Introduction

In this blog post, we will explore how to solve the LeetCode problem "2037" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

There are n availabe seats and n students standing in a room. You are given an array seats of length n, where seats[i] is the position of the ith seat. You are also given the array students of length n, where students[j] is the position of the jth student. You may perform the following move any number of times: Increase or decrease the position of the ith student by 1 (i.e., moving the ith student from position x to x + 1 or x - 1) Return the minimum number of moves required to move each student to a seat such that no two students are in the same seat. Note that there may be multiple seats or students in the same position at the beginning. Example 1: Input: seats = [3,1,5], students = [2,7,4] Output: 4 Explanation: The students are moved as follows: - The first student is moved from position 2 to position 1 using 1 move. - The second student is moved from position 7 to position 5 using 2 moves. - The third student is moved from position 4 to position 3 using 1 move. In total, 1 + 2 + 1 = 4 moves were used. Example 2: Input: seats = [4,1,5,9], students = [1,3,2,6] Output: 7 Explanation: The students are moved as follows: - The first student is not moved. - The second student is moved from position 3 to position 4 using 1 move. - The third student is moved from position 2 to position 5 using 3 moves. - The fourth student is moved from position 6 to position 9 using 3 moves. In total, 0 + 1 + 3 + 3 = 7 moves were used. Example 3: Input: seats = [2,2,6,6], students = [1,3,2,6] Output: 4 Explanation: Note that there are two seats at position 2 and two seats at position 6. The students are moved as follows: - The first student is moved from position 1 to position 2 using 1 move. - The second student is moved from position 3 to position 6 using 3 moves. - The third student is not moved. - The fourth student is not moved. In total, 1 + 3 + 0 + 0 = 4 moves were used. Constraints: n == seats.length == students.length 1 <= n <= 100 1 <= seats[i], students[j] <= 100

Explanation

Here's the solution to the problem:

• Sorting: The core idea is to sort both the seats and students arrays. This allows us to optimally match each student to a seat.
• Minimize moves: By sorting, we ensure that the student closest to a seat (in terms of index after sorting) is assigned to that seat. This minimizes the total number of moves required.

• Calculate moves: Iterate through the sorted arrays and calculate the absolute difference between the position of each student and their assigned seat. Sum these absolute differences to get the minimum total moves.

• Runtime Complexity: O(n log n) due to the sorting operation. Storage Complexity: O(1) if the sorting is done in-place; otherwise, O(n) depending on the sorting algorithm implementation.

Code

    def min_moves_to_seat(seats, students):
    """
    Calculates the minimum number of moves required to move each student to a seat
    such that no two students are in the same seat.

    Args:
        seats: A list of integers representing the positions of the seats.
        students: A list of integers representing the positions of the students.

    Returns:
        The minimum number of moves required.
    """
    seats.sort()
    students.sort()

    moves = 0
    for i in range(len(seats)):
        moves += abs(seats[i] - students[i])

    return moves