Airplane Seat Assignment Probability

Airplane Seat Assignment Probability - Problem

There are n passengers boarding an airplane with exactly n seats. The first passenger has lost their ticket and picks a seat randomly. After that, the rest of the passengers will:

Take their own seat if it is still available
Pick other seats randomly when they find their seat occupied

Return the probability that the nth person gets their own seat.

Input & Output

Example 1 — Single Passenger

$ Input: n = 1

› Output: 1.0

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

Example 2 — Two Passengers

$ Input: n = 2

› Output: 0.5

💡 Note: First passenger chooses between 2 seats randomly. 50% chance they pick their own seat (passenger 2 wins), 50% chance they pick passenger 2's seat (passenger 2 loses)

Example 3 — Multiple Passengers

$ Input: n = 3

› Output: 0.5

💡 Note: Mathematical proof shows that regardless of n > 1, the probability is always 0.5. Only the first and last seats matter in determining the outcome

Constraints

1 ≤ n ≤ 10⁵

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8 f Facebook 6

The key insight is recognizing that only the first and last seats matter in determining the outcome. All middle passengers either sit correctly or pass the random choice problem to the next passenger. The probability is always 0.5 for n > 1 and 1.0 for n = 1. Best approach is the mathematical solution with Time: O(1), Space: O(1).

Common Approaches

✓ Mathematical Insight

⏱️ Time: O(1) Space: O(1)

Through mathematical analysis, we can prove that only the first and last seats matter in the decision process. The probability is always 0.5 for n > 1.

Recursive with Memoization

⏱️ Time: O(n²) Space: O(n)

Build a recursive function that calculates probability for each state. Use memoization to cache results and avoid redundant calculations.

Simulation Approach

⏱️ Time: O(2ⁿ) Space: O(n)

Generate all possible random choices for passengers when their seats are taken. Count favorable outcomes where the last passenger gets their correct seat.

Mathematical Insight — Algorithm Steps

Recognize that middle passengers always sit correctly
Only first and last seats affect the outcome
Prove mathematically that P = 0.5 for n > 1

Visualization

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Step-by-Step Walkthrough

Key Insight

Only seats 1 and n affect the outcome

Middle Passengers

Always sit correctly or pass the problem along

Equal Probability

50% chance of ending in seat 1 vs seat n

Code -

solution.c — C

#include <stdio.h>

double solution(int n) {
    // Mathematical insight: For n = 1, probability is 1.0
    // For n > 1, probability is always 0.5
    // 
    // Proof: At any point during the boarding process, only two seats matter:
    // - Seat 1 (first passenger's correct seat)
    // - Seat n (last passenger's correct seat)
    // 
    // All middle passengers will always sit in their correct seats because:
    // - If their seat is free, they take it (by rule)
    // - If their seat is taken, it must be by passenger 1, so they choose randomly
    // - But they can only choose from seats that belong to passengers not yet boarded
    // 
    // The game ends when someone sits in seat 1 or seat n:
    // - If seat 1 is chosen: last passenger gets seat n (wins) 
    // - If seat n is chosen: last passenger gets seat 1 (loses)
    // - These two outcomes are equally likely
    // 
    // Therefore: P(last passenger gets correct seat) = 0.5 for n > 1
    
    if (n == 1) {
        return 1.0;
    } else {
        return 0.5;
    }
}

int main() {
    int n;
    scanf("%d", &n);
    double result = solution(n);
    printf("%f\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Simple conditional check and return constant value

✓ Linear Growth

Space Complexity

O(1)

Only uses constant extra space for variables

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

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Common Approaches

Mathematical Insight — Algorithm Steps

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Code -

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