LeetCode 849: Maximize Distance to Closest Person

Explanation:

• We can solve this problem by iterating through the array and finding the maximum distance between consecutive occupied seats or the start/end of the array and an occupied seat.
• We calculate the distance by keeping track of the last index of an occupied seat and updating the maximum distance whenever we find a new occupied seat.
• We handle the edge cases of the first and last seat separately to calculate the distance to the closest person from the start and end respectively.
• Finally, we return the maximum distance calculated.

Time complexity: O(n) where n is the number of seats
Space complexity: O(1)

class Solution {
    public int maxDistToClosest(int[] seats) {
        int maxDistance = 0;
        int lastOccupied = -1;
        
        for (int i = 0; i < seats.length; i++) {
            if (seats[i] == 1) {
                if (lastOccupied == -1) {
                    maxDistance = i;
                } else {
                    maxDistance = Math.max(maxDistance, (i - lastOccupied) / 2);
                }
                lastOccupied = i;
            }
        }
        
        return Math.max(maxDistance, seats.length - 1 - lastOccupied);
    }
}