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程序员小墨 2024-03-01 00:47:37 +08:00
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# 力扣题库(完整版) # 力扣题库(完整版)
> 最后更新日期: **2024.02.19** > 最后更新日期: **2024.03.01**
> >
> 使用脚本前请务必仔细完整阅读本 `README.md` 文件 > 使用脚本前请务必仔细完整阅读本 `README.md` 文件

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<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><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>&nbsp;</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>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 &lt;= nums.length &lt;= 100</code></li>
<li><code>nums.length % 2 == 0</code></li>
<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>

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<p>给你两个下标从 <strong>1</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;<code>changeIndices</code>&nbsp;,数组的长度分别为&nbsp;<code>n</code>&nbsp;<code>m</code>&nbsp;</p>
<p>一开始,<code>nums</code>&nbsp;中所有下标都是未标记的,你的任务是标记 <code>nums</code>&nbsp;<strong>所有</strong>&nbsp;下标。</p>
<p>从第 <code>1</code>&nbsp;秒到第 <code>m</code>&nbsp;秒(<b>包括&nbsp;</b>&nbsp;<code>m</code>&nbsp;秒),对于每一秒 <code>s</code>&nbsp;,你可以执行以下操作 <strong>之一</strong>&nbsp;</p>
<ul>
<li>选择范围&nbsp;<code>[1, n]</code>&nbsp;中的一个下标 <code>i</code>&nbsp;,并且将&nbsp;<code>nums[i]</code> <strong>减少</strong>&nbsp;<code>1</code>&nbsp;</li>
<li>如果&nbsp;<code>nums[changeIndices[s]]</code>&nbsp;<strong>等于</strong>&nbsp;<code>0</code>&nbsp;<strong>标记</strong>&nbsp;下标&nbsp;<code>changeIndices[s]</code>&nbsp;</li>
<li>什么也不做。</li>
</ul>
<p>请你返回范围 <code>[1, m]</code>&nbsp;中的一个整数,表示最优操作下,标记&nbsp;<code>nums</code>&nbsp;<strong>所有</strong>&nbsp;下标的 <strong>最早秒数</strong>&nbsp;,如果无法标记所有下标,返回 <code>-1</code>&nbsp;</p>
<p>&nbsp;</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>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 &lt;= n == nums.length &lt;= 2000</code></li>
<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
<li><code>1 &lt;= m == changeIndices.length &lt;= 2000</code></li>
<li><code>1 &lt;= changeIndices[i] &lt;= n</code></li>
</ul>

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<p>给你两个下标从 <strong>1</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;<code>changeIndices</code>&nbsp;,数组的长度分别为&nbsp;<code>n</code>&nbsp;<code>m</code>&nbsp;</p>
<p>一开始,<code>nums</code>&nbsp;中所有下标都是未标记的,你的任务是标记 <code>nums</code>&nbsp;<strong>所有</strong>&nbsp;下标。</p>
<p>从第 <code>1</code>&nbsp;秒到第 <code>m</code>&nbsp;秒(<b>包括&nbsp;</b>&nbsp;<code>m</code>&nbsp;秒),对于每一秒 <code>s</code>&nbsp;,你可以执行以下操作 <strong>之一</strong>&nbsp;</p>
<ul>
<li>选择范围&nbsp;<code>[1, n]</code>&nbsp;中的一个下标 <code>i</code>&nbsp;,并且将&nbsp;<code>nums[i]</code> <strong>减少</strong>&nbsp;<code>1</code>&nbsp;</li>
<li>&nbsp;<code>nums[changeIndices[s]]</code>&nbsp;设置成任意的 <strong>非负</strong>&nbsp;整数。</li>
<li>选择范围&nbsp;<code>[1, n]</code>&nbsp;中的一个下标&nbsp;<code>i</code>&nbsp; 满足&nbsp;<code>nums[i]</code> <strong>等于</strong> <code>0</code>, 并 <strong>标记</strong>&nbsp;下标&nbsp;<code>i</code></li>
<li>什么也不做。</li>
</ul>
<p>请你返回范围 <code>[1, m]</code>&nbsp;中的一个整数,表示最优操作下,标记&nbsp;<code>nums</code>&nbsp;<strong>所有</strong>&nbsp;下标的 <strong>最早秒数</strong>&nbsp;,如果无法标记所有下标,返回 <code>-1</code>&nbsp;</p>
<p>&nbsp;</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>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 &lt;= n == nums.length &lt;= 5000</code></li>
<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
<li><code>1 &lt;= m == changeIndices.length &lt;= 5000</code></li>
<li><code>1 &lt;= changeIndices[i] &lt;= n</code></li>
</ul>

