mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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190 lines
25 KiB
JSON
190 lines
25 KiB
JSON
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"title": "Maximum Segment Sum After Removals",
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"content": "<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums</code> and <code>removeQueries</code>, both of length <code>n</code>. For the <code>i<sup>th</sup></code> query, the element in <code>nums</code> at the index <code>removeQueries[i]</code> is removed, splitting <code>nums</code> into different segments.</p>\n\n<p>A <strong>segment</strong> is a contiguous sequence of <strong>positive</strong> integers in <code>nums</code>. A <strong>segment sum</strong> is the sum of every element in a segment.</p>\n\n<p>Return<em> an integer array </em><code>answer</code><em>, of length </em><code>n</code><em>, where </em><code>answer[i]</code><em> is the <strong>maximum</strong> segment sum after applying the </em><code>i<sup>th</sup></code> <em>removal.</em></p>\n\n<p><strong>Note:</strong> The same index will <strong>not</strong> be removed more than once.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,5,6,1], removeQueries = [0,3,2,4,1]\n<strong>Output:</strong> [14,7,2,2,0]\n<strong>Explanation:</strong> Using 0 to indicate a removed element, the answer is as follows:\nQuery 1: Remove the 0th element, nums becomes [0,2,5,6,1] and the maximum segment sum is 14 for segment [2,5,6,1].\nQuery 2: Remove the 3rd element, nums becomes [0,2,5,0,1] and the maximum segment sum is 7 for segment [2,5].\nQuery 3: Remove the 2nd element, nums becomes [0,2,0,0,1] and the maximum segment sum is 2 for segment [2]. \nQuery 4: Remove the 4th element, nums becomes [0,2,0,0,0] and the maximum segment sum is 2 for segment [2]. \nQuery 5: Remove the 1st element, nums becomes [0,0,0,0,0] and the maximum segment sum is 0, since there are no segments.\nFinally, we return [14,7,2,2,0].</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,2,11,1], removeQueries = [3,2,1,0]\n<strong>Output:</strong> [16,5,3,0]\n<strong>Explanation:</strong> Using 0 to indicate a removed element, the answer is as follows:\nQuery 1: Remove the 3rd element, nums becomes [3,2,11,0] and the maximum segment sum is 16 for segment [3,2,11].\nQuery 2: Remove the 2nd element, nums becomes [3,2,0,0] and the maximum segment sum is 5 for segment [3,2].\nQuery 3: Remove the 1st element, nums becomes [3,0,0,0] and the maximum segment sum is 3 for segment [3].\nQuery 4: Remove the 0th element, nums becomes [0,0,0,0] and the maximum segment sum is 0, since there are no segments.\nFinally, we return [16,5,3,0].\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length == removeQueries.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= 10<sup>9</sup></code></li>\n\t<li><code>0 <= removeQueries[i] < n</code></li>\n\t<li>All the values of <code>removeQueries</code> are <strong>unique</strong>.</li>\n</ul>\n",
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"translatedTitle": "删除操作后的最大子段和",
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"translatedContent": "<p>给你两个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 和 <code>removeQueries</code> ,两者长度都为 <code>n</code> 。对于第 <code>i</code> 个查询,<code>nums</code> 中位于下标 <code>removeQueries[i]</code> 处的元素被删除,将 <code>nums</code> 分割成更小的子段。</p>\n\n<p>一个 <strong>子段</strong> 是 <code>nums</code> 中连续 <strong>正</strong> 整数形成的序列。<strong>子段和</strong> 是子段中所有元素的和。</p>\n\n<p>请你返回一个长度为 <code>n</code> 的整数数组<em> </em><code>answer</code> ,其中<em> </em><code>answer[i]</code>是第 <code>i</code> 次删除操作以后的 <strong>最大</strong> 子段和。</p>\n\n<p><strong>注意:</strong>一个下标至多只会被删除一次。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre><b>输入:</b>nums = [1,2,5,6,1], removeQueries = [0,3,2,4,1]\n<b>输出:</b>[14,7,2,2,0]\n<b>解释:</b>用 0 表示被删除的元素,答案如下所示:\n查询 1 :删除第 0 个元素,nums 变成 [0,2,5,6,1] ,最大子段和为子段 [2,5,6,1] 的和 14 。\n查询 2 :删除第 3 个元素,nums 变成 [0,2,5,0,1] ,最大子段和为子段 [2,5] 的和 7 。\n查询 3 :删除第 2 个元素,nums 变成 [0,2,0,0,1] ,最大子段和为子段 [2] 的和 2 。\n查询 4 :删除第 4 个元素,nums 变成 [0,2,0,0,0] ,最大子段和为子段 [2] 的和 2 。\n查询 5 :删除第 1 个元素,nums 变成 [0,0,0,0,0] ,最大子段和为 0 ,因为没有任何子段存在。\n所以,我们返回 [14,7,2,2,0] 。</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre><b>输入:</b>nums = [3,2,11,1], removeQueries = [3,2,1,0]\n<b>输出:</b>[16,5,3,0]\n<b>解释:</b>用 0 表示被删除的元素,答案如下所示:\n查询 1 :删除第 3 个元素,nums 变成 [3,2,11,0] ,最大子段和为子段 [3,2,11] 的和 16 。\n查询 2 :删除第 2 个元素,nums 变成 [3,2,0,0] ,最大子段和为子段 [3,2] 的和 5 。\n查询 3 :删除第 1 个元素,nums 变成 [3,0,0,0] ,最大子段和为子段 [3] 的和 3 。\n查询 5 :删除第 0 个元素,nums 变成 [0,0,0,0] ,最大子段和为 0 ,因为没有任何子段存在。\n所以,我们返回 [16,5,3,0] 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length == removeQueries.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= 10<sup>9</sup></code></li>\n\t<li><code>0 <= removeQueries[i] < n</code></li>\n\t<li><code>removeQueries</code> 中所有数字 <strong>互不相同</strong> 。</li>\n</ul>\n",
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