Gain insights into dynamic programming, explore recursion basics, and delve into advanced techniques like Bottom-Up optimization. Discover ways to solve complex problems more efficiently with hands-on coding challenges.

Dynamic programming is something every developer should have in their toolkit. It allows you to optimize your algorithm with respect to time and space — a very important concept in real-world applications.

In this course, you’ll start by learning the basics of recursion and work your way to more advanced DP concepts like Bottom-Up optimization. Throughout this course, you will learn various types of DP techniques for solving even the most complex problems. Each section is complete with coding challenges of varying difficulty so you can practice a wide range of problems.

By the time you’ve completed this course, you will be able to utilize dynamic programming in your own projects.

Dynamic Programming in Python: Optimizing Programs for Efficiency

# Problem statement

You have one auditorium, and only one class can take place at the time; however, you have different options for these classes. Each class, *c*, is characterized by its start time, end time, and total utility we can get out of this class. The utility here could be the number of attendants of the class; the more attendants, the more money you can make as an auditorium owner. Here is your problem statement:


Given *n* number of classes, you have to find a schedule that maximizes the utility of an auditorium.

# Input

A list of classes, where each class is a tuple of three numbers. The first number denotes the start time, the second number denotes the end time, and the third number denotes utility value.
 

```
schedule = [(0,2,25), (1,5,40), (6,8,170), (3,7,220)]
```
Note that start and end time are strictly increasing integers.

# Output
Your algorithm will return an integer, denoting the maximum possible utility achievable from the given schedule.  

```
WeightedSchedule(schedule) = 245
```

# Coding challenge

You might find the function `lastConflict()` useful. Given the index of a job and a list of jobs, it finds the last job that does not conflict with the current job given by the index.

Jot down a few examples and try to solve them manually to get a hang of the problem, then build the solution. This is a slightly more difficult problem, so take your time. Best of luck!

In this lesson, you will solve the classic problem of weighted scheduling using dynamic programming.

Chapter 1: From Recursion to Dynamic Programming

Chapter 2: Top-Down Dynamic Programming with Memoization

Chapter 3: Bottom-Up Dynamic Programming with Tabulation

Chapter 4: Practice Problems

Challenge: Weighted Scheduling Problem

Problem statement