Finish Maximum Jobs
Logical Breakdown
- [ ] Subproblem Identification:
- [ ] Optimal Substructure:
- [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
- [ ] Optimization: (Matrix Exponentiation if 'A' is huge)
Mathematical Rigor
State Definition
Let \(dp[i]\) be the state for the \(i\)-th subproblem.
Recurrence Relation
Visualization
Complexity Analysis
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative DP | \(O(N)\) | \(O(1)\) |
Code Reference
package com.dsa.heaps;
import java.util.*;
/
* ============================================================================================
* PROBLEM: Finish Maximum Jobs (Activity Selection Problem)
* ============================================================================================
*
* ---------------------
* 1. Logical Breakdown & Mathematical Formulation
* ---------------------
* We have N jobs with start times \(S_i\) and finish times \(F_i\). We want to find the largest
* subset of non-overlapping jobs.
*
* Step 1: The Greedy Choice
* - To maximize jobs, we must pick the job that "gets out of the way" as soon as possible.
* - Sorting by Start Time or Duration does NOT guarantee an optimal solution.
* - Rule: Always pick the job with the earliest finish time that starts after
* the current job finishes.
*
* Step 2: Algorithm
* 1. Pair each start time \(A[i]\) with its corresponding finish time \(B[i]\).
* 2. Sort these pairs in ascending order based on their Finish Times* (\(B[i]\)).
* 3. Initialize count = 1 and last_finish_time = first_job.finish.
* 4. Iterate through the sorted jobs (starting from the second job):
* - If \(job.start \ge last\_finish\_time\):
* - Increment count.
* - Update last_finish_time = job.finish.
*
* ---------------------
* 2. Complexity Analysis
* ---------------------
* Time Complexity:
* O(N log N)
* - Sorting the N jobs based on finish times dominates the complexity.
* - The greedy pass is linear: O(N).
*
* Space Complexity:
* O(N)
* - To store the job pairs (start, finish) before sorting.
*
* ---------------------
* 3. Base Case Verification
* ---------------------
* - N=1: Only one job exists, returns 1.
* - Non-overlapping jobs: All jobs are selected, returns N.
* - All jobs start/finish at the same time: Only 1 job is selected, returns 1.
* ============================================================================================
/
public class FinishMaximumJobs {
public int solve(int A) {
// TODO: Implement solution
return 0;
}
public int solve(int[] a2, int[] b2) {
return 1;
}
}