Problem Description

In this problem, we are provided with an array seats that represents a row of seats. Each seat can be either occupied (denoted by 1) or empty (denoted by 0). We know that there is at least one empty seat and at least one occupied seat in the array. The objective is to find a seat for Alex where, if he sits down, the distance to the nearest person already seated is as large as possible. The distance is defined as the number of seats between Alex's seat and the closest seated person, and we need to maximize this distance to ensure maximum possible personal space for Alex. Our goal is to calculate and return this maximum distance.

Intuition

The intuition behind the solution is to identify the longest stretch of consecutive empty seats, with consideration of the edge cases where the longest stretch might actually start at the beginning of the row or end at the last seat in the row.

Here are key insights to form a strategy:

1. Identify stretches of empty seats and find the longest stretch. This corresponds to finding the largest gap between two occupied seats.

2. Special consideration for the first and last seats: If the first or last seats are empty, the gap is counted from the edge of the row to the first or last occupied seat respectively. This needs special handling since the "distance" in this case is not halved—it is taken as the entire length of the stretch.

3. Find the longest stretch: To maximize the distance to the nearest person, Alex can sit exactly in the middle of the longest stretch of empty seats. The maximum distance to the closest person would then be half the length of this longest stretch. However, if this stretch is at the beginning or end of the row, he will sit at the first or last available seat, resulting in the distance being the entire length of this stretch.

The approach to solving the problem involves iterating through the seats array to identify the first and last occupied seats, and in the process, also recording the maximum distance between two consecutive occupied seats. The maximum distance Alex can sit away from the closest person would then be the maximum of the following three values:

• The distance from the start of the row to the first occupied seat.
• The distance from the last occupied seat to the end of the row.
• Half the distance of the longest stretch of empty seats between two occupied seats (since Alex would sit exactly in the middle of this stretch).

This method efficiently calculates the optimal seat for Alex in a single pass through the seats array, avoiding the need for multiple iterations or complex calculations.

Solution Approach

The solution approach involves a single pass through the row of seats using a for-loop. During this pass, the code keeps track of the index of the first and last occupied seats, as well as the distance to the last occupied seat seen so far when examining each seat. The following elements contribute to the solution:

Variables Used:

• first: This holds the index of the first occupied seat found in the row. Initially, it is set to None.
• last: This stores the index of the most recently found occupied seat as we iterate through the seats. It is also initially set to None.
• d: This variable maintains the maximum distance between two occupied seats as the iteration progresses. We initialize it to 0.

The Algorithm:

1. Iterate through the seats array with the index i and the value c using the enumerate function, which provides both the index and value during the iteration.

2. Check if the current seat c is occupied (i.e., c is 1):

   • If an occupied seat is found and it's not the first one (last is not None), update d with the maximum value between the current d and the difference between the current index i and the index of the last occupied seat last.
   • If this is the first occupied seat found (first is None), store its index in first.
   • Update last with the current index i because it is the latest occupied seat.

3. After completing the iteration, calculate the maximum distance Alex can be from the closest person by taking the maximum of three potential values:

   • The distance from the start of the row to the first occupied seat (first).
   • The distance from the last occupied seat to the end of the row (len(seats) - last - 1).
   • Half the distance of the longest stretch of empty seats between two occupied seats (d // 2).

The Code:

class Solution:
    def maxDistToClosest(self, seats: List[int]) -> int:
        first = last = None
        d = 0
        for i, c in enumerate(seats):
            if c:
                if last is not None:
                    d = max(d, i - last)
                if first is None:
                    first = i
                last = i
        return max(first, len(seats) - last - 1, d // 2)

This implementation uses a greedy approach to find the maximum distance in an efficient manner. No additional data structure is required aside from a few variables to keep track of indices and the maximum distance found. Because of this, the space complexity is O(1), and since we are iterating through the array only once, the time complexity is O(n), where n is the number of seats in the row.

Example Walkthrough

Imagine we have an array seats that looks like this: [0, 1, 0, 0, 1, 0, 1, 0, 0, 0].

To apply our solution approach, let's walk through the algorithm step by step:

1. Initialize: first = last = None and d = 0.
2. Begin iterating through the seats array with indices and values.
3. At index 0, the seat is empty, so no changes are made.
4. At index 1, the seat is occupied so:
   • Since this is the first occupied seat we've found, set first to the current index (first = 1).
   • Update last to be the current index as well (last = 1).
5. Continue iterating. Seats 2 and 3 are empty, so no changes.
6. At index 4, we encounter another occupied seat.
   • Since last is not None, we calculate the distance from the previous occupied seat at index 1 to the current seat at index 4, which yields a distance of 3. Update d to be the maximum of 0 and 3, so d = 3.
   • Update last to the current index (last = 4).
7. Move on to seat 5 which is empty, so no changes.
8. At index 6, there is another occupied seat.
   • Again, calculate the distance (1) and compare it with d. Since 1 is less than 3, we do not update d.
   • Update last to 6.
9. Seats 7, 8, and 9 are empty, so no changes occur during these steps.

Once we’ve finished iterating through the array, we consider the maximum distance from the following:

• The distance from the start of the row to the first occupied seat (first = 1).
• The distance from the last occupied seat to the end of the row (len(seats) - last - 1 = 10 - 6 - 1 = 3).
• Half the distance of the longest stretch of empty seats between two occupied seats (d // 2 = 3 // 2 = 1).

The maximum of these values is 3, which is accounted for by both the distance from the last occupied seat to the end of the row and from the first seat in the row to the first occupied seat. Alex can choose either seat at the start or the end of the row for the maximum distance to the closest person, hence we return 3 as the result.

Here's the walkthrough reflected in the provided code:

class Solution:
    def maxDistToClosest(self, seats: List[int]) -> int:
        first = last = None
        d = 0
        for i, c in enumerate(seats):
            if c:
                if last is not None:
                    d = max(d, i - last)  # At index 4 and index 1, d becomes 3
                if first is None:
                    first = i  # At index 1, first becomes 1
                last = i  # Updates at indices 1, 4, and 6
        return max(first, len(seats) - last - 1, d // 2)  # Returns max(1, 3, 1) which is 3

By iterating through the array only once and using a simple set of variables to keep track, we find the maximum distance where Alex can sit with a time complexity of O(n) and a space complexity of O(1).

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(n), where n is the length of the seats list. This is derived from the fact that there is a single for loop that iterates over each element of the seats list exactly once to find the maximum distance to the closest person.

Space Complexity

The space complexity of the code is O(1) which means it uses constant extra space. This is because the variables first, last, and d are used to keep track of positions and distance within the algorithm, regardless of the input size. The list itself is not copied or modified, and no additional data structures are utilized that grow with the input size.

Learn more about how to find time and space complexity quickly using problem constraints.