1235. Maximum Profit in Job Scheduling

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1235. Maximum Profit in Job Scheduling

Description

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.

Example 1:

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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:

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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:

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

Constraints:

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

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    vector<int> dp;

    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n = startTime.size();
        dp.resize(n, -1);
        vector<array<int, 3>> jobs;
        for (int i = 0; i < n; i++) {
            jobs.push_back({startTime[i], endTime[i], profit[i]});
        }
        sort(jobs.begin(), jobs.end());
        int res = dfs(0, jobs);
        return res;
    }

    int dfs(int index, vector<array<int, 3>>& jobs) {
        if (index == jobs.size()) {
            return 0;
        }
        if (dp[index] != -1) {
            return dp[index];
        }
        int& res = dp[index];
        // first choice, lets just not pick this item, and continue to the next one
        res = dfs(index + 1, jobs);
        // second option, pick this one, then we need to check the minimum item to start next
        int end = jobs[index][1];
        int left = index, right = jobs.size();
        while (left + 1 < right) {
            int mid = (left + right) / 2;
            if (jobs[mid][0] >= end) {
                right = mid;
            } else {
                left = mid;
            }
        }
        res = max(res, jobs[index][2] + dfs(right, jobs));
        return res;
    }
};