Spirit World Wisdom Pyramid - codepath/compsci_guides GitHub Wiki

Unit 12 Session 1 Advanced (Click for link to problem statements)

Problem Highlights

  • 💡 Difficulty: Easy
  • ⏰ Time to complete: 15 mins
  • 🛠️ Topics: Dynamic Programming, Pascal's Triangle

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

  • Established a set (2-3) of test cases to verify their own solution later.
  • Established a set (1-2) of edge cases to verify their solution handles complexities.
  • Have fully understood the problem and have no clarifying questions.
  • Have you verified any Time/Space Constraints for this problem?

  • What is the goal of the problem?

    • The goal is to return the wisdomLevel-th row of the Wisdom Pyramid where each row is built by combining the two numbers directly above it from the previous row.

  • What are the base cases?

    • The first row (0th row) of the Wisdom Pyramid is [1].
HAPPY CASE
Input: 
    wisdomLevel = 3
Output: 
    [1, 3, 3, 1]
Explanation:
    The 3rd row of the Wisdom Pyramid is built by combining elements of the 2nd row: 
    [1, 2, 1] → [1, (1+2), (2+1), 1] = [1, 3, 3, 1].

EDGE CASE
Input: 
    wisdomLevel = 0
Output: 
    [1]
Explanation:
    The 0th row is simply [1].

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For pyramid row problems, we want to consider the following approaches:

  • Dynamic Programming (DP): This problem is similar to generating Pascal's Triangle, where each row depends on the previous row.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: We will use a dynamic programming approach to generate each row of the Wisdom Pyramid up to the specified wisdomLevel. Each row is built by starting with 1, calculating the middle elements based on the two numbers above, and ending with 1.

Steps:

  1. Base Case:

    • If wisdomLevel is 0, return [1].

  2. DP Array Initialization:

    • Initialize the first row of the pyramid as [1].

  3. Build the Rows:

    • For each row from 1 to wisdomLevel, generate the next row by:
      • Starting with 1.
      • For each middle element j, calculate it as the sum of the two numbers directly above it from the previous row.
      • End the row with 1.

  4. Return the Result:

    • Return the row at the wisdomLevel.

4: I-mplement

Implement the code to solve the algorithm.

def wisdom_pyramid(wisdomLevel):
    if wisdomLevel == 0:
        return [1]
    
    dp = [1](/codepath/compsci_guides/wiki/1)  # Initialize the first row of the pyramid
    
    for i in range(1, wisdomLevel + 1):
        # Start the row with 1
        row = [1]
        
        # Fill in the middle values
        for j in range(1, i):
            row.append(dp[i-1][j-1] + dp[i-1][j])
        
        # End the row with 1
        row.append(1)
        
        dp.append(row)
    
    return dp[wisdomLevel]

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

Example 1:

  • Input: wisdomLevel = 3
  • Expected Output: [1, 3, 3, 1]

Example 2:

  • Input: wisdomLevel = 0
  • Expected Output: [1]

Example 3:

  • Input: wisdomLevel = 5
  • Expected Output: [1, 5, 10, 10, 5, 1]

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume n is the wisdomLevel.

  • Time Complexity: O(n^2) because we need to generate all rows up to wisdomLevel and each row takes linear time.
  • Space Complexity: O(n^2) for storing the DP table containing all the rows of the pyramid.