2286. Booking Concert Tickets in Groups

Explanation:

To solve this problem, we can maintain a data structure to keep track of the available seats in each row. When a group wants to gather together, we can check row by row for the first available consecutive seats that satisfy the constraints. When a group wants to scatter, we can check row by row for the first available non-consecutive seats that satisfy the constraints.

Algorithm:

Initialize a data structure to track available seats in each row.
For the gather method:
- Iterate over each row and find the first row with k consecutive available seats within the maxRow limit.
- If found, mark those seats as occupied and return the row and seat number.
- If no such seats are available, return an empty array.
For the scatter method:
- Iterate over each row and find the first row with k available seats within the maxRow limit.
- If found, mark those seats as occupied and return true.
- If no such seats are available, return false.

Time Complexity:

For the gather method, the time complexity is O(n*m) where n is the number of rows and m is the number of seats per row.
For the scatter method, the time complexity is also O(n*m).

Space Complexity:

The space complexity is O(n*m) to store the status of each seat in the concert hall.

:

View Leetcode Problem

class BookMyShow {
    int rows, seats;
    boolean[][] hall;
    
    public BookMyShow(int n, int m) {
        rows = n;
        seats = m;
        hall = new boolean[n][m];
    }
    
    public int[] gather(int k, int maxRow) {
        for (int i = 0; i <= maxRow; i++) {
            int count = 0;
            int startSeat = -1;
            for (int j = 0; j < seats; j++) {
                if (!hall[i][j]) {
                    if (startSeat == -1) {
                        startSeat = j;
                    }
                    count++;
                } else {
                    startSeat = -1;
                    count = 0;
                }
                if (count == k) {
                    for (int x = startSeat; x < startSeat + k; x++) {
                        hall[i][x] = true;
                    }
                    return new int[]{i, startSeat};
                }
            }
        }
        return new int[]{};
    }
    
    public boolean scatter(int k, int maxRow) {
        for (int i = 0; i <= maxRow; i++) {
            int count = 0;
            for (int j = 0; j < seats; j++) {
                if (!hall[i][j]) {
                    count++;
                    if (count == k) {
                        for (int x = j - k + 1; x <= j; x++) {
                            hall[i][x] = true;
                        }
                        return true;
                    }
                } else {
                    count = 0;
                }
            }
        }
        return false;
    }
}

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