Leetcode 1386: Cinema Seat Allocation
A cinema has multiple rows, each with ten seats arranged in a line. Some seats are already reserved. You need to determine the maximum number of four-person groups that can be seated in the available seats. A four-person group requires four adjacent seats in one row. If an aisle splits a group in the middle, they can still be seated if there are two consecutive seats on both sides of the aisle.
Problem
Approach
Steps
Complexity
Input: The input consists of an integer n representing the number of rows and a list reservedSeats where each element is a pair [row, seat] indicating a reserved seat in the cinema.
Example: Input: n = 5, reservedSeats = [[1, 1], [1, 5], [2, 3], [4, 7], [5, 2], [5, 9]]
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.
Output: The output should be an integer representing the maximum number of four-person groups that can be assigned to the cinema seats.
Example: Output: 6
Constraints:
• The result should be a single integer indicating the number of groups that can be seated.
Goal: Maximize the number of four-person groups that can be seated in the cinema, given the reserved seats and the constraints on adjacency.
Steps:
• Iterate through the reserved seats and mark the occupied positions in each row.
• For each row, check if there are four consecutive available seats either across the entire row or split by the aisle.
• Count the number of valid groups that can be seated and return the total count.
Goal: The solution must handle very large numbers of rows efficiently, and the number of reserved seats is limited by the input constraints.
Steps:
• Handle up to 10^9 rows of cinema efficiently using memory or computational optimizations.
• The number of reserved seats is at most 10^4, making the row iteration feasible.
Assumptions:
• There will always be enough space for at least one four-person group in some row.
• All the reserved seats are valid and within the range of the cinema's capacity.
• Input: Input: n = 5, reservedSeats = [[1, 1], [1, 5], [2, 3], [4, 7], [5, 2], [5, 9]]
• Explanation: This example shows how the rows can be divided into groups based on the reserved seats. In row 1, four seats from 2 to 5 are available for one group. Similarly, the other rows have groups formed based on available seats.
Approach: To solve this problem efficiently, we need to track the reserved seats for each row and calculate the number of available groups in each row while considering aisle splits.
Observations:
• The problem involves checking adjacency of seats across rows and handling aisle splits efficiently.
• We can use bit manipulation or simple checks to determine the available groups in each row.
Steps:
• 1. Initialize a data structure to track reserved seats for each row.
• 2. For each row, calculate the available space by checking the reserved seats and the potential four-person group configurations.
• 3. Count the maximum groups that can be seated and return the result.
Empty Inputs:
• Handle cases where no reserved seats are given (all seats are available).
Large Inputs:
• Handle the scenario where there are a large number of rows but only a few reserved seats.
Special Values:
• Ensure that edge cases such as having just one row or a row with no available seats are handled properly.
Constraints:
• Ensure the solution handles the large values of n efficiently.
int maxNumberOfFamilies(int n, vector<vector<int>>& reservedSeats) {
int ans = n * 2;
unordered_map<int, char> m;
for(auto r: reservedSeats)
if(r[1] > 1 && r[1] < 10)
m[r[0]] |= 1 << (r[1] - 2);
for(auto [row, seats]: m) {
int left = seats & 0b11110000;
int mid = seats & 0b00111100;
int right = seats & 0b00001111;
ans -= left && mid && right ? 2 : 1;
}
return ans;
}
1 : Function Definition
int maxNumberOfFamilies(int n, vector<vector<int>>& reservedSeats) {
Define the function `maxNumberOfFamilies` that takes the number of rows (n) and the reserved seats information (reservedSeats). The function will return the maximum number of families that can be seated.
2 : Initialization
int ans = n * 2;
Initialize `ans` to `n * 2`, as the maximum number of families is initially assumed to be two per row.
3 : Data Structures
unordered_map<int, char> m;
Declare an unordered map `m` to store the seat occupancy for each row. The key is the row number, and the value is a bitmask representing the reserved seats.
4 : Loop
for(auto r: reservedSeats)
Loop through each reserved seat in the `reservedSeats` vector.
5 : Condition
if(r[1] > 1 && r[1] < 10)
Check if the seat in the current row is reserved and falls within the valid seat range (2 to 9).
6 : Bit Manipulation
m[r[0]] |= 1 << (r[1] - 2);
Use bit manipulation to mark the reserved seat in the map for the corresponding row. This sets the appropriate bit in the bitmask.
7 : Loop
for(auto [row, seats]: m) {
Loop through each row and its corresponding seat bitmask stored in `m`.
8 : Bit Manipulation
int left = seats & 0b11110000;
Extract the left block of seats (seats 2 to 5) by using a bitmask operation.
9 : Bit Manipulation
int mid = seats & 0b00111100;
Extract the middle block of seats (seats 4 to 7) using another bitmask operation.
10 : Bit Manipulation
int right = seats & 0b00001111;
Extract the right block of seats (seats 6 to 9) using a final bitmask operation.
11 : Condition
ans -= left && mid && right ? 2 : 1;
If all three blocks (left, middle, and right) are occupied, subtract 2 from `ans` (two families can't be seated). Otherwise, subtract 1 (one family can't be seated).
12 : Return Statement
return ans;
Return the calculated number of families that can be seated after considering all reserved seats.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The solution requires iterating through each row and checking available seats, which takes linear time with respect to the number of rows.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity depends on the number of rows and the reserved seats, requiring storage for each row's reserved seats.
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