{"id":5779,"date":"2019-10-24T07:58:41","date_gmt":"2019-10-24T14:58:41","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=5779"},"modified":"2019-10-24T08:11:01","modified_gmt":"2019-10-24T15:11:01","slug":"leetcode-1235-maximum-profit-in-job-scheduling","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/dynamic-programming\/leetcode-1235-maximum-profit-in-job-scheduling\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1235. Maximum Profit in Job Scheduling"},"content":{"rendered":"\n

We have n<\/code> jobs, where every job is scheduled to be done from startTime[i]<\/code> to endTime[i]<\/code>, obtaining a profit of profit[i]<\/code>.<\/p>\n\n\n\n

You’re given the startTime<\/code> , endTime<\/code> and profit<\/code> arrays, you need to output the maximum profit you can take such that there are no 2 jobs in the subset with overlapping time range.<\/p>\n\n\n\n

If you choose a job that ends at time X<\/code> you will be able to start another job that starts at time X<\/code>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]\nOutput:<\/strong> 120\nExplanation:<\/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\n\n
Example 2:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
\nInput:<\/strong> startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]\nOutput:<\/strong> 150\nExplanation:<\/strong> The subset chosen is the first, fourth and fifth job. \nProfit obtained 150 = 20 + 70 + 60.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]\nOutput:<\/strong> 6\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4<\/code><\/li>
  • 1 <= startTime[i] < endTime[i] <= 10^9<\/code><\/li>
  • 1 <= profit[i] <= 10^4<\/code><\/li><\/ul>\n\n\n\n
    Solution: DP<\/strong> + binary search<\/strong><\/h2>\n\n\n\n
    Sort jobs by ending time.
    dp[t] := max profit by end time t.

    for a job = (s, e, p)
    dp[e] = dp[u] + p, u <= s, and if dp[u] + p > last_element in dp.<\/p>\n\n\n\n
    Time complexity: O(nlogn)
    Space complexity: O(n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int jobScheduling(vector& startTime, vector& endTime, vector& profit) {\n    const int n = startTime.size();  \n    vector> jobs(n);\n    for (int i = 0; i < n; ++i)\n      jobs[i] = {endTime[i], startTime[i], profit[i]};\n    sort(begin(jobs), end(jobs));\n    \n    int ans = 0;\n    map m{{0, 0}}; \/\/ {end_time -> max_profit}\n    for (const auto& job : jobs) {\n      int p = prev(m.upper_bound(job[1]))->second + job[2];\n      if (p > rbegin(m)->second) {\n        m[job[0]] = p;\n      }      \n    }\n    return rbegin(m)->second;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i]. You’re given the startTime , endTime and profit arrays, you need to output the maximum profit you can take…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[46],"tags":[52,18,217],"class_list":["post-5779","post","type-post","status-publish","format-standard","hentry","category-dynamic-programming","tag-binary-search","tag-dp","tag-hard","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5779","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=5779"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5779\/revisions"}],"predecessor-version":[{"id":5781,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5779\/revisions\/5781"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=5779"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=5779"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=5779"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}