When you’re dealing with multiple jobs that overlap and have associated profits, figuring out how to maximize the total profit can be tricky. The “Maximum Profit in Job Scheduling” problem is a classic dynamic programming challenge that requires a strategic approach to solve. In this article, we’ll break down the problem, explore different strategies, and walk through a Python implementation to maximize profits efficiently.

By the end of this guide, you’ll not only understand how to tackle this specific problem but also gain insights into how to approach similar dynamic programming problems.

**Problem Overview**

We are given `n`

jobs, where each job has a start time, end time, and a profit. Some jobs may overlap, so we need to find the maximum profit possible by selecting a subset of jobs that do not overlap.

**Example**

Let’s say we have the following jobs:

Job | Start Time | End Time | Profit |
---|---|---|---|

Job 1 | 1 | 3 | 50 |

Job 2 | 3 | 5 | 20 |

Job 3 | 6 | 19 | 100 |

Job 4 | 2 | 100 | 200 |

In this scenario, you cannot take both **Job 1** and **Job 4** because they **overlap**. The challenge is to select jobs that don’t overlap, maximizing the total profit.

For example, choosing Job 1 and Job 3 gives a total profit of 150. However, choosing Job 4 alone yields a profit of 200, which is higher.

**Approach and Solution**

The first thought might be to go for a **brute force** approach, exploring all possible combinations of jobs. However, that would lead to an exponential time complexity because you’d have to consider all subsets of jobs. A more efficient approach involves using **dynamic programming** with **memoization** and possibly **binary search** for optimization.

Here’s how you can approach the problem step by step:

**1. Sorting the Jobs**

To efficiently handle overlapping jobs, it’s a good idea to sort the jobs based on their start times. This allows us to make decisions sequentially, knowing which jobs come next and how they might affect our choices.

**2. Dynamic Programming with Memoization**

The core idea is to use dynamic programming (DP) to break down the problem into smaller subproblems. For each job, you have two choices:

- Include the job in the optimal set.
- Exclude the job and move to the next.

By memoizing the results of subproblems, we can avoid recalculating the same values repeatedly, which significantly improves efficiency.

**3. Binary Search for Efficiency**

When you decide to include a job, you want to find the next job that doesn’t overlap with it. Instead of scanning through all jobs linearly (which leads to **O(n²)** time complexity), you can use binary search to find the next non-overlapping job, reducing the complexity to **O(n log n)**.

**Code Implementation**

Below is the Python code to solve this problem using dynamic programming and binary search:

```
from bisect import bisect_left
def jobScheduling(startTime, endTime, profit):
# Step 1: Combine all job details into a list of tuples and sort by start time
jobs = sorted(zip(startTime, endTime, profit), key=lambda x: x[0])
# Extract the sorted arrays for easier access later
start_times = [job[0] for job in jobs]
# Step 2: Create a memoization table
n = len(jobs)
dp = [0] * (n + 1) # dp[i] will store the maximum profit for jobs[i:]
# Step 3: Iterate from the last job to the first one
for i in range(n - 1, -1, -1):
# Find the next job that doesn't overlap using binary search
next_job_index = bisect_left(start_times, jobs[i][1])
# Option 1: Take the current job and add its profit to the optimal solution of the next non-overlapping job
take_profit = jobs[i][2] + dp[next_job_index]
# Option 2: Skip the current job
skip_profit = dp[i + 1]
# Store the maximum of taking or skipping the current job
dp[i] = max(take_profit, skip_profit)
# The answer is the maximum profit starting from the first job
return dp[0]
# Example usage
start_times = [1, 2, 3, 4, 6]
end_times = [3, 5, 10, 6, 9]
profits = [50, 10, 40, 70, 60]
print(jobScheduling(start_times, end_times, profits)) # Output: 120
```

**Explanation of the Code**

**Data Preparation:**

We first combine the`startTime`

,`endTime`

, and`profit`

arrays into a list of tuples and sort them by`startTime`

. Sorting helps us efficiently determine which jobs can be taken next.**Binary Search with Bisect:**

The`bisect_left`

function from Python’s`bisect`

module is used to find the index of the first job that starts after the current job ends. This allows us to skip over overlapping jobs without checking them one by one.**Dynamic Programming Array:**

We maintain a`dp`

array where`dp[i]`

represents the maximum profit we can achieve starting from job`i`

. We compute this value by considering two options: taking the current job and adding the profit from the next valid job, or skipping the current job entirely.**Final Answer:**

The final result is stored in`dp[0]`

, which gives us the maximum profit achievable by scheduling non-overlapping jobs from the start.

**Time and Space Complexity**

**Time Complexity:**Sorting the jobs takes O(n log n). For each job, we perform a binary search which takes O(log n). Thus, the overall time complexity is O(n log n).**Space Complexity:**The space complexity is O(n) for the`dp`

array and the sorted job list.

**Conclusion**

The “Maximum Profit in Job Scheduling” problem is an excellent example of how sorting, binary search, and dynamic programming can be combined to solve complex problems efficiently. By breaking down the problem into smaller subproblems, using memoization, and optimizing the search process with binary search, we can handle larger inputs without running into performance issues.

Understanding this approach will not only help you solve this problem but also equip you with strategies for similar challenges involving scheduling, optimization, and dynamic programming.

If you found this article helpful, keep practicing and apply these techniques to other coding problems to reinforce your learning!