1386

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 families you can allocate on the cinema seats. A four-person family occupies fours seats in one row, that are next to each other. Seats across an aisle (such as [3,3] and [3,4]) are not considered to be next to each other, however, It is permissible for the four-person family to be separated by an aisle, but in that case, exactly two people have to sit on each side of the aisle.

 

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 families, where seats mark with blue are already reserved and contiguous seats mark with orange are for one family. 

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.

Related Topics:
Array, Greedy

Solution 1.

// OJ: https://leetcode.com/problems/cinema-seat-allocation/
// Author: github.com/lzl124631x
// Time: O(SlogS)
// Space: O(1)
class Solution {
public:
    int maxNumberOfFamilies(int n, vector<vector<int>>& A) {
        sort(A.begin(), A.end());
        int ans = 2 * n, i = 0, N = A.size();
        while (i < N) {
            vector<int> v(4,1);
            int j = i;
            for (;j < N && A[i][0] == A[j][0];++j) {
                if (A[j][1] == 1 || A[j][1] == 10) continue;
                v[A[j][1]/2-1] = 0;
            }
            int c = 0;
            if (v[0] && v[1]) ++c;
            if (v[2] && v[3]) ++c;
            if (!c && v[1] && v[2]) ++c;
            i = j;
            ans -= 2 - c;
        }
        return ans;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/cinema-seat-allocation/
// Author: github.com/lzl124631x
// Time: O(S)
// Space: O(S)
class Solution {
public:
    int maxNumberOfFamilies(int n, vector<vector<int>>& A) {
        int ans = 2 * n;
        unordered_map<int, char> m;
        for (auto &s : A) {
            if (s[1] == 1 || s[1] == 10) continue;
            m[s[0]] |= 1 << (s[1] - 2);
        }
        for (auto &p : m) {
            bool a = !(p.second & 0b11110000);
            bool b = !(p.second & 0b00111100);
            bool c = !(p.second & 0b00001111);
            if (a && c) continue;
            if (a || b || c) --ans;
            else ans -= 2;
        }
        return ans;
    }
};