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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.
Solution 1.
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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; } } } ############ class Solution { public int maxNumberOfFamilies(int n, int[][] reservedSeats) { Map<Integer, Integer> m = new HashMap<>(); for (int[] e : reservedSeats) { int i = e[0], j = 10 - e[1]; int v = m.getOrDefault(i, 0); v |= 1 << j; m.put(i, v); } int[] masks = {0b0111100000, 0b0000011110, 0b0001111000}; int ans = (n - m.size()) << 1; for (int v : m.values()) { for (int mask : masks) { if ((v & mask) == 0) { v |= mask; ++ans; } } } return ans; } }
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// 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; } };
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class Solution: def maxNumberOfFamilies(self, n: int, reservedSeats: List[List[int]]) -> int: m = defaultdict(int) for i, j in reservedSeats: m[i] = m[i] | (1 << (10 - j)) masks = (0b0111100000, 0b0000011110, 0b0001111000) ans = (n - len(m)) << 1 for v in m.values(): for mask in masks: if (v & mask) == 0: v |= mask ans += 1 return ans
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func maxNumberOfFamilies(n int, reservedSeats [][]int) int { m := map[int]int{} for _, e := range reservedSeats { i, j := e[0], 10-e[1] m[i] |= 1 << j } masks := []int{0b0111100000, 0b0000011110, 0b0001111000} ans := (n - len(m)) << 1 for _, v := range m { for _, mask := range masks { if (v & mask) == 0 { v |= mask ans++ } } } return ans }
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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) { for (const mask of masks) { if ((x & mask) === 0) { x |= mask; ++ans; } } } return ans; }