Introduction

In this blog post, we will explore how to solve the LeetCode problem "1665" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array tasks where tasks[i] = [actuali, minimumi]: actuali is the actual amount of energy you spend to finish the ith task. minimumi is the minimum amount of energy you require to begin the ith task. For example, if the task is [10, 12] and your current energy is 11, you cannot start this task. However, if your current energy is 13, you can complete this task, and your energy will be 3 after finishing it. You can finish the tasks in any order you like. Return the minimum initial amount of energy you will need to finish all the tasks. Example 1: Input: tasks = [[1,2],[2,4],[4,8]] Output: 8 Explanation: Starting with 8 energy, we finish the tasks in the following order: - 3rd task. Now energy = 8 - 4 = 4. - 2nd task. Now energy = 4 - 2 = 2. - 1st task. Now energy = 2 - 1 = 1. Notice that even though we have leftover energy, starting with 7 energy does not work because we cannot do the 3rd task. Example 2: Input: tasks = [[1,3],[2,4],[10,11],[10,12],[8,9]] Output: 32 Explanation: Starting with 32 energy, we finish the tasks in the following order: - 1st task. Now energy = 32 - 1 = 31. - 2nd task. Now energy = 31 - 2 = 29. - 3rd task. Now energy = 29 - 10 = 19. - 4th task. Now energy = 19 - 10 = 9. - 5th task. Now energy = 9 - 8 = 1. Example 3: Input: tasks = [[1,7],[2,8],[3,9],[4,10],[5,11],[6,12]] Output: 27 Explanation: Starting with 27 energy, we finish the tasks in the following order: - 5th task. Now energy = 27 - 5 = 22. - 2nd task. Now energy = 22 - 2 = 20. - 3rd task. Now energy = 20 - 3 = 17. - 1st task. Now energy = 17 - 1 = 16. - 4th task. Now energy = 16 - 4 = 12. - 6th task. Now energy = 12 - 6 = 6. Constraints: 1 <= tasks.length <= 105 1 <= actual​i <= minimumi <= 104

Explanation

Here's the solution to the problem:

• Key Idea: The optimal order of tasks is determined by the difference between the minimum energy required and the actual energy spent. Tasks with a larger difference (minimumi - actuali) should be done earlier. This is because doing these tasks early frees up more energy for later tasks.

• Algorithm: Sort the tasks in descending order of the difference (minimumi - actuali). Iterate through the sorted tasks, keeping track of the current energy level. If the current energy is insufficient to start a task, increase the initial energy to cover the deficit.

• Runtime & Storage Complexity:

  • Runtime: O(n log n) due to the sorting step.
  • Storage: O(1) excluding the input array as we're sorting in place. If we are not allowed to sort in place then the storage complexity would be O(n).

Code

    def minimum_initial_energy(tasks):
    """
    Calculates the minimum initial energy needed to complete all tasks.

    Args:
        tasks: A list of tasks, where each task is a list [actual_energy, min_energy].

    Returns:
        The minimum initial energy required.
    """

    # Sort tasks in descending order of (minimum_energy - actual_energy)
    tasks.sort(key=lambda x: x[1] - x[0], reverse=True)

    initial_energy = 0
    current_energy = 0

    for actual, minimum in tasks:
        if current_energy < minimum:
            initial_energy += (minimum - current_energy)
            current_energy = minimum
        current_energy -= actual

    return initial_energy