Aang's Meditation for Energy Balance

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

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Dynamic Programming

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 calculate the total energy Aang gains on the nth day, where the energy is the sum of the energy gained on the two previous days.
• What are the base cases?
  • On days 1 and 2, Aang gains 1 unit of energy.

HAPPY CASE
Input: 
    n = 5
Output: 
    5
Explanation:
    The energy gained on day 5 is the sum of the energy on days 3 and 4 (2 + 3 = 5).

EDGE CASE
Input: 
    n = 1
Output: 
    1
Explanation:
    On the first day, Aang gains 1 unit of energy.

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 energy balance problems, we want to consider the following approaches:

• Dynamic Programming (DP): This is a classic Fibonacci-like problem where the energy on day n depends on the sum of the energy from the two previous days.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use dynamic programming to compute the energy Aang gains on day n. We start with base cases for days 1 and 2, where Aang gains 1 unit of energy, and for each subsequent day, we calculate the energy as the sum of the previous two days.

Steps:

1. Base Case:

   • On days 1 and 2, Aang gains 1 unit of energy.

2. Dynamic Programming Array:

   • Create a DP array dp where dp[i] stores the energy gained on day i.

3. Recurrence Relation:

   • For each day i from 3 to n, calculate dp[i] as the sum of dp[i - 1] and dp[i - 2].

4. Return the Result:

   • The value at dp[n] will give the energy gained on day n.

4: I-mplement

Implement the code to solve the algorithm.

def energy_on_nth_day(n):
    # Base case for days 1 and 2, Aang gains 1 unit of energy
    if n == 1 or n == 2:
        return 1
    
    # Create a dp array to store the energy Aang gains on each day
    dp = [0] * (n + 1)
    
    # Initial energy gains for day 1 and day 2
    dp[1] = 1
    dp[2] = 1
    
    # For each day from 3 to n, calculate energy as the sum of the previous two days
    for i in range(3, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]
    
    # Return the energy gained on day n
    return dp[n]

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: n = 5
• Expected Output: 5

Example 2:

• Input: n = 1
• Expected Output: 1

6: E-valuate

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

Assume n is the input number of days.

• Time Complexity: O(n) because we calculate the energy for each day up to n.
• Space Complexity: O(n) to store the DP array with n elements.