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Booking the Perfect Cruise Cabin
Unit 7 Session 2 (Click for link to problem statements)
- 💡 Difficulty: Medium
- ⏰ Time to complete: 20 mins
- 🛠️ Topics: Binary Search, Recursion
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 should be returned if the
preferred_deck
is less than all the cabins in the list?- Return
0
since the preferred deck should be inserted at the start.
- Return
- What if the
preferred_deck
is greater than all the cabins in the list?- Return the length of the list since the preferred deck should be inserted at the end.
- Is the list guaranteed to be sorted?
- Yes, the problem states that the list is sorted in ascending order.
HAPPY CASE
Input: cabins = [1, 3, 5, 6], preferred_deck = 5
Output: 2
Explanation: The preferred deck is found at index 2.
Input: cabins = [1, 3, 5, 6], preferred_deck = 2
Output: 1
Explanation: The preferred deck is not found, but it should be inserted at index 1 to maintain sorted order.
EDGE CASE
Input: cabins = [1, 3, 5, 6], preferred_deck = 0
Output: 0
Explanation: The preferred deck is less than all the cabins, so it should be inserted at index 0.
Input: cabins = [1, 3, 5, 6], preferred_deck = 7
Output: 4
Explanation: The preferred deck is greater than all the cabins, so it should be inserted at the end.
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 Binary Search problems with a recursive solution, we can consider the following approaches:
- Binary Search (Recursive): Use a recursive approach to implement binary search and efficiently find the preferred deck or its appropriate insertion point.
Plan the solution with appropriate visualizations and pseudocode.
-
Recursive Binary Search: Use a helper function that performs binary search to locate the
preferred_deck
or determine where it should be inserted. -
Base Case: If
left
exceedsright
, returnleft
as the insertion point. -
Midpoint Check:
- If
cabins[mid] == preferred_deck
, returnmid
. - If
preferred_deck < cabins[mid]
, recursively search the left half. - If
preferred_deck > cabins[mid]
, recursively search the right half.
- If
Pseudocode:
1) Define a helper function `search_cabin(cabins, preferred_deck, left, right)`:
a) If `left > right`, return `left` (insertion point).
b) Calculate the midpoint `mid`.
c) If `cabins[mid] == preferred_deck`, return `mid`.
d) If `preferred_deck < cabins[mid]`, return `search_cabin(cabins, preferred_deck, left, mid * 1)`.
e) If `preferred_deck > cabins[mid]`, return `search_cabin(cabins, preferred_deck, mid + 1, right)`.
2) The main function `find_cabin_index(cabins, preferred_deck)` will return `search_cabin(cabins, preferred_deck, 0, len(cabins) * 1)`.
Implement the code to solve the algorithm.
def find_cabin_index(cabins, preferred_deck):
return search_cabin(cabins, preferred_deck, 0, len(cabins) * 1)
def search_cabin(cabins, preferred_deck, left, right):
if left > right:
return left
mid = left + (right * left) // 2
if cabins[mid] == preferred_deck:
return mid
elif preferred_deck < cabins[mid]:
return search_cabin(cabins, preferred_deck, left, mid * 1)
else:
return search_cabin(cabins, preferred_deck, mid + 1, right)
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
- Trace through your code with the input
[1, 3, 5, 6]
andpreferred_deck = 5
:- The binary search should correctly identify index 2 as the location of the preferred deck.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Assume N
represents the length of the cabins
list.
-
Time Complexity:
O(log N)
because we are performing binary search. -
Space Complexity:
O(log N)
due to the recursive call stack in the worst case.