LeetCode #1109 — MEDIUM

Corporate Flight Bookings

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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] = [first_i, last_i, seats_i] represents a booking for flights first_i through last_i (inclusive) with seats_i seats reserved for each flight in the range.

Return an array answer of length n, where answer[i] is the total number of seats reserved for flight i.

Example 1:

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]

Example 2:

Input: bookings = [[1,2,10],[2,2,15]], n = 2
Output: [10,25]
Explanation:
Flight labels:        1   2
Booking 1 reserved:  10  10
Booking 2 reserved:      15
Total seats:         10  25
Hence, answer = [10,25]

Constraints:

1 <= n <= 2 * 10⁴
1 <= bookings.length <= 2 * 10⁴
bookings[i].length == 3
1 <= first_i <= last_i <= n
1 <= seats_i <= 10⁴

Step 01

Brute Force Baseline

Problem summary: 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. Return an array answer of length n, where answer[i] is the total number of seats reserved for flight i.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[1,2,10],[2,3,20],[2,5,25]]
5

Example 2

[[1,2,10],[2,2,15]]
2

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1109: Corporate Flight Bookings
class Solution {
    public int[] corpFlightBookings(int[][] bookings, int n) {
        int[] ans = new int[n];
        for (var e : bookings) {
            int first = e[0], last = e[1], seats = e[2];
            ans[first - 1] += seats;
            if (last < n) {
                ans[last] -= seats;
            }
        }
        for (int i = 1; i < n; ++i) {
            ans[i] += ans[i - 1];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1109: Corporate Flight Bookings
func corpFlightBookings(bookings [][]int, n int) []int {
	ans := make([]int, n)
	for _, e := range bookings {
		first, last, seats := e[0], e[1], e[2]
		ans[first-1] += seats
		if last < n {
			ans[last] -= seats
		}
	}
	for i := 1; i < n; i++ {
		ans[i] += ans[i-1]
	}
	return ans
}

# Accepted solution for LeetCode #1109: Corporate Flight Bookings
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))

// Accepted solution for LeetCode #1109: Corporate Flight Bookings
impl Solution {
    #[allow(dead_code)]
    pub fn corp_flight_bookings(bookings: Vec<Vec<i32>>, n: i32) -> Vec<i32> {
        let mut ans = vec![0; n as usize];

        // Build the difference vector first
        for b in &bookings {
            let (l, r) = ((b[0] as usize) - 1, (b[1] as usize) - 1);
            ans[l] += b[2];
            if r < (n as usize) - 1 {
                ans[r + 1] -= b[2];
            }
        }

        // Build the prefix sum vector based on the difference vector
        for i in 1..n as usize {
            ans[i] += ans[i - 1];
        }

        ans
    }
}

// Accepted solution for LeetCode #1109: Corporate Flight Bookings
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1109: Corporate Flight Bookings
// class Solution {
//     public int[] corpFlightBookings(int[][] bookings, int n) {
//         int[] ans = new int[n];
//         for (var e : bookings) {
//             int first = e[0], last = e[1], seats = e[2];
//             ans[first - 1] += seats;
//             if (last < n) {
//                 ans[last] -= seats;
//             }
//         }
//         for (int i = 1; i < n; ++i) {
//             ans[i] += ans[i - 1];
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.