Climbing Stairs - Dynamic Programming - Leetcode 70 - Python
NeetCode
18 min, 8 sec
An in-depth explanation of solving the Climbing Stairs problem using dynamic programming.
Summary
- The video explains solving the Climbing Stairs problem, which is a basic dynamic programming challenge.
- The presenter explains the problem statement, demonstrates how to approach it with brute force, and then optimizes with memoization.
- The Climbing Stairs problem is related to the Fibonacci sequence, and the presenter shows how to solve it using a bottom-up approach.
- The final solution requires only a linear time complexity and minimal space, achieving an optimized algorithm.
Chapter 1
Introducing the Climbing Stairs dynamic programming problem and its relevance to the Blind 75 questions list.
- The presenter introduces the topic of writing neat code and today's focus on the Climbing Stairs problem.
- The Climbing Stairs problem is described as an easier dynamic programming problem compared to others previously solved.
- It is a good problem for understanding the fundamentals of dynamic programming, brute force, memoization, and true dynamic programming solutions.
- The problem is part of the Blind 75 questions, which the presenter is tracking on a spreadsheet.
Chapter 2
A clear explanation of the Climbing Stairs problem statement and the goal to find the number of unique ways to reach the top of a staircase.
- The Climbing Stairs problem involves climbing a staircase with 'n' steps to reach the top.
- The challenge is to find how many distinct ways there are to reach the top by taking either one step or two steps at a time.
- Examples are given for n=2 and n=3, demonstrating the different paths that can be taken to reach the top of the staircase.
Chapter 3
Visualizing the problem with a decision tree and identifying the brute force approach's inefficiencies.
- The problem is further explained by visualizing a decision tree that represents possible steps at each stage of climbing the staircase.
- The presenter illustrates that for n=5, there are several paths to take and each decision leads to a subtree with more decisions.
- It is shown that many subtrees are repeated in the decision tree, which is inefficient in a brute force approach.
Chapter 4
Introducing memoization to optimize the brute force solution by storing and reusing results of subproblems.
- The presenter introduces memoization as a technique to store the results of subproblems to prevent recalculating them, optimizing the brute force approach.
- The idea is to cache the results of the subproblems in a dynamic programming (DP) table or array.
- By storing solutions to subproblems, the time complexity is reduced from exponential to linear.
Chapter 5
Demonstrating the bottom-up approach to dynamic programming for solving the Climbing Stairs problem.
- The presenter explains the bottom-up dynamic programming approach, which starts with solving the base case and works up to the original problem.
- A DP array is used to store the results of subproblems, starting with the base case where the number of ways to reach the last step is one.
- The values in the DP array are filled by summing the results of the next two subproblems until the result for the starting point is found.
- This technique aligns with the Fibonacci sequence, where each number is the sum of the two preceding ones.
Chapter 6
Reducing space complexity by eliminating the need for a full DP array and using two variables instead.
- The presenter shows that since each DP value only depends on the two subsequent values, there is no need for a full DP array.
- By using two variables instead of an array, the space complexity is optimized while maintaining a linear time complexity.
- The two variables are updated iteratively, shifting along the sequence until the final result is reached.
Chapter 7
The presenter walks through the efficient code solution for the Climbing Stairs problem.
- The final code solution is presented, which consists of only a few lines due to the optimizations made.
- Two variables are initialized to one, representing the base cases, and a loop updates these variables n-1 times to compute the result.
- The final value of one of the variables, after completing the loop, is returned as the total number of ways to climb the stairs.
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