leetcode 1386 - Cinema Seat Allocation

https://leetcode.com/problems/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.

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class Solution {
public:
int maxNumberOfFamilies(int n, vector<vector<int>>& reservedSeats) {
unordered_map<int, int> maps;

for (auto v : reservedSeats) {
maps[v[0]] |= (1 << v[1]);
}

int total = 0;
for (auto v : maps) {
int num = v.second;
if ((num & (15 << 2)) == 0) {
total++;
if ((num & (15 << 6)) == 0)
total++;
} else if ((num & (15 << 4)) == 0) {
total++;
} else if ((num & (15 << 6)) == 0) {
total++;
}
}

total += (n - maps.size()) * 2;

return total;
}
};