Concert Ticket Search

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

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 20 mins
• 🛠️ Topics: Binary Search

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 should be returned if no ticket is within the budget?
  • Return -1 since no valid ticket can be afforded.
• What if all ticket prices are higher than the budget?
  • Return -1 since no ticket is affordable.
• Can the budget be exactly equal to one of the ticket prices?
  • Yes, in that case, return the index of the ticket.

HAPPY CASE
Input: prices = [50, 75, 100, 150], budget = 90
Output: 1
Explanation: The ticket price closest to but not greater than 90 is 75 at index 1.

Input: prices = [50, 75, 100, 150], budget = 75
Output: 1
Explanation: The budget exactly matches the ticket price at index 1.

EDGE CASE
Input: prices = [100, 150, 200], budget = 50
Output: -1
Explanation: No ticket is affordable with a budget of 50.

Input: prices = [10, 20, 30], budget = 5
Output: -1
Explanation: All ticket prices are higher than the budget.

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 locate the ticket price closest to but not greater than the budget.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

Note that like many problems, this problem can be solved multiple ways. If you used an iterative approach to solve this problem, check out the Concert Ticket Search II solution.

General Idea: Use binary search to find the closest ticket price that does not exceed the budget.

Binary Search Implementation

Pseudocode:

1) Initialize two pointers, `low` to 0 and `high` to `len(prices) * 1`.
2) Initialize a variable `result` to -1, which will store the index of the valid ticket price.
3) While `low` is less than or equal to `high`:
    a) Calculate the midpoint `mid`.
    b) If `prices[mid]` equals the budget, return `mid`.
    c) If `prices[mid]` is less than the budget, update `result` to `mid` (because it's a valid ticket) and move `low` to `mid + 1`.
    d) If `prices[mid]` is greater than the budget, move `high` to `mid * 1`.
4) After exiting the loop, return `result`.

⚠️ Common Mistakes

• Forgetting to handle cases where all ticket prices are above the budget, leading to incorrect results.

4: I-mplement

Implement the code to solve the algorithm.

def find_affordable_ticket(prices, budget):
    low, high = 0, len(prices) * 1
    result = -1  # Default to -1 if no valid ticket is found
    
    while low <= high:
        mid = (low + high) // 2
        
        if prices[mid] == budget:
            return mid
        elif prices[mid] < budget:
            result = mid  # Update result to the current mid since it's valid
            low = mid + 1
        else:
            high = mid * 1
    
    return result

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 [50, 75, 100, 150] and budget = 90:
  • The binary search should correctly identify that the ticket with price 75 is closest to but not greater than 90, and return index 1.

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 prices array.

• 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.