{"id":8775,"date":"2021-11-26T08:29:15","date_gmt":"2021-11-26T16:29:15","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=8775"},"modified":"2021-11-26T08:33:06","modified_gmt":"2021-11-26T16:33:06","slug":"leetcode-2008-maximum-earnings-from-taxi","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/dynamic-programming\/leetcode-2008-maximum-earnings-from-taxi\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2008. Maximum Earnings From Taxi"},"content":{"rendered":"\n

There are n<\/code> points on a road you are driving your taxi on. The n<\/code> points on the road are labeled from 1<\/code> to n<\/code> in the direction you are going, and you want to drive from point 1<\/code> to point n<\/code> to make money by picking up passengers. You cannot change the direction of the taxi.<\/p>\n\n\n\n

The passengers are represented by a 0-indexed<\/strong> 2D integer array rides<\/code>, where rides[i] = [starti<\/sub>, endi<\/sub>, tipi<\/sub>]<\/code> denotes the ith<\/sup><\/code> passenger requesting a ride from point starti<\/sub><\/code> to point endi<\/sub><\/code> who is willing to give a tipi<\/sub><\/code> dollar tip.<\/p>\n\n\n\n

For each <\/strong>passenger i<\/code> you pick up, you earn<\/strong> endi<\/sub> - starti<\/sub> + tipi<\/sub><\/code> dollars. You may only drive at most one <\/strong>passenger at a time.<\/p>\n\n\n\n

Given n<\/code> and rides<\/code>, return the maximum<\/strong> number of dollars you can earn by picking up the passengers optimally.<\/em><\/p>\n\n\n\n

Note:<\/strong> You may drop off a passenger and pick up a different passenger at the same point.<\/p>\n\n\n\n

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

Input:<\/strong> n = 5, rides = [[2,5,4],[1,5,1]]\nOutput:<\/strong> 7\nExplanation:<\/strong> We can pick up passenger 0 to earn 5 - 2 + 4 = 7 dollars.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> n = 20, rides = [[1,6,1],[3,10,2],[10,12,3],[11,12,2],[12,15,2],[13,18,1]]\nOutput:<\/strong> 20\nExplanation:<\/strong> We will pick up the following passengers:\n- Drive passenger 1 from point 3 to point 10 for a profit of 10 - 3 + 2 = 9 dollars.\n- Drive passenger 2 from point 10 to point 12 for a profit of 12 - 10 + 3 = 5 dollars.\n- Drive passenger 5 from point 13 to point 18 for a profit of 18 - 13 + 1 = 6 dollars.\nWe earn 9 + 5 + 6 = 20 dollars in total.<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= n <= 105<\/sup><\/code><\/li>
  • 1 <= rides.length <= 3 * 104<\/sup><\/code><\/li>
  • rides[i].length == 3<\/code><\/li>
  • 1 <= starti<\/sub> < endi<\/sub> <= n<\/code><\/li>
  • 1 <= tipi<\/sub> <= 105<\/sup><\/code><\/li><\/ul>\n\n\n\n
    Solution: DP<\/strong><\/h2>\n\n\n\n
    dp[i] := max earnings we can get at position i and the taxi is empty.<\/p>\n\n\n\n
    dp[i] = max(dp[i – 1], dp[s] + gain) where e = i, gain = e – s + tips<\/p>\n\n\n\n
    For each i, we check all the rides that end at i and find the best one (which may have different starting points), otherwise the earning will be the same as previous position (i – 1).<\/p>\n\n\n\n
    answer = dp[n] <\/p>\n\n\n\n
    Time complexity: O(m + n)
    Space complexity: O(m + n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  long long maxTaxiEarnings(int n, vector>& rides) {    \n    vector dp(n + 1);\n    vector>> m(n + 1);\n    for (const auto& r : rides)\n      m[r[1]].emplace_back(r[0], r[1] - r[0] + r[2]);\n    for (int i = 1; i <= n; ++i) {\n      dp[i] = dp[i - 1];\n      for (const auto [s, g] : m[i])\n        dp[i] = max(dp[i], dp[s] + g);\n    }\n    return dp[n];\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    There are n points on a road you are driving your taxi on. The n points on the road are labeled from 1 to n in the direction you are going, and you…<\/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":[18,177],"class_list":["post-8775","post","type-post","status-publish","format-standard","hentry","category-dynamic-programming","tag-dp","tag-medium","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8775","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=8775"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8775\/revisions"}],"predecessor-version":[{"id":8778,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8775\/revisions\/8778"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=8775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=8775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=8775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}