LeetCode-in-Java

1386. Cinema Seat Allocation

Medium

A cinema has n rows of seats, numbered from 1 to n and there are ten seats in each row, labelled from 1 to 10 as shown in the figure above.

Given the array reservedSeats containing the numbers of seats already reserved, for example, reservedSeats[i] = [3,8] means the seat located in row 3 and labelled with 8 is already reserved.

Return the maximum number of four-person groups you can assign on the cinema seats. A four-person group occupies four adjacent seats in one single row. Seats across an aisle (such as [3,3] and [3,4]) are not considered to be adjacent, but there is an exceptional case on which an aisle split a four-person group, in that case, the aisle split a four-person group in the middle, which means to have two people on each side.

Example 1:

Input: n = 3, reservedSeats = [[1,2],[1,3],[1,8],[2,6],[3,1],[3,10]]

Output: 4

Explanation: The figure above shows the optimal allocation for four groups, where seats mark with blue are already reserved and contiguous seats mark with orange are for one group.

Example 2:

Input: n = 2, reservedSeats = [[2,1],[1,8],[2,6]]

Output: 2

Example 3:

Input: n = 4, reservedSeats = [[4,3],[1,4],[4,6],[1,7]]

Output: 4

Constraints:

• 1 <= n <= 10^9
• 1 <= reservedSeats.length <= min(10*n, 10^4)
• reservedSeats[i].length == 2
• 1 <= reservedSeats[i][0] <= n
• 1 <= reservedSeats[i][1] <= 10
• All reservedSeats[i] are distinct.

Solution

import java.util.HashMap;
import java.util.Map;

public class Solution {
    public int maxNumberOfFamilies(int n, int[][] reservedSeats) {
        Map<Integer, int[]> occupiedFamilySeats = new HashMap<>();
        for (int[] reservedSeat : reservedSeats) {
            int row = reservedSeat[0];
            int col = reservedSeat[1];
            if (col == 1 || col == 10) {
                continue;
            }
            int[] rowFamilySeats = occupiedFamilySeats.getOrDefault(row, new int[3]);
            if (col == 2 || col == 3) {
                // mark left family seating as occupied
                rowFamilySeats[0] = 1;
                occupiedFamilySeats.put(row, rowFamilySeats);
            }
            if (col == 8 || col == 9) {
                // mark right family seating as occupied
                rowFamilySeats[2] = 1;
                occupiedFamilySeats.put(row, rowFamilySeats);
            }
            if (col == 4 || col == 5) {
                // mark left family seating as occupied
                rowFamilySeats[0] = 1;
                // mark min family seating as occupied
                rowFamilySeats[1] = 1;
                occupiedFamilySeats.put(row, rowFamilySeats);
            }
            if (col == 6 || col == 7) {
                // mark min family seating as occupied
                rowFamilySeats[1] = 1;
                // mark right family seating as occupied
                rowFamilySeats[2] = 1;
                occupiedFamilySeats.put(row, rowFamilySeats);
            }
        }
        // max number of family seats per row is 2, so we start that minus the rows for which we
        // have reservations
        int count = n * 2 - 2 * occupiedFamilySeats.size();
        // for each row with reservations, count remaining family seatings
        for (int[] familySeats : occupiedFamilySeats.values()) {
            if (familySeats[0] == 0) {
                count++;
            }
            if (familySeats[2] == 0) {
                count++;
            }
            if (familySeats[0] != 0 && familySeats[2] != 0 && familySeats[1] == 0) {
                count++;
            }
        }
        return count;
    }
}

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