Minimum Number of Moves to Seat Everyone

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ArraysGreedy Algorithms

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 i^th seat. You are also given the array students of length n, where students[j] is the position of the j^th student.

You may perform the following move any number of times:

Increase or decrease the position of the i^th student by 1 (i.e., moving the i^th 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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the values in the `seats` and `students` arrays? Can they be negative?
Can the `seats` and `students` arrays have different lengths? If so, what should I return if the arrays have different lengths?
Are the elements in the `seats` and `students` arrays guaranteed to be integers?
Can there be duplicate values within the `seats` array or the `students` array, and if so, how should those be handled?
What should I return if either `seats` or `students` are null or empty?

Brute Force Solution

Approach

The brute force approach to this seating problem involves trying every single possible way to match students to seats. We figure out the total moves needed for each possible seating arrangement. Then we pick the seating arrangement that needs the fewest moves.

Here's how the algorithm would work step-by-step:

First, consider one specific arrangement of students in seats.
Calculate how many moves each student needs to make to get to their assigned seat in that particular arrangement.
Add up all of those moves to get the total number of moves for that one specific arrangement.
Now, consider a completely different arrangement of students in the seats.
Again, calculate the total number of moves for this new arrangement.
Keep doing this for absolutely every possible way to arrange the students.
Once you have tried every possible seating arrangement and calculated the total moves for each one, compare all the totals.
The smallest total number of moves represents the best possible seating arrangement, and that's your answer.

Code Implementation

import itertools

def minimum_moves_to_seat_everyone_brute_force(student_positions, seat_positions):
    number_of_students = len(student_positions)
    number_of_seats = len(seat_positions)

    if number_of_students != number_of_seats:
        return -1

    minimum_total_moves = float('inf')

    # Iterate through all possible permutations of student seating arrangements
    for student_arrangement in itertools.permutations(student_positions):
        total_moves_for_arrangement = 0

        # Calculate the moves needed for this specific arrangement
        for student_index in range(number_of_students):

            # Find the number of moves for one student
            total_moves_for_arrangement += abs(student_arrangement[student_index] - seat_positions[student_index])

        # Keep track of the best seating arrangement
        minimum_total_moves = min(minimum_total_moves, total_moves_for_arrangement)

    return minimum_total_moves

Big(O) Analysis

Time Complexity

O(n!) – The described brute force approach considers every possible arrangement of students in seats. This means we are dealing with permutations. For n students, there are n! (n factorial) possible permutations. Calculating the moves for each permutation involves iterating through all n students to sum up their individual moves. Therefore, the overall time complexity is dominated by generating all n! permutations, making the runtime O(n!).

Space Complexity

O(N!) – The brute force approach explores every possible permutation of students to seats. To generate these permutations, a recursive algorithm implicitly uses a call stack. In the worst-case scenario, where all permutations are generated and stored (even implicitly on the stack for processing), the depth of recursion could reach N, where N is the number of students (or seats). However, the number of possible arrangements that are generated is N! (N factorial), dominating the space requirement. Therefore, this algorithm uses space to hold these permutations, and in the worst case, the space required grows factorially with the input size.

Optimal Solution

Approach

The best way to solve this problem is to match each person to the closest available seat. This minimizes the total distance people need to move. To do this, we'll sort both the people and the seats and match them up in order.

Here's how the algorithm would work step-by-step:

First, arrange the people in order of where they are currently located.
Next, arrange the seats in order of their positions.
Now, match each person to the seat that's in the same position in the sorted lists. This means the first person goes to the first seat, the second person to the second seat, and so on.
For each person-seat pair, calculate the distance between the person's starting position and the seat's location. This distance is how many moves that person needs to make.
Add up all the individual distances to find the total number of moves needed to seat everyone. This total is the minimum number of moves required.

Code Implementation

def min_moves_to_seat(seats, students):
    seats.sort()
    students.sort()

    total_moves = 0

    # Matching students to seats by index
    for i in range(len(seats)):
        # Calculate moves for each student to their assigned seat.
        moves_required = abs(seats[i] - students[i])
        total_moves += moves_required

    return total_moves

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operations are sorting the 'people' and 'seats' arrays, each of size n, which takes O(n log n) time. Iterating through the sorted arrays to calculate the absolute difference between corresponding elements takes O(n) time. Since O(n log n) + O(n) simplifies to O(n log n), the overall time complexity is determined by the sorting step. Therefore, the algorithm's time complexity is O(n log n).

Space Complexity

O(1) – The provided solution sorts the input arrays 'people' and 'seats' in place. While sorting algorithms themselves may have space complexity (e.g., mergesort's O(N)), the description implies using an in-place sorting algorithm like heapsort or quicksort (in the best or average case, after potential optimization to handle worst-case scenarios) or that the language's built-in sort does so. Therefore, the auxiliary space consists primarily of a few variables for the loop index and calculating distances, which are independent of the input size N (where N is the number of people or seats). The space used remains constant.

Edge Cases

Case	How to Handle
Null or empty seats array	Return 0 or throw an IllegalArgumentException if either `seats` or `students` is null or empty, depending on the requirements and language conventions.
Null or empty students array	Return 0 or throw an IllegalArgumentException if either `seats` or `students` is null or empty, depending on the requirements and language conventions.
Seats and students arrays have different lengths	Throw an IllegalArgumentException if the arrays have different lengths as each student needs exactly one seat.
Arrays with a very large number of elements (performance)	Sorting algorithm must be efficient (e.g., O(n log n)) to handle large inputs without exceeding time limits.
Seats or students have the same position (zero move)	The absolute difference calculation correctly accounts for zero moves when student and seat positions are identical.
Large absolute differences between seat and student positions (potential overflow)	Use a 64-bit integer type (long in Java, int64_t in C++) to store the sum of absolute differences to avoid potential integer overflow.
Negative seat or student positions	The absolute difference calculation works correctly with negative numbers, ensuring proper distance calculation.
Extreme values for seat or student positions (near integer limits)	Ensure no intermediate calculations exceed the maximum or minimum values of the used integer type (e.g., check before performing additions to avoid potential overflow).

Null or empty seats array

How to Handle:

Return 0 or throw an IllegalArgumentException if either `seats` or `students` is null or empty, depending on the requirements and language conventions.

Null or empty students array

How to Handle:

Return 0 or throw an IllegalArgumentException if either `seats` or `students` is null or empty, depending on the requirements and language conventions.

Seats and students arrays have different lengths

How to Handle:

Throw an IllegalArgumentException if the arrays have different lengths as each student needs exactly one seat.

Arrays with a very large number of elements (performance)

How to Handle:

Sorting algorithm must be efficient (e.g., O(n log n)) to handle large inputs without exceeding time limits.

Seats or students have the same position (zero move)

How to Handle:

The absolute difference calculation correctly accounts for zero moves when student and seat positions are identical.

Large absolute differences between seat and student positions (potential overflow)

How to Handle:

Use a 64-bit integer type (long in Java, int64_t in C++) to store the sum of absolute differences to avoid potential integer overflow.

Negative seat or student positions

How to Handle:

The absolute difference calculation works correctly with negative numbers, ensuring proper distance calculation.

Extreme values for seat or student positions (near integer limits)

How to Handle:

Ensure no intermediate calculations exceed the maximum or minimum values of the used integer type (e.g., check before performing additions to avoid potential overflow).