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.

Constraints

• 1 <= startTime.length == endTime.length == profit.length <= 5 * 104

• 1 <= startTime[i] < endTime[i] <= 109

• 1 <= profit[i] <= 104

Approach

Links

• Binarysearch

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

// Time Limit Error
class Solution {
    private class Job implements Comparable<Job> {
        int startTime;
        int endTime;
        int profit;
        
        Job(int startTime, int endTime, int profit) {
            this.startTime = startTime;
            this.endTime = endTime;
            this.profit = profit;
        }
        
        public int compareTo(Job otherJob) {
            int d = this.endTime-otherJob.endTime;
            return d == 0? (this.startTime-otherJob.startTime): d;
        }
    }
    
    public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
        int n = profit.length;
        Job[] jobs = new Job[n];
        
        for(int i = 0; i < n; i++) {
            jobs[i] = new Job(startTime[i], endTime[i], profit[i]);
        }
        Arrays.sort(jobs);
        
        int[] dp = new int[n];
        int maxProfit = 0;
        
        for(int i = 0; i < n; i++) {
            dp[i] = jobs[i].profit;
            for(int j = 0; j < i; j++) {
                if(jobs[j].endTime <= jobs[i].startTime) {
                    dp[i] = Math.max(dp[i], jobs[i].profit+dp[j]);
                }
            }
            maxProfit = Math.max(maxProfit, dp[i]);
        }
        
        return maxProfit;
    }
}

Solution 2

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    private class Job implements Comparable<Job> {
        int startTime;
        int endTime;
        int profit;
        
        Job(int startTime, int endTime, int profit) {
            this.startTime = startTime;
            this.endTime = endTime;
            this.profit = profit;
        }
        
        public int compareTo(Job otherJob) {
            return this.endTime-otherJob.endTime;
        }
    }
    
    public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
        int n = profit.length;
        Job[] jobs = new Job[n];
        
        for(int i = 0; i < n; i++) {
            jobs[i] = new Job(startTime[i], endTime[i], profit[i]);
        }
        Arrays.sort(jobs);
        
        int[] dp = new int[n];
        dp[0] = jobs[0].profit;
        
        for(int i = 1; i < n; i++) {
            dp[i] = Math.max(jobs[i].profit, dp[i-1]);
            for(int j = i-1; j >= 0; j--) {
                if(jobs[j].endTime <= jobs[i].startTime) {
                    dp[i] = Math.max(dp[i], jobs[i].profit+dp[j]);
                    break;
                }
            }
        }
        
        return dp[n-1];
    }
}

Follow up