Problem
There are n
seats and n
students 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 by1
(i.e., moving theith
student from positionx
tox + 1
orx - 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.
Examples
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 from position 2 to position 1 using 1 move.
- The second student is moved from from position 7 to position 5 using 2 moves.
- The third student is moved from 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 from position 3 to position 4 using 1 move.
- The third student is moved from from position 2 to position 5 using 3 moves.
- The fourth student is moved from 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 from position 1 to position 2 using 1 move.
- The second student is moved from from position 3 to position 6 using 3 moves.
- The fourth student is not moved.
In total, 1 + 3 + 0 + 0 = 4 moves were used.
Solution
Method 1 - Sorting
In each step we want to find a student , closest to a position , so that student have to move minimum positions.
We sort both the arrays , so that we can get the minimum positions as well as the students nearest to the minimum position.
Here is the video explanation:
Code
Java
class Solution {
public int minMovesToSeat(int[] seats, int[] students) {
Arrays.sort(seats);
Arrays.sort(students);
int ans = 0;
for (int i = 0; i < seats.length; i++) {
ans += Math.abs(seats[i] - students[i]);
}
return ans;
}
}
Complexity
- ⏰ Time complexity:
O(n log n)
- 🧺 Space complexity:
O(1)