2037. Minimum Number of Moves to Seat Everyone - Explanation

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Description

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 = [2,1,4], students = [3,1,3]

Output: 2

Explanation:

• The first student is moved from position 3 to position 2 using 1 move.
• The second student at position 1 is not moved.
• The third student at position 3 to position 4 using 1 move.
  In total, 1 + 1 = 2 moves are used.

Example 2:

Input: seats = [4,1,5,9], students = [1,3,2,6]

Output: 7

Explanation:

• 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.

Constraints:

• n == seats.length == students.length
• 1 <= n <= 100
• 1 <= seats[i], students[j] <= 100

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1. Greedy + Sorting

class Solution:
    def minMovesToSeat(self, seats: List[int], students: List[int]) -> int:
        seats.sort()
        students.sort()

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

Time & Space Complexity

• Time complexity: $O(n \log n)$
• Space complexity: $O(1)$ or $O(n)$ depending on the sorting algorithm.

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2. Counting Sort

class Solution:
    def minMovesToSeat(self, seats: List[int], students: List[int]) -> int:
        max_index = max(max(seats), max(students)) + 1
        count_seats = [0] * max_index
        count_students = [0] * max_index

        for seat in seats:
            count_seats[seat] += 1
        for student in students:
            count_students[student] += 1
        
        i = j = res = 0
        remain = len(seats)
        while remain:
            if count_seats[i] == 0:
                i += 1
            if count_students[j] == 0:
                j += 1
            if count_seats[i] and count_students[j]:
                res += abs(i - j)
                count_seats[i] -= 1
                count_students[j] -= 1
                remain -= 1
        return res

Time & Space Complexity

• Time complexity: $O(n + m1 + m2)$
• Space complexity: $O(m1 + m2)$

Where $n$ is the size of the input arrays, $m1$ is the maximum value in the array $seats$, and $m2$ is the maximum value in the array $students$.

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3. Counting Sort (Optimal)

class Solution:
    def minMovesToSeat(self, seats: List[int], students: List[int]) -> int:
        count_seats = [0] * (max(seats) + 1)
        count_students = [0] * (max(students) + 1)

        def count_sort(arr, count):
            for num in arr:
                count[num] += 1

        count_sort(seats, count_seats)
        count_sort(students, count_students)

        remain = len(seats)
        i = j = res = 0
        while remain:
            if count_seats[i] == 0:
                i += 1
            if count_students[j] == 0:
                j += 1
            if count_seats[i] and count_students[j]:
                tmp = min(count_seats[i], count_students[j])
                res += abs(i - j) * tmp
                count_seats[i] -= tmp
                count_students[j] -= tmp
                remain -= tmp
        return res

Time & Space Complexity

• Time complexity: $O(n + m1 + m2)$
• Space complexity: $O(m1 + m2)$

Where $n$ is the size of the input arrays, $m1$ is the maximum value in the array $seats$, and $m2$ is the maximum value in the array $students$.