Status of Flight Tickets - Practice Coding Problems

Status of Flight Tickets - Problem

Database Hard

Flight Ticket Status Tracker

You're building a flight reservation system that needs to track passenger ticket statuses in real-time. When passengers book tickets, they get confirmed if seats are available, or placed on a waitlist if the flight is full.

The Challenge: Given flight capacities and passenger booking times, determine each passenger's current ticket status.

Key Rules:

Passengers are processed in chronological order of booking time
If seats available → ticket Confirmed
If flight full → passenger goes on Waitlist
Each flight has a fixed capacity limit

Input: Two tables - Flights (flight_id, capacity) and Passengers (passenger_id, flight_id, booking_time)

Output: Each passenger's status ordered by passenger_id

Input & Output

example_1.py — Basic Flight Booking

$ Input: Flights: [[1, 2]], Passengers: [[1, 1, '2023-07-10 16:30:00'], [2, 1, '2023-07-10 17:00:00'], [3, 1, '2023-07-10 18:00:00']]

› Output: [[1, 'Confirmed'], [2, 'Confirmed'], [3, 'Waitlist']]

💡 Note: Flight 1 has capacity 2. Passengers 1 and 2 book first (confirmed), passenger 3 books last (waitlisted)

example_2.py — Multiple Flights

$ Input: Flights: [[1, 1], [2, 2]], Passengers: [[1, 1, '2023-07-10 16:00:00'], [2, 2, '2023-07-10 17:00:00'], [3, 1, '2023-07-10 18:00:00'], [4, 2, '2023-07-10 19:00:00']]

› Output: [[1, 'Confirmed'], [2, 'Confirmed'], [3, 'Waitlist'], [4, 'Confirmed']]

💡 Note: Flight 1 (capacity 1): P1 confirmed, P3 waitlisted. Flight 2 (capacity 2): P2 and P4 both confirmed

example_3.py — Same Booking Time Edge Case

$ Input: Flights: [[1, 1]], Passengers: [[1, 1, '2023-07-10 16:00:00'], [2, 1, '2023-07-10 16:00:00']]

› Output: [[1, 'Confirmed'], [2, 'Waitlist']]

💡 Note: When booking times are identical, passenger with lower ID gets priority (deterministic ordering)

Visualization

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Understanding the Visualization

Group Bookings by Flight

Separate passengers by their destination flight

Sort by Booking Time

Within each flight, arrange passengers by when they booked

Assign Sequential Numbers

Give each passenger a number (1st, 2nd, 3rd, etc.) in their flight

Compare with Capacity

If passenger number ≤ flight capacity: Confirmed, otherwise Waitlist

Key Takeaway

🎯 Key Insight: By processing passengers in chronological order within each flight and comparing their position with flight capacity, we efficiently determine ticket status in a single pass per flight.

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Sorting passengers by booking_time within each flight partition

⚡ Linearithmic

Space Complexity

O(n)

Temporary space for sorting and window function processing

⚡ Linearithmic Space

Constraints

1 ≤ flights.length ≤ 10⁴
1 ≤ passengers.length ≤ 10⁵
1 ≤ flight_id, passenger_id, capacity ≤ 10⁹
booking_time format: 'YYYY-MM-DD HH:MM:SS'
All passenger_id values are distinct
All booking_time values are distinct
Each passenger books exactly one flight

Asked in

E Expedia 45 B Booking.com 38 A Airbnb 32 AA American Airlines 28

The optimal solution uses SQL window functions with ROW_NUMBER() OVER (PARTITION BY flight_id ORDER BY booking_time) to rank passengers chronologically within each flight, then compares ranks with flight capacity to determine Confirmed or Waitlist status in O(n log n) time.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Query)	O(n²)	O(1)	For each passenger, count all earlier bookings on same flight
Window Function Approach (Optimal)	O(n log n)	O(n)	Use ROW_NUMBER() window function to rank passengers by booking time

Brute Force (Nested Query) — Algorithm Steps

For each passenger, find their booking time
Count passengers who booked the same flight earlier
Compare count with flight capacity
Assign status based on comparison

Visualization

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Step-by-Step Walkthrough

Select Passenger

Choose passenger to evaluate

Scan All Bookings

Count earlier bookings for same flight

Compare with Capacity

Determine status based on count vs capacity

Repeat

Repeat for every passenger

Code -

solution.c — C

// Brute Force C Implementation
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    int passenger_id;
    int flight_id;
    char booking_time[20];
} Passenger;

typedef struct {
    int flight_id;
    int capacity;
} Flight;

typedef struct {
    int passenger_id;
    char status[10];
} Result;

int compareResults(const void* a, const void* b) {
    Result* ra = (Result*)a;
    Result* rb = (Result*)b;
    return ra->passenger_id - rb->passenger_id;
}

Result* getFlightStatus(Flight* flights, int flightCount, 
                       Passenger* passengers, int passengerCount) {
    Result* result = malloc(passengerCount * sizeof(Result));
    
    for (int i = 0; i < passengerCount; i++) {
        // Count earlier bookings
        int earlierBookings = 0;
        for (int j = 0; j < passengerCount; j++) {
            if (passengers[j].flight_id == passengers[i].flight_id &&
                strcmp(passengers[j].booking_time, passengers[i].booking_time) <= 0) {
                earlierBookings++;
            }
        }
        
        // Find capacity
        int capacity = 0;
        for (int k = 0; k < flightCount; k++) {
            if (flights[k].flight_id == passengers[i].flight_id) {
                capacity = flights[k].capacity;
                break;
            }
        }
        
        result[i].passenger_id = passengers[i].passenger_id;
        strcpy(result[i].status, 
               earlierBookings <= capacity ? "Confirmed" : "Waitlist");
    }
    
    qsort(result, passengerCount, sizeof(Result), compareResults);
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n passengers, we scan all n passengers to count earlier bookings

⚠ Quadratic Growth

Space Complexity

O(1)

Only using temporary variables for counting

✓ Linear Space

Constraints

1 ≤ flights.length ≤ 10⁴
1 ≤ passengers.length ≤ 10⁵
1 ≤ flight_id, passenger_id, capacity ≤ 10⁹
booking_time format: 'YYYY-MM-DD HH:MM:SS'
All passenger_id values are distinct
All booking_time values are distinct
Each passenger books exactly one flight

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Flight Ticket Status Tracker

Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Nested Query) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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