README.md

1386. 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.

Solutions (Rust)

1. Solution

use std::collections::HashMap;

impl Solution {
    pub fn max_number_of_families(n: i32, reserved_seats: Vec<Vec<i32>>) -> i32 {
        let mut seats_map = HashMap::new();
        let mut ret = 0;

        for seat in &reserved_seats {
            seats_map
                .entry(seat[0])
                .and_modify(|x| *x |= 1 << seat[1])
                .or_insert(1 << seat[1]);
        }

        for row in seats_map.values() {
            if row & 0x3fc == 0 {
                ret += 2;
            } else if row & 0x3c == 0 {
                ret += 1;
            } else if row & 0xf0 == 0 {
                ret += 1;
            } else if row & 0x3c0 == 0 {
                ret += 1;
            }
        }

        ret += (n - seats_map.len() as i32) * 2;

        ret
    }
}