LeetCode in Kotlin

2037. Minimum Number of Moves to Seat Everyone

Easy

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 by 1 (i.e., moving the ith 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 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 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

Solution

class Solution {
    fun minMovesToSeat(seats: IntArray, students: IntArray): Int {
        var ans = 0
        seats.sort()
        students.sort()
        for (i in seats.indices) {
            ans += Math.abs(seats[i] - students[i])
        }
        return ans
    }
}

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