How do you solve Minimum Time to Complete All Tasks on LeetCode?

Minimum Time to Complete All Tasks (LeetCode #2589) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting tasks by end time allows for efficient scheduling.

What is the brute force solution for Minimum Time to Complete All Tasks?

The brute force approach for Minimum Time to Complete All Tasks is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible time point to see if a task can be scheduled at that time. It runs through the entire range of time for each task, which can be inefficient but is straightforward to understand. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Time to Complete All Tasks?

Minimum Time to Complete All Tasks uses the following concepts: Array, Binary Search, Stack, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Minimum Time to Complete All Tasks?

Always clarify the problem constraints and requirements. Think about edge cases, such as overlapping tasks. Explain your thought process clearly while coding.

What are common mistakes when solving Minimum Time to Complete All Tasks?

Not considering idle time between tasks. Failing to sort tasks before processing.

#2589

Minimum Time to Complete All Tasks

Hard

ArrayBinary SearchStackGreedySortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

The optimal solution sorts the tasks by their end times and uses a greedy approach to allocate the required durations efficiently. This minimizes the total time the computer needs to be on by ensuring tasks are scheduled in the most compact way possible.

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Algorithm

3 steps

1Step 1: Sort the tasks by their end time.
2Step 2: Initialize a variable to track the current time and another for the total active time.
3Step 3: For each task, check if it can fit into the available time slots and allocate the required duration, updating the total active time accordingly.

solution.py13 lines

1# Full working Python code
2
3def minTime(tasks):
4    tasks.sort(key=lambda x: x[1])
5    total_time = 0
6    current_time = 0
7    for start, end, duration in tasks:
8        if current_time < start:
9            current_time = start
10        available_time = min(end, current_time + duration) - current_time
11        total_time += available_time
12        current_time += available_time
13    return total_time

ℹ

Complexity note: The sorting step takes O(n log n) time, and the subsequent iteration through the tasks takes O(n), leading to an overall complexity dominated by the sort operation.

1Sorting tasks by end time allows for efficient scheduling.
2Greedy allocation of time minimizes total active time.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.