Cinema Seat Allocation
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:
class Solution {
public int maxNumberOfFamilies(int n, int[][] reservedSeats) {
Map<Integer, Set<Integer>> reserved = new HashMap();
for (int[] r : reservedSeats) {
reserved.computeIfAbsent(r[0], k -> new HashSet());
reserved.get(r[0]).add(r[1]);;
}
// System.out.println(reserved);
int result = 0;
for (Map.Entry<Integer, Set<Integer>> entry : reserved.entrySet()) {
Set<Integer> r = entry.getValue();
boolean one = true, two = true, three = true;
for (int j = 2; j <= 5; j ++) {
if (r.contains(j)) {
one = false;
}
}
for (int j = 4; j <= 7; j ++) {
if (r.contains(j)) {
two = false;
}
}
for (int j = 6; j <= 9; j ++) {
if (r.contains(j)) {
three = false;
}
}
// System.out.println(i + ", " + one + ", " + two + ", " + three);
if (one && three) {
result += 2;
} else if (two) {
result += 1;
} else if (one || three) {
result += 1;
}
}
result += (n - reserved.size()) * 2;
return result;
}
}