# 1235. Maximum Profit in Job Scheduling

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 startTimeendTime and profit arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.

If you choose a job that ends at time X you will be able to start another job that starts at time X.

Example 1:

Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
Output: 120
Explanation: The subset chosen is the first and fourth job.
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.

Example 2:

Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
Output: 150
Explanation: The subset chosen is the first, fourth and fifth job.
Profit obtained 150 = 20 + 70 + 60.

Example 3:

Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
Output: 6

Constraints:

• 1 <= startTime.length == endTime.length == profit.length <= 5 * 104
• 1 <= startTime[i] < endTime[i] <= 10^9
• 1 <= profit[i] <= 10^4

class Solution {
public:
int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
vector<vector<int>> jobs;
map<int, int> dp{{0, 0}};
for (int i = 0; i < startTime.size(); ++i) {
jobs.push_back({endTime[i], startTime[i], profit[i]});
}
sort(jobs.begin(), jobs.end());
for (auto &job : jobs) {
int cur = prev(dp.upper_bound(job[1]))->second + job[2];
if (cur > dp.rbegin()->second) dp[job[0]] = cur;
}
return dp.rbegin()->second;
}
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1235

Maximum Earnings From Taxi

https://leetcode.com/problems/maximum-profit-in-job-scheduling/

https://leetcode.com/problems/maximum-profit-in-job-scheduling/discuss/409188/C%2B%2B-with-picture

https://leetcode.com/problems/maximum-profit-in-job-scheduling/discuss/409009/JavaC%2B%2BPython-DP-Solution

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