Airplane Seat Assignment Probability - Practice Coding Problems

Airplane Seat Assignment Probability - Problem

The Airplane Seat Assignment Probability Problem

Imagine you're on an airplane with exactly n seats numbered 1 through n, and n passengers are boarding. Here's where it gets interesting:

🎫 The first passenger has lost their boarding pass and doesn't know which seat is theirs, so they pick a seat completely at random.

👥 Each subsequent passenger follows this simple rule:
• If their assigned seat is available, they sit in it
• If their assigned seat is occupied, they pick any available seat randomly

🎯 Your mission: Calculate the probability that the last passenger (passenger n) ends up in their own assigned seat.

This is a classic probability puzzle that appears deceptively complex but has an elegant mathematical solution. The key insight lies in understanding that only certain seats matter in determining the final outcome!

Input & Output

example_1.py — Single Passenger

$ Input: n = 1

› Output: 1.00000

💡 Note: With only one passenger, they must sit in their own seat (seat 1), so the probability is 1.0

example_2.py — Two Passengers

$ Input: n = 2

› Output: 0.50000

💡 Note: First passenger randomly chooses between seat 1 and seat 2. If they choose seat 1, passenger 2 gets seat 2. If they choose seat 2, passenger 2 gets seat 1. Probability = 1/2

example_3.py — Large Case

$ Input: n = 1000

› Output: 0.50000

💡 Note: Regardless of the number of passengers, the mathematical principle holds: only seats 1 and n matter, and they are equally likely to be chosen, so the probability remains 1/2

Visualization

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Understanding the Visualization

Setup

Hotel has n rooms, first guest has no key

Chain Reaction

Each displaced guest continues the random selection

Two Outcomes

Process ends when someone picks room 1 or room n

Equal Probability

By symmetry, both endings are equally likely

Key Takeaway

🎯 Key Insight: The elegant solution emerges from recognizing that this complex-seeming problem reduces to a simple symmetry argument - only two seats matter, and they're equally likely to be chosen!

Time & Space Complexity

Time Complexity

⏱️

O(1)

The answer is always 1/2 for n > 1, so no computation needed

✓ Linear Growth

Space Complexity

O(1)

No additional space required beyond storing the result

✓ Linear Space

Constraints

1 ≤ n ≤ 10⁶
Output format: Return probability as decimal with exactly 5 decimal places
For n = 1, answer is 1.00000; for n > 1, answer is 0.50000

Asked in

G Google 23 a Amazon 18 f Meta 15 ⊞ Microsoft 12 🍎 Apple 8

This classic probability puzzle has an elegant O(1) mathematical solution. The key insight is that only two seats matter: seat 1 (first passenger's) and seat n (last passenger's). The process always terminates when someone sits in either of these seats, and by symmetry, both outcomes are equally likely. Therefore, the probability is 1/2 for n > 1 and 1 for n = 1.

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Analysis (Optimal)	O(1)	O(1)	Solve using pure mathematical reasoning and probability theory

Mathematical Analysis (Optimal) — Algorithm Steps

Recognize that passengers 2 through n-1 either sit in their correct seat (continuing the game) or end the game
Identify that only seats 1 and n can end the game when chosen randomly
Apply symmetry: seats 1 and n are equally likely to be chosen in any random selection
Conclude that P(last passenger gets seat n) = P(someone sits in seat n before seat 1) = 1/2

Visualization

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Step-by-Step Walkthrough

Key Observation

Only seats 1 and n matter for the final outcome

Game Rules

Process continues until someone sits in seat 1 or n

Symmetry

When choosing randomly, seats 1 and n are equally likely

Conclusion

Probability = 1/2 for any n > 1

Code -

solution.c — C

#include <stdio.h>

/**
 * Mathematical solution to airplane seat assignment probability.
 * 
 * Key insight: At any point, only seats 1 and n matter.
 * - If someone sits in seat 1, last passenger gets seat n (probability 1)
 * - If someone sits in seat n, last passenger doesn't get seat n (probability 0)
 * - Passengers 2 to n-1 either:
 *   * Sit in their correct seat (game continues)
 *   * Sit in seat 1 or n (game ends)
 * 
 * By symmetry, when a random choice is made, seats 1 and n are equally likely.
 * Therefore, P(last passenger gets seat n) = 1/2
 */
double numPointsInCircle(int n) {
    if (n == 1) {
        return 1.0; // Only one passenger, they get their seat
    } else {
        return 0.5; // For n > 1, probability is always 1/2
    }
}

int main() {
    int n;
    scanf("%d", &n);
    double result = numPointsInCircle(n);
    if (result == 1.0) {
        printf("1.00000\n");
    } else {
        printf("0.50000\n");
    }
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

The answer is always 1/2 for n > 1, so no computation needed

✓ Linear Growth

Space Complexity

O(1)

No additional space required beyond storing the result

✓ Linear Space

Constraints

1 ≤ n ≤ 10⁶
Output format: Return probability as decimal with exactly 5 decimal places
For n = 1, answer is 1.00000; for n > 1, answer is 0.50000

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Mathematical Analysis (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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