Leetcode #2037: Minimum Number of Moves to Seat Everyone
In this guide, we solve Leetcode #2037 Minimum Number of Moves to Seat Everyone in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Greedy, Array, Counting Sort, Sorting
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
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.
Python Solution
class Solution:
def minMovesToSeat(self, seats: List[int], students: List[int]) -> int:
seats.sort()
students.sort()
return sum(abs(a - b) for a, b in zip(seats, students))
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.