How do you solve Airplane Seat Assignment Probability on LeetCode?

Airplane Seat Assignment Probability (LeetCode #1227) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: The probability for the nth passenger to sit in their own seat stabilizes at 0.5 for n >= 2.

What is the time complexity of Airplane Seat Assignment Probability?

The optimal solution for Airplane Seat Assignment Probability runs in O(1) time and uses O(1) space. The optimal solution runs in constant time O(1) and uses constant space O(1) because it only checks the value of n and returns a fixed result.

What is the brute force solution for Airplane Seat Assignment Probability?

The brute force approach for Airplane Seat Assignment Probability is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating the seating arrangement for all passengers. We can track the choices of each passenger and calculate the probability of the nth passenger sitting in their own seat based on the random choices made by the first passenger. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Airplane Seat Assignment Probability?

Airplane Seat Assignment Probability uses the following concepts: Math, Dynamic Programming, Brainteaser, Probability and Statistics. The recommended patterns to study are: Probability and Statistics, Dynamic Programming (in terms of recognizing patterns).

What are the interview tips for Airplane Seat Assignment Probability?

Always start with a base case when dealing with recursive or probabilistic problems. Look for patterns in the problem that can simplify your calculations. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Airplane Seat Assignment Probability?

Assuming the probability changes with larger n without recognizing the underlying pattern. Not considering edge cases like n = 1 and n = 2 properly.

#1227

Airplane Seat Assignment Probability

Medium

MathDynamic ProgrammingBrainteaserProbability and StatisticsProbability and StatisticsDynamic Programming (in terms of recognizing patterns)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

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Intuition

Time O(1)Space O(1)

The optimal solution leverages the observation that the probability of the nth passenger sitting in their own seat is always 1/2 for n >= 2. This is because the first passenger's choice creates a 50/50 chance for the last passenger, regardless of the number of passengers.

⚙️

Algorithm

3 steps

1Step 1: If n = 1, return 1.0 (the only passenger sits in their seat).
2Step 2: If n = 2, return 0.5 (the first passenger has an equal chance of taking either seat).
3Step 3: For n >= 3, return 0.5 (the nth passenger's chance remains constant).

solution.py4 lines

1def optimal_probability(n):
2    return 1.0 if n == 1 else 0.5
3
4print(optimal_probability(2))

ℹ

Complexity note: The optimal solution runs in constant time O(1) and uses constant space O(1) because it only checks the value of n and returns a fixed result.

1The probability for the nth passenger to sit in their own seat stabilizes at 0.5 for n >= 2.
2The first passenger's random choice creates a chain of dependencies that ultimately leads to a 50/50 chance for the last passenger.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.