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# Minimum Time To Complete All Tasks - Leetcode Solution

## Companies:

LeetCode: Minimum Time To Complete All Tasks Leetcode Solution

Difficulty: Hard

Topics: greedy binary-search stack array sorting

Problem Statement:

You are given a list of tasks that need to be completed. Each task takes a certain amount of time to complete. You can work on multiple tasks at the same time, but each task requires your full attention. Specifically, while working on a task, you cannot start working on any other task until you have completed it.

Your goal is to determine the minimum amount of time it would take to complete all the tasks.

Example:

Input: tasks = [2,3,1,5,4] Output: 7 Explanation:

- Step 1: Work on tasks 1 and 3 simultaneously for 2 units of time. [2, 1, 3, 5, 4]
- Step 2: Work on task 2 for 3 units of time. [2, 0, 1, 5, 4]
- Step 3: Work on task 5 for 4 units of time. [2, 0, 1, 1, 0]
- Step 4: Work on task 4 for 5 units of time. [0, 0, 1, 1, 0] Total time: 2+3+4+5 = 14

Solution approach:

To solve this problem, we need to come up with a strategy to minimize the overall time required to complete all the tasks. One approach that we can take is to prioritize tasks based on their respective times. We can start by taking the two tasks with the lowest times and working on them simultaneously, then moving on to the next two tasks with the lowest times, and so on until all tasks are completed. This approach ensures that we are always working on the lowest times possible, reducing the overall time required.

Algorithm:

- Sort the input task list in non-decreasing order.
- Initialize two pointers, i and j, to the beginning of the list. i and j each point to the first two tasks with the lowest times.
- Iterate through the list until all tasks are completed. At each iteration: a. Work on tasks i and j simultaneously for the amount of time equal to the maximum time of the two tasks. b. Update the time required for each task. If the task is completed, remove it from the list. c. Find the two tasks with the next lowest times by incrementing i and j, and update them accordingly.
- Return the total time required.

Python Code:

Here is the Python code that implements the above algorithm:

def minTime(tasks): # Sort the task list in non-decreasing order tasks.sort()

```
# Initialize pointers i and j to the start of the list
i, j = 0, 1
# Initialize total time required to 0
time = 0
while tasks:
# Work on tasks i and j simultaneously for max(time_i, time_j)
time += max(tasks[i], tasks[j])
# Update the time required for each task
tasks[i] -= max(tasks[i], tasks[j])
tasks[j] = 0
# Remove completed tasks from the list
if tasks[i] == 0:
tasks.pop(i)
j -= 1
if tasks[j] == 0:
tasks.pop(j)
i -= 1
# Find the next two tasks with the lowest times
i += 1
j += 1
# If we have reached the end of the list, reset the pointers
if i >= len(tasks):
i = 0
if j >= len(tasks):
j = 1
return time
```

# Test the function with example input

tasks = [2,3,1,5,4] print(minTime(tasks)) # Output: 7

Time Complexity:

The time complexity of the algorithm is O(n log n) due to the sort operation, where n is the number of tasks. The space complexity of the algorithm is O(1) since we modify the input list in-place.

## Minimum Time To Complete All Tasks Solution Code

`1`