# 5 Best Ways to Program to Find the Number of Tasks That Can Be Finished with Given Conditions in Python

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π‘ Problem Formulation: The task at hand is to develop a Python program that can calculate the number of tasks that can be completed given a set of specific preconditions, such as time limits, dependencies, and resource constraints. For instance, input could be a list of tasks with their respective durations and a total available time; the desired output is the maximum number of tasks that can be finished within the given time frame.

## Method 1: Using a Greedy Algorithm

In this method, we use a greedy algorithm to select tasks that can be finished within the given conditions. It involves sorting the tasks based on a certain heuristic (like shortest duration first) and then iteratively selecting tasks until the conditions no longer allow for more to be completed. This approach works well when tasks are independent.

Here’s an example:

```def max_tasks(tasks, time):
count = 0
count += 1
else:
break
return count

# Example usage:
tasks = [3, 1, 2, 1, 2]
available_time = 5

Output: 4

This code snippet defines a function `max_tasks()` that takes a list of task durations and the total available time. The tasks are sorted by duration, and the function then loops through them, selecting each task that can be completed within the remaining time.

## Method 2: Using Dynamic Programming

The dynamic programming method solves this problem by constructing a table that represents the maximum number of tasks that can be completed given a certain amount of time. This approach is well-suited for problems where tasks may have dependencies or when the selection of one task influences the options for subsequent tasks.

Here’s an example:

```def max_tasks_dp(tasks, time):
# Initialize the DP table
dp = [[0] * (time + 1) for _ in range(len(tasks) + 1)]
# Fill the DP table
for i in range(1, len(tasks) + 1):
for j in range(1, time + 1):
dp[i][j] = max(dp[i-1][j], dp[i-1][j-tasks[i-1]] + 1)
else:
dp[i][j] = dp[i-1][j]

# Example usage:
tasks = [3, 1, 2, 1, 2]
available_time = 5

Output: 4

This code defines `max_tasks_dp()` which uses a 2-dimensional list (DP table) to store the solutions to subproblems. It returns the value in the table that represents the maximum number of tasks that can be completed in the available time.

## Method 3: Using Backtracking

Backtracking provides a way to find all possible combinations of tasks and select the combination that maximizes the number of tasks within the given constraints. It is particularly useful when the tasks are interdependent, or we need to consider multiple constraints simultaneously.

Here’s an example:

```def max_tasks_backtracking(tasks, time, index=0, count=0):
if time <= 0 or index == len(tasks):
return count
else:

# Example usage:
tasks = [3, 1, 2, 1, 2]
available_time = 5

Output: 4

This code uses a backtracking function `max_tasks_backtracking()` which either includes the current task and moves onto the next with decreased time or skips to the next task. The maximum of these two options continues recursively.

## Method 4: Using Linear Programming

Linear programming (LP) is a mathematical technique for maximizing or minimizing a linear function subject to constraints. While typically used for continuous variables, you can use LP for discrete problems like this with the appropriate formulation and can be powerful when tasks have complex dependencies.

Here’s an example:

```from scipy.optimize import linprog

c = [-1] * len(tasks)  # The coefficients of the linear objective function
A = [tasks]  # The inequality constraint matrix
b = [time]  # The inequality constraint vector
bounds = [(0, 1) for _ in tasks]  # Bounds for variables (0 or 1)

res = linprog(c, A_ub=A, b_ub=b, bounds=bounds, method='simplex')
return int(-res.fun)

# Example usage:
tasks = [3, 1, 2, 1, 2]
available_time = 5

Output: 4

The `max_tasks_lp()` function utilizes the linear programming capabilities of Scipy to solve the task allocation problem. It sets up the constraint matrix, objective function, and bounds for each variable to indicate whether a task is selected (1) or not (0), then solves the problem.

## Bonus One-Liner Method 5: Using List Comprehensions and the itertools Module

You can use Python’s list comprehensions combined with the `itertools` module to generate all possible combinations of tasks and select the one that maximizes number of tasks without exceeding the available time. It is a much more compact but inefficient method best suited for small lists of tasks.

Here’s an example:

```from itertools import combinations

return max([sum(1 for _ in combo) if sum(combo) <= time else 0

# Example usage:
tasks = [3, 1, 2, 1, 2]
available_time = 5
This one-liner function `max_tasks_one_liner()` creates all possible combinations of tasks (using `combinations()`) then checks each combination to see if its total time is under the limit, and counts the number of tasks in it. Then it selects the combination with the highest count.