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Leetcode #1109: Corporate Flight Bookings

In this guide, we solve Leetcode #1109 Corporate Flight Bookings in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

There are n flights that are labeled from 1 to n. You are given an array of flight bookings bookings, where bookings[i] = [firsti, lasti, seatsi] represents a booking for flights firsti through lasti (inclusive) with seatsi seats reserved for each flight in the range.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Array, Prefix Sum

Intuition

Range queries become simple once we precompute cumulative sums.

We can transform subarray conditions into prefix comparisons.

Approach

Compute prefix sums and use a map to find matching prefixes.

This avoids nested loops while keeping the logic clear.

Steps:

  • Compute prefix sums.
  • Use a map to find valid ranges.
  • Update the answer.

Example

Input: bookings = [[1,2,10],[2,3,20],[2,5,25]], n = 5 Output: [10,55,45,25,25] Explanation: Flight labels: 1 2 3 4 5 Booking 1 reserved: 10 10 Booking 2 reserved: 20 20 Booking 3 reserved: 25 25 25 25 Total seats: 10 55 45 25 25 Hence, answer = [10,55,45,25,25]

Python Solution

class Solution: def corpFlightBookings(self, bookings: List[List[int]], n: int) -> List[int]: ans = [0] * n for first, last, seats in bookings: ans[first - 1] += seats if last < n: ans[last] -= seats return list(accumulate(ans))

Complexity

The time complexity is O(n)O(n)O(n), where nnn is the number of flights. The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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