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 want to drive from point 1 to point n to make money by picking up passengers. You cannot change the direction of the taxi.
The passengers are represented by a 0-indexed 2D integer array rides, where rides[i] = [starti, endi, tipi] denotes the ith passenger requesting a ride from point starti to point endi who is willing to give a tipi dollar tip.
For each passenger i you pick up, you earn endi - starti + tipi dollars. You may only drive at most one passenger at a time.
Given n and rides, return the maximum number of dollars you can earn by picking up the passengers optimally.
Note: You may drop off a passenger and pick up a different passenger at the same point.
\n\n\n
Example 1:
\n\n\nInput: n = 5, rides = [[2,5,4],[1,5,1]]\nOutput: 7\nExplanation: We can pick up passenger 0 to earn 5 - 2 + 4 = 7 dollars.\n\n\n
Example 2:
\n\n\nInput: n = 20, rides = [[1,6,1],[3,10,2],[10,12,3],[11,12,2],[12,15,2],[13,18,1]]\nOutput: 20\nExplanation: 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.\n\n
\n
Constraints:
\n\n1 <= n <= 1051 <= rides.length <= 3 * 104rides[i].length == 31 <= starti < endi <= n1 <= tipi <= 105Compiled with clang 11 using the latest C++ 17 standard.
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