mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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188 lines
18 KiB
JSON
188 lines
18 KiB
JSON
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"title": "Maximum Profit in Job Scheduling",
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"content": "<p>We have <code>n</code> jobs, where every job is scheduled to be done from <code>startTime[i]</code> to <code>endTime[i]</code>, obtaining a profit of <code>profit[i]</code>.</p>\n\n<p>You're given the <code>startTime</code>, <code>endTime</code> and <code>profit</code> arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.</p>\n\n<p>If you choose a job that ends at time <code>X</code> you will be able to start another job that starts at time <code>X</code>.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png\" style=\"width: 380px; height: 154px;\" /></strong></p>\n\n<pre>\n<strong>Input:</strong> startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]\n<strong>Output:</strong> 120\n<strong>Explanation:</strong> The subset chosen is the first and fourth job. \nTime range [1-3]+[3-6] , we get profit of 120 = 50 + 70.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png\" style=\"width: 600px; height: 112px;\" /> </strong></p>\n\n<pre>\n<strong>Input:</strong> startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]\n<strong>Output:</strong> 150\n<strong>Explanation:</strong> The subset chosen is the first, fourth and fifth job. \nProfit obtained 150 = 20 + 70 + 60.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png\" style=\"width: 400px; height: 112px;\" /></strong></p>\n\n<pre>\n<strong>Input:</strong> startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]\n<strong>Output:</strong> 6\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= startTime.length == endTime.length == profit.length <= 5 * 10<sup>4</sup></code></li>\n\t<li><code>1 <= startTime[i] < endTime[i] <= 10<sup>9</sup></code></li>\n\t<li><code>1 <= profit[i] <= 10<sup>4</sup></code></li>\n</ul>\n",
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"Think on DP.",
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"Sort the elements by starting time, then define the dp[i] as the maximum profit taking elements from the suffix starting at i.",
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"Use binarySearch (lower_bound/upper_bound on C++) to get the next index for the DP transition."
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