LeetCode 603: Consecutive Available Seats Solution

Master LeetCode problem 603 (Consecutive Available Seats), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

603. Consecutive Available Seats

Easy

Database

Problem Explanation

Explanation:

To solve this problem, we can iterate through the reserved seats and track the available seats between each reserved seat. We can then calculate the maximum number of consecutive available seats in each row.

Sort the reserved seats by row and seat number.
For each row, calculate the number of consecutive available seats between reserved seats.
Update the maximum consecutive available seats seen so far.

Time complexity: O(nlogn) where n is the number of reserved seats. Sorting the reserved seats takes O(nlogn) time. Space complexity: O(n) for storing the reserved seats.

: :

Solution Code

class Solution {
    public int maxNumberOfFamilies(int n, int[][] reservedSeats) {
        Arrays.sort(reservedSeats, (a, b) -> a[0] - b[0]);
        
        int maxConsecutiveSeats = 2 * n;
        HashMap<Integer, Integer> rowMap = new HashMap<>();
        
        for (int[] seat : reservedSeats) {
            int row = seat[0];
            int seatNum = seat[1];
            
            int mask = (1 << (seatNum - 1));
            rowMap.put(row, rowMap.getOrDefault(row, 0) | mask);
        }
        
        for (int row : rowMap.keySet()) {
            int occupied = rowMap.get(row);
            int consecutiveSeats = 0;
            
            if ((occupied & 30) == 0) {
                consecutiveSeats = 2;
            } else if ((occupied & 120) == 0 || (occupied & 480) == 0 || (occupied & 1920) == 0) {
                consecutiveSeats = 1;
            }
            
            maxConsecutiveSeats -= consecutiveSeats;
        }
        
        return maxConsecutiveSeats;
    }
}

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