1235.Maximum Profit in Job Scheduling

tags: Hard,Array,DP,Sorting,Binary Search

1235. Maximum Profit in Job Scheduling

題目描述

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, 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.

範例

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 <=
  $5\ast {10}^4$
• 1 <= startTime[i] < endTime[i] <=
  ${10}^9$
• 1 <= profit[i] <=
  ${10}^4$

解答

C++

struct Job {
    int start, end, profit;
    bool operator<(const Job &rhs) const {return end < rhs.end;}
};

class Solution {
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n = startTime.size();
        vector<Job> jobs(n);
        for (int i = 0; i < n; i++) {
            jobs[i].start = startTime[i];
            jobs[i].end = endTime[i];
            jobs[i].profit = profit[i];
        }
        sort(jobs.begin(), jobs.end());

        vector<int> dp(n+1, 0);
        int best = 0;
        for (int i = 0; i < n; i++) {
            auto job = jobs[i];
            Job fake;
            fake.end = job.start + 1;
            int k = lower_bound(jobs.begin(), jobs.end(), fake) - jobs.begin();
            int value = dp[k] + job.profit;
            if (value > best) {
                dp[i+1] = value;
                best = value;
            } else {
                dp[i+1] = best;
            }
        }
        return best;
    }
};

Yen-Chi ChenSun, Nov 26, 2022 10:08 PM

Python

class Solution:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        jobs = sorted(zip(startTime, endTime, profit), key=lambda job: job[1])

        best = 0
        dp = [[0, best]]
        for start, end, profit in jobs:
            k = bisect.bisect(dp, [start+1])
            v = dp[k-1][1] + profit
            if v > best:
                best = v
                dp.append([end, best])
        return best

Yen-Chi ChenSun, Nov 26, 2022 10:08 PM

Time:

$O(n\log n)$
Extra Space:

$O(n)$

Reference

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