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This approach relies on sorting both the seats and students arrays. After sorting them, we pair the i-th seat with the i-th student. The sum of absolute differences between corresponding elements gives us the minimum number of moves required.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) since sorting is in-place.
1import java.util.Arrays;
2
3public class Solution {
4 public int minMovesToSeat(int[] seats, int[] students) {
5 Arrays.sort(seats);
6 Arrays.sort(students);
7 int result = 0;
8 for (int i = 0; i < seats.length; i++) {
9 result += Math.abs(seats[i] - students[i]);
10 }
11 return result;
12 }
13 public static void main(String[] args) {
14 Solution sol = new Solution();
15 int[] seats = {3, 1, 5};
16 int[] students = {2, 7, 4};
17 System.out.println(sol.minMovesToSeat(seats, students));
18 }
19}We use Arrays.sort() to sort both arrays, then calculate the sum of absolute differences between paired elements of the arrays, giving us the minimum moves required.
This approach involves creating frequency arrays or maps for both seats and students positions. Align students to seats by finding the closest available seat iteratively, simulating a greedy approach.
Time Complexity: O(n + m), where m is the range of possible positions (maximum 100).
Space Complexity: O(m) due to frequency arrays.
1
The function mobilizes frequency arrays and aligns adjacent iterables to resolve optimally close seat-to-student exchanges, ensuring minimal repositionment steps are delineated.