Program to find maximum number of courses we can take based on interval time in Python?

Suppose we have a list of intervals in the form [start, end], representing the start and end time of courses. We need to find the maximum number of courses we can take, assuming we can only take one course at a time and the start of a course must be after the end of the previous course.

So, if the input is like times = [[3, 6], [6, 9], [7, 8], [9, 11]], then the output will be 3, as we can take courses [[3, 6], [7, 8], [9, 11]].

Algorithm

To solve this, we will follow these steps ?

• Sort intervals based on ending time
• Initialize counter = 0 and end = -1
• For each interval, if start time is greater than the last end time, select it
• Update counter and end time

Implementation

class Solution:
    def solve(self, times):
        times.sort(key=lambda x: x[1])
        
        counter = 0
        end = -1
        
        for i in range(len(times)):
            if times[i][0] > end:
                counter += 1
                end = times[i][1]
        return counter

ob = Solution()
times = [
    [3, 6],
    [6, 9], 
    [7, 8],
    [9, 11]
]
print(ob.solve(times))

3

How It Works

The algorithm uses a greedy approach ?

1. Sort by end time: [[3, 6], [7, 8], [6, 9], [9, 11]]
2. Select [3, 6]: counter = 1, end = 6
3. Select [7, 8]: 7 > 6, so counter = 2, end = 8
4. Skip [6, 9]: 6 < 8, overlaps with previous
5. Select [9, 11]: 9 > 8, so counter = 3, end = 11

Alternative Approach

Here's a more readable version using direct iteration ?

def max_courses(times):
    # Sort by end time
    times.sort(key=lambda x: x[1])
    
    courses_taken = 0
    last_end_time = -1
    
    for start, end in times:
        if start > last_end_time:
            courses_taken += 1
            last_end_time = end
    
    return courses_taken

# Test the function
times = [[3, 6], [6, 9], [7, 8], [9, 11]]
result = max_courses(times)
print(f"Maximum courses: {result}")

Maximum courses: 3

Conclusion

This problem is solved using a greedy algorithm by sorting intervals by end time and selecting non-overlapping courses. The time complexity is O(n log n) due to sorting, and space complexity is O(1).