Recap of key concepts
Lesson 14 of 15 in Coddy's Dynamic Programming 101 course.
In this lesson, we will review the key concepts covered in the Dynamic Programming 101 course.
- Dynamic Programming: Dynamic programming is a technique for solving optimization problems by breaking them down into simpler subproblems and solving each subproblem only once.
- Memoization: Memoization is a technique for avoiding redundant calculations by storing the results of expensive function calls and returning the cached result when the same inputs occur again.
- Tabulation: Tabulation is a technique for solving dynamic programming problems by iteratively filling out a table or array of solutions, building up to the final solution.
- Optimal substructure: A problem exhibits optimal substructure if the optimal solution to the problem contains within it optimal solutions to its subproblems.
- Overlapping subproblems: A problem exhibits overlapping subproblems if it can be broken down into subproblems that share sub-subproblems.
- Space optimization: Space optimization is a technique for reducing the memory requirements of a dynamic programming algorithm by only keeping track of the necessary state.
- Bit masking: Bit masking is a technique for using bit manipulation to represent a set of elements as a binary number.
- Pruning: Pruning is a technique used to reduce the number of computations required by a dynamic programming algorithm by avoiding unnecessary computations.
Challenge
MediumRecap Challenge: minimum cost path
You are given an n x n grid representing a map of the city. Each cell in the grid represents a street intersection, and the values in the cells represent the cost of traversing that intersection. You want to travel from the top-left corner of the grid to the bottom-right corner, and you can only move down or right at each intersection.
Write a function min_cost_path(grid) that takes in the grid as an input and returns the minimum cost of traversing the grid from the top-left to the bottom-right corner.
For example:
grid = [
[1, 3, 1],
[1, 5, 1],
[4, 2, 1]
]
min_cost_path(grid) => 7Explanation: The minimum cost path is 1 -> 3 -> 1 -> 1 -> 1, which has a total cost of 7.
Try it yourself
def min_cost_path(grid):
# Write code hereAll lessons in Dynamic Programming 101
3Dynamic programming algorithms
Longest common subsequenceKnapsack problemCoin change problemEdit distance