Minimum Initial Energy to Finish Tasks

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ArraysGreedy Algorithms

You are given an array tasks where tasks[i] = [actual_i, minimum_i]:

actual_i is the actual amount of energy you spend to finish the i^th task.
minimum_i is the minimum amount of energy you require to begin the i^th 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 <= minimum_i <= 10⁴

Try coding it in LeetCode

Minimum Initial Energy to Finish Tasks

Hard

Asked by:

11 views

Topics:

ArraysGreedy Algorithms

You are given an array tasks where tasks[i] = [actual_i, minimum_i]:

actual_i is the actual amount of energy you spend to finish the i^th task.
minimum_i is the minimum amount of energy you require to begin the i^th 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 <= minimum_i <= 10⁴

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the `actualEnergy` and `requiredEnergy` values for each task? Can they be negative, zero, or only positive integers?
What is the maximum number of tasks I should expect in the input array?
If it's impossible to finish all tasks with any initial energy, what should I return? Should I return -1, throw an exception, or is there a specific error code?
Are the tasks independent, or does completing one task affect the energy required for subsequent tasks?
If multiple initial energy values are sufficient to complete all tasks, should I return the minimum, maximum, or any valid value?

Brute Force Solution

Approach

The goal is to figure out the smallest amount of energy you need to start with to complete a series of tasks. The brute force method involves trying every possible starting energy level until we find one that works.

Here's how the algorithm would work step-by-step:

Start by guessing a low initial energy level.
Go through the tasks one by one, subtracting the energy required for each task from your current energy level.
If at any point your energy level drops below zero, the initial energy level you guessed was too low.
Increase your guessed initial energy level and try again.
Keep increasing the initial energy and repeating the process until you find a starting energy level that allows you to complete all the tasks without your energy dropping below zero at any point.
The first initial energy level that allows you to complete all tasks is the minimum initial energy required.

Code Implementation

def minimum_initial_energy_brute_force(tasks):
    initial_energy_guess = 0

    while True:
        current_energy = initial_energy_guess
        task_completion_possible = True

        # Iterate through each task to check energy sufficiency
        for task_energy_requirement in tasks:

            # If current energy dips below zero, we need more initial
            if current_energy < task_energy_requirement:
                task_completion_possible = False
                break

            current_energy -= task_energy_requirement

        # We found the minimum energy.
        if task_completion_possible:
            return initial_energy_guess

        # If initial energy insufficient, increment and retry.
        initial_energy_guess += 1

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm involves an outer loop that iteratively increments the guessed initial energy (let's say it goes up to m). For each guess in the outer loop, the inner loop iterates through all n tasks to simulate the energy consumption. In the worst-case scenario, m could be proportional to the sum of the task energies or a similar value dependent on the task values. Thus, the time complexity is O(n*m), where n is the number of tasks and m is the maximum initial energy tested.

Space Complexity

O(1) – The provided algorithm attempts different initial energy levels and simulates the task execution. It doesn't create any auxiliary data structures like lists, arrays, or hash maps to store intermediate results or track visited states. The algorithm primarily uses a constant number of variables to keep track of the current energy level and whether the current initial energy level is sufficient. Therefore, the space complexity is O(1), representing constant extra space, independent of the number of tasks (N).

Optimal Solution

Approach

To find the minimum starting energy needed, we'll work backward, focusing on the trickiest tasks first. We want to make sure we can complete all tasks without running out of energy, even if it means starting with a bit more than the energy each task needs individually.

Here's how the algorithm would work step-by-step:

Imagine you're doing the tasks in reverse order. This helps see which tasks are most likely to cause you to run out of energy.
As you go through the tasks in reverse, keep track of the lowest energy you'd have at any point if you started with zero energy and did the tasks backwards.
Convert this negative lowest energy into the positive value for calculating the initial energy needed.
Consider the needed value from the previous step along with the cost and actual energy needed for the task at hand and calculate a new value. Keep choosing the higher value of the two.
When you reach the first task, the value you have is the minimum initial energy you must start with to successfully complete all tasks in the correct order.

Code Implementation

def minimum_initial_energy(tasks):
    minimum_energy_needed = 0
    current_energy = 0

    for cost, required in reversed(tasks):

        if current_energy < required:
            # Need to increase initial energy to complete this task.
            minimum_energy_needed += (required - current_energy)
            current_energy = required

        current_energy -= cost

    # Ensures we start with enough energy to never go negative.
    if current_energy < 0:
        minimum_energy_needed -= current_energy

    return minimum_energy_needed

Big(O) Analysis

Time Complexity

O(n) – The provided strategy involves iterating through the 'tasks' list once, but in reverse order, to simulate completing tasks backward and calculate the minimum initial energy required. Inside the loop, operations like tracking the minimum energy and updating the initial energy needed take constant time. Since we perform constant-time operations for each of the n tasks, the overall time complexity is directly proportional to n.

Space Complexity

O(1) – The algorithm operates primarily with scalar variables to track the minimum energy needed. No auxiliary data structures that scale with the input size (N, representing the number of tasks) are created or used. The algorithm keeps track of the lowest energy level and the initial energy, both of which are stored in constant space. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty tasks array	Return 0 as no energy is needed to complete no tasks.
Null tasks array	Throw an IllegalArgumentException to indicate invalid input.
Tasks with extremely large values leading to integer overflow	Use long data type to store energy and task values to prevent overflow.
Tasks array with a single task [cost, required]	Return the required energy for that single task, Math.max(cost, required).
Tasks where cost is always greater than required energy.	The minimum initial energy is equal to the maximum cost value in the tasks array
Tasks where required energy is always greater than cost.	The minimum initial energy needed is the sum of energy differences (required - cost) plus the largest cost.
Tasks are sorted in increasing order of 'required' energy	The standard algorithm should still work correctly, iterating through the sorted tasks adjusting current energy.
Tasks array contains negative numbers for cost or required energy.	Throw an IllegalArgumentException as cost and required energy are expected to be non-negative.

Empty tasks array

How to Handle:

Return 0 as no energy is needed to complete no tasks.

Null tasks array

How to Handle:

Throw an IllegalArgumentException to indicate invalid input.

Tasks with extremely large values leading to integer overflow

How to Handle:

Use long data type to store energy and task values to prevent overflow.

Tasks array with a single task [cost, required]

How to Handle:

Return the required energy for that single task, Math.max(cost, required).

Tasks where cost is always greater than required energy.

How to Handle:

The minimum initial energy is equal to the maximum cost value in the tasks array

Tasks where required energy is always greater than cost.

How to Handle:

The minimum initial energy needed is the sum of energy differences (required - cost) plus the largest cost.

Tasks are sorted in increasing order of 'required' energy

How to Handle:

The standard algorithm should still work correctly, iterating through the sorted tasks adjusting current energy.

Tasks array contains negative numbers for cost or required energy.

How to Handle:

Throw an IllegalArgumentException as cost and required energy are expected to be non-negative.