How do you solve Exchange Seats on LeetCode?

Exchange Seats (LeetCode #626) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Swapping pairs can be done in a single pass for efficiency.

What is the brute force solution for Exchange Seats?

The brute force approach for Exchange Seats is Brute Force, with O(n²) time complexity and O(n) space complexity. In this approach, we will iterate through the list of students and swap every two consecutive students' IDs. This is straightforward but inefficient as it requires multiple passes over the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Exchange Seats?

Exchange Seats uses the following concepts: Database. The recommended patterns to study are: Array, In-place swapping.

What are the interview tips for Exchange Seats?

Always think about the simplest approach first. Discuss your thought process out loud to clarify your understanding. Consider edge cases like empty lists or lists with one student.

What are common mistakes when solving Exchange Seats?

Not handling the case of an odd number of students correctly. Overcomplicating the solution by creating unnecessary data structures.

#626

Exchange Seats

Medium

DatabaseArrayIn-place swapping

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

In the optimal solution, we can directly swap the IDs in a single pass through the list. This avoids the overhead of creating a new list and sorting it, making it more efficient.

⚙️

Algorithm

3 steps

1Step 1: Iterate through the list of students in steps of 2.
2Step 2: For each pair, swap their IDs directly in the original list.
3Step 3: Return the modified list, which is already in the correct order.

solution.py4 lines

1def exchangeSeats(seats):
2    for i in range(0, len(seats) - 1, 2):
3        seats[i], seats[i + 1] = seats[i + 1], seats[i]  # Swap
4    return seats

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the list. The space complexity is O(1) since we are swapping in place without using additional data structures.

1Swapping pairs can be done in a single pass for efficiency.
2Understanding how to manipulate lists directly can save time and space.

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