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<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><code>topRight</code>,两个数组的大小都是 <code>n x 2</code> ,其中 <code>bottomLeft[i]</code><code>topRight[i]</code> 分别代表第 <code>i</code> 个矩形的<strong> 左下角 </strong><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>&nbsp;形成。你需要找出能够放入该区域 <strong></strong><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>&nbsp;</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>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>n == bottomLeft.length == topRight.length</code></li>
<li><code>2 &lt;= n &lt;= 10<sup>3</sup></code></li>
<li><code>bottomLeft[i].length == topRight[i].length == 2</code></li>
<li><code>1 &lt;= bottomLeft[i][0], bottomLeft[i][1] &lt;= 10<sup>7</sup></code></li>
<li><code>1 &lt;= topRight[i][0], topRight[i][1] &lt;= 10<sup>7</sup></code></li>
<li><code>bottomLeft[i][0] &lt; topRight[i][0]</code></li>
<li><code>bottomLeft[i][1] &lt; topRight[i][1]</code></li>
</ul>

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<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>&nbsp;</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>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 &lt;= nums.length &lt;= 100</code></li>
<li><code>nums.length % 2 == 0 </code></li>
<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>

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<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>&nbsp;</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&#39;t in changeIndices.
Hence, the answer is -1.
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 &lt;= n == nums.length &lt;= 2000</code></li>
<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
<li><code>1 &lt;= m == changeIndices.length &lt;= 2000</code></li>
<li><code>1 &lt;= changeIndices[i] &lt;= n</code></li>
</ul>

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<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>&nbsp;</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&#39;t have enough seconds.
Hence, the answer is -1.
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 &lt;= n == nums.length &lt;= 5000</code></li>
<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
<li><code>1 &lt;= m == changeIndices.length &lt;= 5000</code></li>
<li><code>1 &lt;= changeIndices[i] &lt;= n</code></li>
</ul>

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<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&nbsp;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>&nbsp;</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>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>n == bottomLeft.length == topRight.length</code></li>
<li><code>2 &lt;= n &lt;= 10<sup>3</sup></code></li>
<li><code>bottomLeft[i].length == topRight[i].length == 2</code></li>
<li><code>1 &lt;= bottomLeft[i][0], bottomLeft[i][1] &lt;= 10<sup>7</sup></code></li>
<li><code>1 &lt;= topRight[i][0], topRight[i][1] &lt;= 10<sup>7</sup></code></li>
<li><code>bottomLeft[i][0] &lt; topRight[i][0]</code></li>
<li><code>bottomLeft[i][1] &lt; topRight[i][1]</code></li>
</ul>

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<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>&nbsp;</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&#39;t in changeIndices.
Hence, the answer is -1.
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
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<li><code>1 &lt;= n == nums.length &lt;= 2000</code></li>
<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
<li><code>1 &lt;= m == changeIndices.length &lt;= 2000</code></li>
<li><code>1 &lt;= changeIndices[i] &lt;= n</code></li>
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<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>&nbsp;</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&#39;t have enough seconds.
Hence, the answer is -1.
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 &lt;= n == nums.length &lt;= 5000</code></li>
<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
<li><code>1 &lt;= m == changeIndices.length &lt;= 5000</code></li>
<li><code>1 &lt;= changeIndices[i] &lt;= n</code></li>
</ul>

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<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&nbsp;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>&nbsp;</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>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>n == bottomLeft.length == topRight.length</code></li>
<li><code>2 &lt;= n &lt;= 10<sup>3</sup></code></li>
<li><code>bottomLeft[i].length == topRight[i].length == 2</code></li>
<li><code>1 &lt;= bottomLeft[i][0], bottomLeft[i][1] &lt;= 10<sup>7</sup></code></li>
<li><code>1 &lt;= topRight[i][0], topRight[i][1] &lt;= 10<sup>7</sup></code></li>
<li><code>bottomLeft[i][0] &lt; topRight[i][0]</code></li>
<li><code>bottomLeft[i][1] &lt; topRight[i][1]</code></li>
</ul>

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<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>&nbsp;</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>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 &lt;= nums.length &lt;= 100</code></li>
<li><code>nums.length % 2 == 0 </code></li>
<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>