Doubling the Power of a Spell - codepath/compsci_guides GitHub Wiki
TIP102 Unit 7 Session 1 Standard (Click for link to problem statements)
Problem 5: Doubling the Power of a Spell
The court magician is practicing a spell that doubles its power with each incantation. Given an integer initial_power
and a non-negative integer n
, write a recursive function that doubles initial_power
n
times.
Problem Highlights
- 💡 Difficulty: Easy
- ⏰ Time to complete: 10-15 mins
- 🛠️ Topics: Recursion, Mathematical Operations
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?
- Q: What is the main task in this problem?
- A: The task is to recursively double an initial power
n
times.
- A: The task is to recursively double an initial power
- Q: What should the function return if
n
is 0?- A: The function should return the
initial_power
without any changes.
- A: The function should return the
HAPPY CASE
Input: initial_power = 5, n = 3
Output: 40
Explanation: 5 doubled 3 times: 5 -> 10 -> 20 -> 40
Input: initial_power = 7, n = 2
Output: 28
Explanation: 7 doubled 2 times: 7 -> 14 -> 28
EDGE CASE
Input: initial_power = 10, n = 0
Output: 10
Explanation: If no doubling is required, the initial power remains 10.
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 Exponentiation via Doubling, we want to consider the following approaches:
- Recursive Doubling: Recursively double the power by multiplying the current power by 2 and decrementing the count
n
untiln
reaches 0.
3: P-lan
Plan the solution with appropriate visualizations and pseudocode.
General Idea:
- To double the power
n
times, multiply theinitial_power
by 2 and call the function recursively withn-1
.
Recursive Approach:
1) Base case: If `n` is 0, return the `initial_power` as no further doubling is needed.
2) Recursive case:
a) Multiply the `initial_power` by 2.
b) Return the result of the function called with the updated `initial_power` and `n-1`.
⚠️ Common Mistakes
- Not handling the base case correctly, which can result in incorrect results or infinite recursion.
- Incorrectly updating the value of
initial_power
in each recursive call.
4: I-mplement
Implement the code to solve the algorithm.
def double_power(initial_power, n):
# Base case: No more doubling needed
if n == 0:
return initial_power
# Recursive case: Double the power and reduce n
return double_power(initial_power * 2, n - 1)
5: R-eview
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
- Trace through the
double_power
function with the input(5, 3)
. The function should return40
after doubling5
three times. - Test the function with edge cases like
n = 0
. The function should return theinitial_power
unchanged.
6: E-valuate
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
- Time Complexity:
O(N)
whereN
is the number of times the power needs to be doubled. The function performsN
recursive calls. - Space Complexity:
O(N)
due to the recursion stack. The depth of recursion is proportional ton
.