Exchange Seats — LeetCode 626 Python Solution

The problem

Swap the seats of every pair of neighbouring students, so seat 1 trades with seat 2, seat 3 with seat 4, and so on. When the number of students is odd the last one keeps their seat. Report the new seating ordered by id. The Seat table has one row per student: id (int, the primary key) and student (varchar). The ids start at 1 and increase without gaps.

Example

Seat table:

| id | student |
| -- | ------- |
| 1  | Nadia   |
| 2  | Oscar   |
| 3  | Petra   |
| 4  | Rui     |
| 5  | Tomas   |

Result:

| id | student |
| -- | ------- |
| 1  | Oscar   |
| 2  | Nadia   |
| 3  | Rui     |
| 4  | Petra   |
| 5  | Tomas   |

The first two students trade seats and so do the next two, while Tomas is the odd one out at the last seat and stays where he is.

Python solution

import pandas as pd

def exchange_seats(seat: pd.DataFrame) -> pd.DataFrame:
    df = seat.copy()
    max_id = df['id'].max()
    def swap_id(x):
        if x % 2 == 1 and x != max_id:
            return x + 1
        if x % 2 == 0:
            return x - 1
        return x
    df['new_id'] = df['id'].apply(swap_id)
    res = df.sort_values('new_id').rename(columns={'new_id': 'id'})
    return res[['id', 'student']]

Complexity

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