# 1386. Cinema Seat Allocation

## Description

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.

## Solutions

Solution 1: Hash Table + Bit Manipulation

We use a hash table $d$ to store all the reserved seats, where the key is the row number, and the value is the state of the reserved seats in that row, i.e., a binary number. The $j$-th bit being $1$ means the $j$-th seat is reserved, and $0$ means the $j$-th seat is not reserved.

We traverse $reservedSeats$, for each seat $(i, j)$, we add the state of the $j$-th seat (corresponding to the $10-j$ bit in the lower bits) to $d[i]$.

For rows that do not appear in the hash table $d$, we can arrange $2$ families arbitrarily, so the initial answer is $(n - len(d)) \times 2$.

Next, we traverse the state of each row in the hash table. For each row, we try to arrange the situations $1234, 5678, 3456$ in turn. If a situation can be arranged, we add $1$ to the answer.

After the traversal, we get the final answer.

The time complexity is $O(m)$, and the space complexity is $O(m)$. Where $m$ is the length of $reservedSeats$.

• class Solution {
public int maxNumberOfFamilies(int n, int[][] reservedSeats) {
Map<Integer, Integer> d = new HashMap<>();
for (var e : reservedSeats) {
int i = e[0], j = e[1];
d.merge(i, 1 << (10 - j), (x, y) -> x | y);
}
int[] masks = {0b0111100000, 0b0000011110, 0b0001111000};
int ans = (n - d.size()) * 2;
for (int x : d.values()) {
if ((x & mask) == 0) {
++ans;
}
}
}
return ans;
}
}

• class Solution {
public:
int maxNumberOfFamilies(int n, vector<vector<int>>& reservedSeats) {
unordered_map<int, int> d;
for (auto& e : reservedSeats) {
int i = e[0], j = e[1];
d[i] |= 1 << (10 - j);
}
int masks[3] = {0b0111100000, 0b0000011110, 0b0001111000};
int ans = (n - d.size()) * 2;
for (auto& [_, x] : d) {
if ((x & mask) == 0) {
++ans;
}
}
}
return ans;
}
};

• class Solution:
def maxNumberOfFamilies(self, n: int, reservedSeats: List[List[int]]) -> int:
d = defaultdict(int)
for i, j in reservedSeats:
d[i] |= 1 << (10 - j)
ans = (n - len(d)) * 2
for x in d.values():
if (x & mask) == 0:
ans += 1
return ans


• func maxNumberOfFamilies(n int, reservedSeats [][]int) int {
d := map[int]int{}
for _, e := range reservedSeats {
i, j := e[0], e[1]
d[i] |= 1 << (10 - j)
}
ans := (n - len(d)) * 2
for _, x := range d {
ans++
}
}
}
return ans
}

• function maxNumberOfFamilies(n: number, reservedSeats: number[][]): number {
const d: Map<number, number> = new Map();
for (const [i, j] of reservedSeats) {
d.set(i, (d.get(i) ?? 0) | (1 << (10 - j)));
}
let ans = (n - d.size) << 1;
const masks = [0b0111100000, 0b0000011110, 0b0001111000];
for (let [_, x] of d) {
if ((x & mask) === 0) {