Maximize Distance to Closest Person

Tested Python solution for LeetCode 849 with 20 pytest cases. Generate a practice environment with lcpy.

LeetCode 849, Medium. Topics: Array. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 20 parametrized pytest cases, and a playground notebook:

lcpy gen -n 849   # by problem number
lcpy gen -s maximize_distance_to_closest_person   # by problem name

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Problem

You are given an array representing a row of seats where seats[i] = 1 represents a person sitting in the i<sup>th</sup> seat, and seats[i] = 0 represents that the i<sup>th</sup> seat is empty (0-indexed).

There is at least one empty seat, and at least one person sitting.

Alex wants to sit in the seat such that the distance between him and the closest person to him is maximized.

Return that maximum distance to the closest person.

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Examples

Input: seats = [1,0,0,0,1,0,1]
Output: 2
Explanation:
If Alex sits in the second open seat (i.e. seats[2]), then the closest person has distance 2.
If Alex sits in any other open seat, the closest person has distance 1.
Thus, the maximum distance to the closest person is 2.

Input: seats = [1,0,0,0]
Output: 3
Explanation:
If Alex sits in the last seat (i.e. seats[3]), the closest person is 3 seats away.
This is the maximum distance possible, so the answer is 3.

Input: seats = [0,1]
Output: 1

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Constraints

• 2 <= seats.length <= 2 * 10^4
• seats[i] is 0 or 1.
• At least one seat is empty.
• At least one seat is occupied.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(n)
    # Space: O(1)
    def max_dist_to_closest(self, seats: list[int]) -> int:
        best = 0
        prev = -1
        n = len(seats)
        for i, occupied in enumerate(seats):
            if occupied:
                best = i if prev < 0 else max(best, (i - prev) // 2)
                prev = i
        return max(best, n - 1 - prev)

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Complexity

Time	Space
O(n)	O(1)

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