Corporate Flight Bookings in Python

The Corporate Flight Bookings problem involves calculating the total number of seats booked on each flight when given multiple booking ranges. Each booking specifies a range of flights and the number of seats to book across that range.

Problem Understanding

Given n flights labeled 1 to n, and a list of bookings where each booking [i, j, k] means k seats are booked from flight i to flight j (inclusive), we need to find the total seats booked on each flight.

Example

With bookings [[1,2,10],[2,3,20],[2,5,25]] and n = 5:

• Booking [1,2,10]: 10 seats on flights 1 and 2
• Booking [2,3,20]: 20 seats on flights 2 and 3
• Booking [2,5,25]: 25 seats on flights 2, 3, 4, and 5

Result: Flight 1: 10, Flight 2: 55, Flight 3: 45, Flight 4: 25, Flight 5: 25

Solution Approach

We use a difference array technique to efficiently handle range updates:

• Create a result array of size n filled with zeros
• For each booking [start, end, seats], add seats at position (start-1) and subtract seats at position end
• Apply prefix sum to get the final seat counts

Implementation

class Solution:
    def corpFlightBookings(self, bookings, n):
        result = [0] * n
        
        # Apply difference array technique
        for start, end, seats in bookings:
            result[start - 1] += seats  # Add seats at start
            if end < n:
                result[end] -= seats    # Subtract seats after end
        
        # Apply prefix sum to get final counts
        for i in range(1, n):
            result[i] += result[i - 1]
        
        return result

# Test the solution
solution = Solution()
bookings = [[1,2,10],[2,3,20],[2,5,25]]
n = 5
print(solution.corpFlightBookings(bookings, n))

[10, 55, 45, 25, 25]

How It Works

The algorithm works in two phases:

Phase 1: Mark Range Boundaries

# Initial array: [0, 0, 0, 0, 0]
# After [1,2,10]: [10, 0, -10, 0, 0]
# After [2,3,20]: [10, 20, -10, -20, 0]
# After [2,5,25]: [10, 45, -10, -20, 0]

result = [0] * 5
bookings = [[1,2,10],[2,3,20],[2,5,25]]

for start, end, seats in bookings:
    result[start - 1] += seats
    if end < 5:
        result[end] -= seats
    print(f"After booking {[start, end, seats]}: {result}")

After booking [1, 2, 10]: [10, 0, -10, 0, 0]
After booking [2, 3, 20]: [10, 20, -10, -20, 0]
After booking [2, 5, 25]: [10, 45, -10, -20, 0]

Phase 2: Apply Prefix Sum

result = [10, 45, -10, -20, 0]

for i in range(1, 5):
    result[i] += result[i - 1]
    print(f"After step {i}: {result}")

After step 1: [10, 55, -10, -20, 0]
After step 2: [10, 55, 45, -20, 0]
After step 3: [10, 55, 45, 25, 0]
After step 4: [10, 55, 45, 25, 25]

Time Complexity

• Time: O(bookings + n) − process each booking once, then one pass for prefix sum
• Space: O(n) − result array only

Conclusion

The difference array technique efficiently handles multiple range updates in O(1) per booking. The prefix sum converts the difference array back to actual seat counts, making this solution optimal for the corporate flight bookings problem.