Count Checked In Passengers

Unit 7 Session 2 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Binary Search

1: U-nderstand

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

• Q: What should be returned if no passengers have checked in?
  • A: Return 0 since there are no 1s in the list.
• Q: Is the list guaranteed to be sorted?
  • A: Yes, the problem states that the list is sorted, with all 0s appearing before any 1s.
• Q: Can the list be empty?
  • A: The problem does not specify, but if the list is empty, return 0.

HAPPY CASE
Input: rooms = [0, 0, 0, 1, 1, 1, 1]
Output: 4
Explanation: The total number of checked-in passengers is 4, corresponding to the four `1`s.

Input: rooms = [0, 0, 0, 0, 0, 1]
Output: 1
Explanation: The total number of checked-in passengers is 1, corresponding to the one `1`.

EDGE CASE
Input: rooms = [0, 0, 0, 0, 0, 0]
Output: 0
Explanation: No passengers have checked in, so return 0.

Input: rooms = [1, 1, 1, 1]
Output: 4
Explanation: All passengers have checked in, so return 4.

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 Binary Search problems, we want to consider the following approaches:

• Binary Search: Use binary search to efficiently find the first occurrence of 1, and then calculate the number of 1s based on the list's length.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

Plan

1. Binary Search: Perform binary search to locate the first occurrence of 1 in the list.
2. Calculate Result: Once the first occurrence is found, the number of 1s in the list is simply the total length of the list minus the index of the first 1.

Binary Search Implementation

Pseudocode:

1) Initialize `left` to 0 and `right` to `len(rooms) - 1`.
2) Initialize `first_one_index` to `len(rooms)` as a default (assuming no `1`s).
3) While `left` is less than or equal to `right`:
    a) Calculate the midpoint `mid`.
    b) If `rooms[mid] == 1`, update `first_one_index` to `mid` and move `right` to `mid - 1` (search the left side).
    c) If `rooms[mid] == 0`, move `left` to `mid + 1` (search the right side).
4) Return the total number of `1`s as `len(rooms) - first_one_index`.

4: I-mplement

Implement the code to solve the algorithm.

def count_checked_in_passengers(rooms):
    n = len(rooms)
    
    # Binary search to find the first occurrence of 1
    left, right = 0, n - 1
    first_one_index = n  # Assume there are no 1's initially
    
    while left <= right:
        mid = left + (right - left) // 2
        if rooms[mid] == 1:
            first_one_index = mid
            right = mid - 1  # Look on the left side for the first 1
        else:
            left = mid + 1  # Look on the right side
    
    # Number of 1's is the total length minus the index of the first 1
    return n - first_one_index

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 your code with the input [0, 0, 0, 1, 1, 1, 1]:
  • The binary search should correctly identify that the first 1 appears at index 3, resulting in a total of 4 checked-in passengers.

6: E-valuate

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

Assume N represents the length of the rooms list.

• Time Complexity: O(log N) because we are performing binary search.
• Space Complexity: O(1) because we are using a constant amount of extra space.