1 <= topRight[i][0], topRight[i][1] <= 10<sup>7</sup></code></li>
|
||||
<li><code>bottomLeft[i][0] < topRight[i][0]</code></li>
|
||||
<li><code>bottomLeft[i][1] < topRight[i][1]</code></li>
|
||||
</ul>
|
||||
35
leetcode/problem/split-the-array.html
Normal file
35
leetcode/problem/split-the-array.html
Normal file
@ -0,0 +1,35 @@
|
||||
<p>You are given an integer array <code>nums</code> of <strong>even</strong> length. You have to split the array into two parts <code>nums1</code> and <code>nums2</code> such that:</p>
|
||||
|
||||
<ul>
|
||||
<li><code>nums1.length == nums2.length == nums.length / 2</code>.</li>
|
||||
<li><code>nums1</code> should contain <strong>distinct </strong>elements.</li>
|
||||
<li><code>nums2</code> should also contain <strong>distinct</strong> elements.</li>
|
||||
</ul>
|
||||
|
||||
<p>Return <code>true</code><em> if it is possible to split the array, and </em><code>false</code> <em>otherwise</em><em>.</em></p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,1,2,2,3,4]
|
||||
<strong>Output:</strong> true
|
||||
<strong>Explanation:</strong> One of the possible ways to split nums is nums1 = [1,2,3] and nums2 = [1,2,4].
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,1,1,1]
|
||||
<strong>Output:</strong> false
|
||||
<strong>Explanation:</strong> The only possible way to split nums is nums1 = [1,1] and nums2 = [1,1]. Both nums1 and nums2 do not contain distinct elements. Therefore, we return false.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= nums.length <= 100</code></li>
|
||||
<li><code>nums.length % 2 == 0 </code></li>
|
||||
<li><code>1 <= nums[i] <= 100</code></li>
|
||||
</ul>
|
||||
Loading…
Reference in New Issue
Block a user