mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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196 lines
24 KiB
JSON
196 lines
24 KiB
JSON
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"categoryTitle": "Algorithms",
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"boundTopicId": 303612,
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"title": "Max Value of Equation",
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"content": "<p>You are given an array <code>points</code> containing the coordinates of points on a 2D plane, sorted by the x-values, where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> such that <code>x<sub>i</sub> < x<sub>j</sub></code> for all <code>1 <= i < j <= points.length</code>. You are also given an integer <code>k</code>.</p>\n\n<p>Return <em>the maximum value of the equation </em><code>y<sub>i</sub> + y<sub>j</sub> + |x<sub>i</sub> - x<sub>j</sub>|</code> where <code>|x<sub>i</sub> - x<sub>j</sub>| <= k</code> and <code>1 <= i < j <= points.length</code>.</p>\n\n<p>It is guaranteed that there exists at least one pair of points that satisfy the constraint <code>|x<sub>i</sub> - x<sub>j</sub>| <= k</code>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> points = [[1,3],[2,0],[5,10],[6,-10]], k = 1\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> The first two points satisfy the condition |x<sub>i</sub> - x<sub>j</sub>| <= 1 and if we calculate the equation we get 3 + 0 + |1 - 2| = 4. Third and fourth points also satisfy the condition and give a value of 10 + -10 + |5 - 6| = 1.\nNo other pairs satisfy the condition, so we return the max of 4 and 1.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> points = [[0,0],[3,0],[9,2]], k = 3\n<strong>Output:</strong> 3\n<strong>Explanation: </strong>Only the first two points have an absolute difference of 3 or less in the x-values, and give the value of 0 + 0 + |0 - 3| = 3.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 <= points.length <= 10<sup>5</sup></code></li>\n\t<li><code>points[i].length == 2</code></li>\n\t<li><code>-10<sup>8</sup> <= x<sub>i</sub>, y<sub>i</sub> <= 10<sup>8</sup></code></li>\n\t<li><code>0 <= k <= 2 * 10<sup>8</sup></code></li>\n\t<li><code>x<sub>i</sub> < x<sub>j</sub></code> for all <code>1 <= i < j <= points.length</code></li>\n\t<li><code>x<sub>i</sub></code> form a strictly increasing sequence.</li>\n</ul>\n",
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"translatedTitle": "满足不等式的最大值",
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"translatedContent": "<p>给你一个数组 <code>points</code> 和一个整数 <code>k</code> 。数组中每个元素都表示二维平面上的点的坐标，并按照横坐标 x 的值从小到大排序。也就是说 <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> ，并且在 <code>1 <= i < j <= points.length</code> 的前提下， <code>x<sub>i</sub> < x<sub>j</sub></code> 总成立。</p>\n\n<p>请你找出<em> </em><code>y<sub>i</sub> + y<sub>j</sub> + |x<sub>i</sub> - x<sub>j</sub>|</code> 的 <strong>最大值</strong>，其中 <code>|x<sub>i</sub> - x<sub>j</sub>| <= k</code> 且 <code>1 <= i < j <= points.length</code>。</p>\n\n<p>题目测试数据保证至少存在一对能够满足 <code>|x<sub>i</sub> - x<sub>j</sub>| <= k</code> 的点。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>points = [[1,3],[2,0],[5,10],[6,-10]], k = 1\n<strong>输出：</strong>4\n<strong>解释：</strong>前两个点满足 |x<sub>i</sub> - x<sub>j</sub>| <= 1 ，代入方程计算，则得到值 3 + 0 + |1 - 2| = 4 。第三个和第四个点也满足条件，得到值 10 + -10 + |5 - 6| = 1 。\n没有其他满足条件的点，所以返回 4 和 1 中最大的那个。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>points = [[0,0],[3,0],[9,2]], k = 3\n<strong>输出：</strong>3\n<strong>解释：</strong>只有前两个点满足 |x<sub>i</sub> - x<sub>j</sub>| <= 3 ，代入方程后得到值 0 + 0 + |0 - 3| = 3 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 <= points.length <= 10^5</code></li>\n\t<li><code>points[i].length == 2</code></li>\n\t<li><code>-10^8 <= points[i][0], points[i][1] <= 10^8</code></li>\n\t<li><code>0 <= k <= 2 * 10^8</code></li>\n\t<li>对于所有的<code>1 <= i < j <= points.length</code> ，<code>points[i][0] < points[j][0]</code> 都成立。也就是说，<code>x<sub>i</sub></code> 是严格递增的。</li>\n</ul>\n",
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"Use a priority queue to store for each point i, the tuple [yi-xi, xi]",
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"Loop through the array and pop elements from the heap if the condition xj - xi > k, where j is the current index and i is the point on top the queue.",
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"After popping elements from the queue. If the queue is not empty, calculate the equation with the current point and the point on top of the queue and maximize the answer."
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