Formatted question description: https://leetcode.ca/all/1386.html

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/

// 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/

// 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;
    }
};

Java

class Solution {
    public int maxNumberOfFamilies(int n, int[][] reservedSeats) {
        Arrays.sort(reservedSeats, new Comparator<int[]>() {
            public int compare(int[] reservedSeat1, int[] reservedSeat2) {
                return reservedSeat1[0] - reservedSeat2[0];
            }
        });
        int families = 0;
        int prevRow = 0;
        Set<Integer> set = new HashSet<Integer>();
        int length = reservedSeats.length;
        for (int i = 0; i < length; i++) {
            int[] reservedSeat = reservedSeats[i];
            int row = reservedSeat[0], column = reservedSeat[1];
            if (column == 1 || column == 10)
                continue;
            if (row == prevRow)
                set.add(column);
            else {
                if (prevRow > 0) {
                    families += maxNumberOfFamilies(set);
                    set.clear();
                }
                set.add(column);
                for (int j = prevRow + 1; j < row; j++)
                    families += 2;
                prevRow = row;
            }
        }
        families += maxNumberOfFamilies(set);
        for (int j = prevRow + 1; j <= n; j++)
            families += 2;
        return families;
    }

    public int maxNumberOfFamilies(Set<Integer> set) {
        if (set.isEmpty())
            return 2;
        boolean flag1 = set.contains(4) || set.contains(5);
        boolean flag2 = set.contains(6) || set.contains(7);
        if (flag1 && flag2)
            return 0;
        else if (!flag1 && !flag2)
            return 1;
        else if (flag1) {
            if (!set.contains(8) && !set.contains(9))
                return 1;
            else
                return 0;
        } else {
            if (!set.contains(2) && !set.contains(3))
                return 1;
            else
                return 0;
        }
    }
}

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