The “Minimum Number of Moves to Seat Everyone” problem is a fascinating question that revolves around array manipulation and optimization. While the problem may appear simple, achieving the most efficient solution requires a deep understanding of sorting algorithms and mathematical intuition. In this article, we will dissect the problem, look at different ways to solve it, and evaluate their complexities.
1. Problem Statement
The problem asks us to find the minimum number of moves needed to move students so that they can all sit on the seats. Both the
seats are represented as arrays of integers, and the distance between the
i-th student and the
i-th seat is
abs(students[i] - seats[i]).
students = [3,1,5], seats = [2,7,4]
2. Assumptions and Constraints
- The lengths of
seatsarrays are equal, between 1 and 100.
- The position of the students and seats will be integers between 1 and 100.
3. Solution Approaches
A. Brute-Force Approach
One approach is to consider all possible permutations of the
seats array and calculate the moves for each permutation, then return the minimum number of moves. This is, however, highly inefficient with a time complexity of O(n!).
B. Sorting Approach
A more efficient approach is to sort both the
seats arrays first. Then, calculate the absolute difference for the corresponding elements from the two sorted arrays.
def minMovesToSeat(students, seats): students.sort() seats.sort() return sum(abs(students[i] - seats[i]) for i in range(len(students)))
4. Time Complexity Analysis
- Brute-Force Approach: O(n!), where n is the length of the arrays. This is inefficient and not recommended for this problem.
- Sorting Approach: O(n log n), where n is the length of the arrays. This is because the sorting step dominates the overall time complexity.
The “Minimum Number of Moves to Seat Everyone” problem might appear straightforward but requires attention to detail and understanding of sorting algorithms for an efficient solution. By comparing the brute-force approach and the sorting approach, we see a clear winner in terms of efficiency. This problem serves as a great exercise for those looking to deepen their understanding of sorting and optimization problems.