What is the time complexity of Maximize Distance to Closest Person?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximize Distance to Closest Person?

This problem uses Array. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximize Distance to Closest Person?

Insight 1: The maximum distance could be at the beginning of the row (before the first person), in the middle of the row (between two people), or at the end of the row (after the last person). Insight 2: We need to keep track of the position of the last person we saw.

Which companies ask LeetCode 849. Maximize Distance to Closest Person in interviews?

This problem is frequently asked at Roblox, Samsung, Snap, Snowflake, Tinkoff and 1 more companies.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Maximize Distance to Closest Person - LeetCode 849 Solution

Maximize Distance to Closest Person - Complete Solution Guide

Maximize Distance to Closest Person is LeetCode problem 849, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to find the maximum distance that Alex can sit from the closest person in a row of seats. The input is an array `seats` where `seats[i] = 1` represents a person sitting in the i-th seat, and `seats[i] = 0` represents an empty seat. We need to find an empty seat for Alex such that the distance to the nearest occupied seat is maximized. There is guaranteed to be at least one empty seat and at least one occupied seat.

Solution Approach

The solution iterates through the `seats` array, keeping track of the index of the last person encountered. When it finds a person, it calculates the distance from the last person (if any) and updates the maximum distance found so far. Special consideration is given to the beginning and the end of the array.

Step-by-Step Algorithm

Step 1: Initialize `max_dist` to 0 and `last_person` to -1 (indicating no person seen yet).
Step 2: Iterate through the `seats` array from left to right.
Step 3: If `seats[i]` is 1 (a person is sitting), check if it's the first person encountered (`last_person == -1`).
Step 4: If it's the first person, the distance is `i` (distance from the beginning). Update `max_dist` with `i` if it's larger than the current `max_dist`.
Step 5: If it's not the first person, calculate the distance between the current person and the last person: `(i - last_person) // 2`. Update `max_dist` if this distance is larger.
Step 6: Update `last_person` to the current index `i`.
Step 7: After the loop, calculate the distance from the last person to the end of the array: `n - 1 - last_person`. Update `max_dist` if this distance is larger.
Step 8: Return `max_dist`.

Key Insights

Insight 1: The maximum distance could be at the beginning of the row (before the first person), in the middle of the row (between two people), or at the end of the row (after the last person).
Insight 2: We need to keep track of the position of the last person we saw.
Insight 3: The distance between two people is divided by 2 (integer division) to find the maximum distance Alex can sit in between them.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: Roblox, Samsung, Snap, Snowflake, Tinkoff, VK.

Maximize Distance to Closest Person | LeetCode 849

Medium

Array

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maxDistToClosest(int[] seats) {
        int n = seats.length;
        int max_dist = 0;
        int last_person = -1;

        for (int i = 0; i < n; i++) {
            if (seats[i] == 1) {
                if (last_person == -1) {
                    max_dist = i;
                } else {
                    int dist = (i - last_person) / 2;
                    max_dist = Math.max(max_dist, dist);
                }
                last_person = i;
            }
        }
        
        int dist_at_end = n - 1 - last_person;
        max_dist = Math.max(max_dist, dist_at_end);

        return max_dist;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 849: Maximize Distance to Closest Person - Detailed explanation and solution approaches.

Topics

Array

Companies

RobloxSamsungSnapSnowflakeTinkoffVK

Stats

Difficulty:Medium

Languages:4

Companies:6

Problem #:849

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `max_dist` to 0 and `last_person` to -1 (indicating no person seen yet).
2
Step 2: Iterate through the `seats` array from left to right.
3
Step 3: If `seats[i]` is 1 (a person is sitting), check if it's the first person encountered (`last_person == -1`).
4
Step 4: If it's the first person, the distance is `i` (distance from the beginning). Update `max_dist` with `i` if it's larger than the current `max_dist`.
5
Step 5: If it's not the first person, calculate the distance between the current person and the last person: `(i - last_person) // 2`. Update `max_dist` if this distance is larger.
6
Step 6: Update `last_person` to the current index `i`.
7
Step 7: After the loop, calculate the distance from the last person to the end of the array: `n - 1 - last_person`. Update `max_dist` if this distance is larger.
8
Step 8: Return `max_dist`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is the single loop through the input array, giving a time complexity of O(n). The space used is independent of the input size because we only use a few integer variables.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Two Pointers

Use two pointers to track the start and end of a contiguous block of empty seats. Iterate through the array and update the maximum distance whenever a block of empty seats is found. This approach still has O(n) time complexity but might be slightly more complex to implement correctly.

Time: O(n)Space: O(1)

Trade-offs: The two-pointer approach offers a different way of thinking about the problem, but in terms of efficiency, it's similar to the provided solution. Might be considered less readable or more error prone for some people.

✨

Key Insights

1Insight 1: The maximum distance could be at the beginning of the row (before the first person), in the middle of the row (between two people), or at the end of the row (after the last person).
2Insight 2: We need to keep track of the position of the last person we saw.
3Insight 3: The distance between two people is divided by 2 (integer division) to find the maximum distance Alex can sit in between them.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: When all empty seats are at the beginning of the row (e.g., [0,0,0,1]). The algorithm correctly identifies the distance as the index of the first person.
Edge case 2: When all empty seats are at the end of the row (e.g., [1,0,0,0]). The algorithm correctly calculates the distance from the last person to the end of the row.
Edge case 3: When there are multiple empty seats between two people. The algorithm correctly calculates half the distance between the people to determine the maximum distance Alex can sit.
Edge case 4: When input is [0,1] or [1,0]. Algorithm returns 1

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge cases at the beginning and end of the array correctly. Forgetting to consider that Alex could sit at either extreme.

✗Common mistake 2: Using floating-point division instead of integer division when calculating the distance between two people. This could lead to incorrect results due to rounding errors or type mismatches.

✗Common mistake 3: Incorrectly initializing `last_person`. It must be initialized to -1 to handle the case where the empty seats are at the beginning of the array.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing placement of emergency exits in a building to maximize the distance from any given point to the nearest exit.

🏦Real-world scenario 2: Determining the best location for a new cell tower to maximize coverage while ensuring no area is too far from a tower.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximize Distance to Closest Person"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?