5 steps to solve any Dynamic Programming problem

Summary

In this video, the presenter explains dynamic programming (DP), a problem-solving technique often found in coding interviews, and emphasizes its difficulty and importance. Using the Longest Increasing Subsequence (LIS) problem, the presenter outlines the steps to solve DP problems, starting with making assumptions and progressing to recursive rules and base cases. Through this process, they demonstrate how assumptions help solve DP problems efficiently, especially by focusing on subsequences. The video aims to simplify DP for viewers, offering insights into tackling common interview challenges and offering hope to those struggling with coding problems.

Takeaways

• 😀 Dynamic Programming (DP) is often considered one of the hardest topics in coding interviews because it is both challenging and frequently asked.
• 😀 DP is not an algorithm, but a problem-solving technique, commonly used to break problems into manageable subproblems.
• 😀 The first step in solving DP problems is to make an assumption, rather than breaking the problem into subproblems right away.
• 😀 In the Longest Increasing Subsequence (LIS) problem, the assumption made is that we already know the LIS for all numbers except the last one.
• 😀 Understanding how to expand a subsequence by adding the last number is crucial for solving DP problems like LIS.
• 😀 In DP, the original problem and the assumed problem should differ in some way for the technique to work effectively.
• 😀 The key to solving LIS is knowing where a subsequence ends and whether it can be extended.
• 😀 The recursive rule in DP is used to formalize the process of expanding subsequences and updating their lengths.
• 😀 The base case in DP is often initializing the solution with a simple value, such as 1 for the length of the LIS of any single number.
• 😀 In LIS, we need to compute the LIS for each index in the array, considering all previous indices that can extend the current subsequence.
• 😀 To implement the solution, use loops to calculate the LIS for every index, then return the maximum subsequence length from the array.

Q & A

Why is dynamic programming (DP) considered one of the hardest algorithms in coding interviews?

DP is considered difficult because it requires a deep understanding of problem-solving techniques, often involving breaking problems into subproblems. Additionally, DP problems are frequently asked in interviews, adding to their perceived difficulty.

Is dynamic programming an algorithm or a technique?

Dynamic programming is not an algorithm; it's a problem-solving technique. It helps break complex problems into simpler subproblems that can be solved independently and then combined to solve the larger problem.

What makes the Longest Increasing Subsequence (LIS) problem a good example for explaining DP?

The LIS problem is a balanced example—it is not too hard nor too easy, allowing for a thorough exploration of DP concepts. It also has an interesting history for the speaker, making it relatable for the audience.

What assumption is crucial for solving DP problems like the LIS problem?

The key assumption in DP is that you already know the answer to a similar but smaller problem. For the LIS problem, this means knowing the length of the longest increasing subsequence for all numbers except the last one.

How do you expand a subsequence in the LIS problem?

To expand a subsequence, you compare the current number to the previous numbers in the array. If the number is larger than a previous number, the subsequence can be extended by adding this number, increasing its length.

Why is it important to know where a subsequence ends in DP problems like LIS?

Knowing where a subsequence ends is essential because it helps determine whether the subsequence can be expanded by adding the next number. This is a critical step in constructing longer subsequences.

How do you ensure that you can find the LIS efficiently using DP?

By storing the length of the longest increasing subsequence at every index in the array, you can efficiently calculate the LIS. You iterate through each element and compare it with previous elements to update the subsequence length.

What is the base case for the LIS problem in DP?

The base case for the LIS problem is that the length of the longest increasing subsequence ending at each index is initially 1, since each number by itself can be seen as an increasing subsequence of length 1.

What does the recursive rule for the LIS problem involve?

The recursive rule involves iterating through all the previous indices for each number in the array. If the number at a previous index is smaller than the current number, the subsequence can be extended, and its length is updated.

How is the final answer to the LIS problem derived?

The final answer is the maximum length found in the array of subsequence lengths. Since the LIS can end at any index, the longest subsequence will be the maximum value in the array storing subsequence lengths.