LeetCode #849 — MEDIUM

Maximize Distance to Closest Person

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array representing a row of seats where seats[i] = 1 represents a person sitting in the i^th seat, and seats[i] = 0 represents that the i^th 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.

Example 1:

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.

Example 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.

Example 3:

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

Constraints:

2 <= seats.length <= 2 * 10⁴
seats[i] is 0 or 1.
At least one seat is empty.
At least one seat is occupied.

Step 01

Brute Force Baseline

Problem summary: You are given an array representing a row of seats where seats[i] = 1 represents a person sitting in the ith seat, and seats[i] = 0 represents that the ith 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[1,0,0,0,1,0,1]

Example 2

[1,0,0,0]

Example 3

[0,1]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #849: Maximize Distance to Closest Person
class Solution {
    public int maxDistToClosest(int[] seats) {
        int first = -1, last = -1;
        int d = 0, n = seats.length;
        for (int i = 0; i < n; ++i) {
            if (seats[i] == 1) {
                if (last != -1) {
                    d = Math.max(d, i - last);
                }
                if (first == -1) {
                    first = i;
                }
                last = i;
            }
        }
        return Math.max(d / 2, Math.max(first, n - last - 1));
    }
}

// Accepted solution for LeetCode #849: Maximize Distance to Closest Person
func maxDistToClosest(seats []int) int {
	first, last := -1, -1
	d := 0
	for i, c := range seats {
		if c == 1 {
			if last != -1 {
				d = max(d, i-last)
			}
			if first == -1 {
				first = i
			}
			last = i
		}
	}
	return max(d/2, max(first, len(seats)-last-1))
}

# Accepted solution for LeetCode #849: Maximize Distance to Closest Person
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)

// Accepted solution for LeetCode #849: Maximize Distance to Closest Person
struct Solution;

impl Solution {
    fn max_dist_to_closest(seats: Vec<i32>) -> i32 {
        let mut first: Option<usize> = None;
        let mut last: Option<usize> = None;
        let mut prev: Option<usize> = None;
        let n = seats.len();
        let mut max = 0;
        for i in 0..n {
            if seats[i] == 1 {
                if first.is_none() {
                    first = Some(i);
                }
                if let Some(j) = prev {
                    max = usize::max((i - j) / 2, max);
                }
                prev = Some(i);
                last = Some(i);
            }
        }
        max = usize::max(first.unwrap(), max);
        max = usize::max(n - 1 - last.unwrap(), max);
        max as i32
    }
}

#[test]
fn test() {
    let seats = vec![1, 0, 0, 0, 1, 0, 1];
    assert_eq!(Solution::max_dist_to_closest(seats), 2);
    let seats = vec![1, 0, 0, 0];
    assert_eq!(Solution::max_dist_to_closest(seats), 3);
}

// Accepted solution for LeetCode #849: Maximize Distance to Closest Person
function maxDistToClosest(seats: number[]): number {
    let first = -1,
        last = -1;
    let d = 0,
        n = seats.length;
    for (let i = 0; i < n; ++i) {
        if (seats[i] === 1) {
            if (last !== -1) {
                d = Math.max(d, i - last);
            }
            if (first === -1) {
                first = i;
            }
            last = i;
        }
    }
    return Math.max(first, n - last - 1, d >> 1);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.