The Instructor's Seat Mrs. Geeta Rao manages the ...

Question

The Instructor's Seat

Mrs. Geeta Rao manages the UNIVERSITY classroom seating arrangements. The classroom has an infinite number of seats.

One day, a finite number of students come to attend a lecture.

The students consist of:

An Instructor.

→An unknown number of groups consisting of S members per group where S≠1.

The Instructor was given a separate seat, and the rest were assigned one seat per group.

Mrs. Geeta has a list of randomly arranged seat entries. The seat numbers will appear S times per group except for the Instructor's seat (which will occur only once). She needs you to help her find the Instructor's seat number.
Input Format

The first line contains an integer n, representing the number of students.

The second line consists of an integer, S, the size of each group.

Next n lines contains one integer per line representing seät number.

Output Format

Print the Instructor's seat number.

write code in python 3.9

Answer 1

Here’s an efficient and concise Python 3.9 solution.
It follows the trick used in HackerRank’s Captain’s Room problem: compare the sum of the unique seat numbers (multiplied by S) to the total sum. The difference isolates the Instructor’s seat.

python
# The Instructor's Seat
# Python 3.9

import sys

def main() -> None:
    data = sys.stdin.read().strip().split()
    if not data:
        return                          # no input

    n = int(data[0])                    # number of students (not actually needed)
    S = int(data[1])                    # group size
    seats = list(map(int, data[2:]))

    unique_seats = set(seats)
    # Formula: (S * sum(unique) - sum(all)) // (S - 1)
    instructor_seat = (S * sum(unique_seats) - sum(seats)) // (S - 1)
    print(instructor_seat)

if __name__ == "__main__":
    main()

Why this works

Each group seat number appears exactly S times, so
S × (sum of unique group seats) double-counts every group seat compared with the real total.
Subtracting the actual total sum(seats) leaves (S – 1) copies of every group seat plus one copy of the Instructor’s seat.
Dividing by (S – 1) cancels out the extra multiplicity, yielding the Instructor’s seat number in O(n) time and O(u) memory (u = number of distinct seats).

Answer 2

Here’s the same solution wrapped in a main() function, ready for most online-judge runners:

python
# Instructor's Seat
# Python 3.9+

import sys

def main() -> None:
    """
    Reads:
        n  – total number of students (ignored after reading)
        S  – size of each group (S ≠ 1)
        n seat numbers, one per line
    Prints:
        The Instructor's seat number.
    """
    data = sys.stdin.read().strip().split()
    if not data:
        return                    # no input provided

    n = int(data[0])              # total students (not needed for logic)
    S = int(data[1])              # group size
    seats = list(map(int, data[2:]))

    unique_seats = set(seats)
    # Each group seat appears S times; instructor seat appears once.
    instructor_seat = (S * sum(unique_seats) - sum(seats)) // (S - 1)
    print(instructor_seat)

if __name__ == "__main__":
    main()

Explanation of the formula

For every group seat number x, the expression S × sum(unique_seats) counts x exactly S times.
Subtracting the true total sum(seats) leaves (S – 1) copies of each group seat plus one copy of the Instructor’s seat.
Dividing by (S – 1) removes the excess multiplicity, isolating the Instructor’s seat number in O(n) time and O(u) memory (u = number of distinct seat numbers).

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