How do you solve Corporate Flight Bookings on LeetCode?

Corporate Flight Bookings (LeetCode #1109) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a difference array allows for efficient range updates.

What is the brute force solution for Corporate Flight Bookings?

The brute force approach for Corporate Flight Bookings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through each booking and updating the seat counts for each flight in the specified range. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Corporate Flight Bookings?

Corporate Flight Bookings uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Difference Array, Prefix Sum.

What are the interview tips for Corporate Flight Bookings?

Explain your thought process clearly while coding. Consider edge cases and constraints before finalizing your approach. Practice similar problems to recognize patterns.

What are common mistakes when solving Corporate Flight Bookings?

Not considering the edge case when last_i equals n. Failing to initialize the answer array correctly.

#1109

Corporate Flight Bookings

Medium

ArrayPrefix SumDifference ArrayPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a difference array technique to efficiently manage seat counts. This allows us to apply the bookings in constant time and then compute the final seat counts in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array 'answer' of size n with all elements set to 0.
2Step 2: For each booking, increment answer[first_i - 1] by seats_i and decrement answer[last_i] by seats_i (if last_i < n).
3Step 3: Iterate through the answer array and compute the prefix sum to get the total seats for each flight.

solution.py11 lines

1# Full working Python code
2
3def corpFlightBookings(bookings, n):
4    answer = [0] * (n + 1)
5    for first_i, last_i, seats_i in bookings:
6        answer[first_i - 1] += seats_i
7        if last_i < n:
8            answer[last_i] -= seats_i
9    for i in range(1, n):
10        answer[i] += answer[i - 1]
11    return answer[:-1]

ℹ

Complexity note: The time complexity is O(n) because we process each booking in constant time and then compute the prefix sum in a single pass. The space complexity is O(n) due to the additional array used for the difference array technique.

1Using a difference array allows for efficient range updates.
2Prefix sums can be used to derive final results from incremental changes.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.