In algorithmic problem-solving, finding an optimal solution is often about understanding the given scenario and applying the right logic. One such popular problem is “Maximize Distance to Closest Person.” This problem is commonly encountered on coding platforms like LeetCode, and it challenges programmers to implement a solution that simulates real-life scenarios of sitting in a row of seats.

The essence of this problem is to select a seat in such a way that the person sitting in that seat has the maximum possible distance from the closest occupied seat. This simple-sounding problem becomes trickier when you start considering various cases like empty seats at the beginning or end of the row or large gaps between occupied seats.

In this blog, we will break down the problem, provide an in-depth analysis, and walk through a solution step-by-step. We will also provide a reference to the corresponding LeetCode problem and discuss optimization techniques.

Problem Statement

You are given an array seats where seats[i] is 0 if the seat is empty, and seats[i] is 1 if the seat is occupied. Your task is to determine the maximum possible distance to the closest person if you choose an empty seat.

LeetCode Problem: Maximize Distance to Closest Person

Understanding the Problem

The problem is set up as a row of seats where some seats are occupied and others are empty. The goal is to find an empty seat that maximizes the distance to the closest person. If there are multiple empty seats that satisfy this condition, any of them will be an acceptable solution.

Real-World Scenario

Imagine you walk into a movie theater, and there are already people sitting in some of the seats. You want to sit down in such a way that you are as far away from other people as possible. This problem can be viewed in this context, where finding the best seat becomes a matter of maximizing your personal space.

Examples

Example 1:

Input: seats = [1, 0, 0, 0, 1, 0, 1]
Output: 2
Explanation: 
- If you sit in seat index 2, you will be 2 seats away from both occupied seats at indices 0 and 4. 
- Any other seat choice results in a smaller distance to the closest person.

Example 2:

Input: seats = [1, 0, 0, 0]
Output: 3
Explanation: 
- You can sit at index 3, where you will be 3 seats away from the person at index 0.

Example 3:

Input: seats = [0, 1]
Output: 1
Explanation: 
- If you sit in seat index 0, the closest person is at index 1, and the distance is 1 seat.

Breaking Down the Problem

To solve the problem, we need to focus on three key areas:

  1. Empty Seats at the Beginning of the Row:
  • If there are empty seats at the start of the row before the first occupied seat, we can choose the seat that’s farthest from the first occupied seat.
  1. Empty Seats at the End of the Row:
  • Similarly, if there are empty seats at the end of the row after the last occupied seat, the distance to the closest person is the number of empty seats before the end.
  1. Empty Seats Between Two Occupied Seats:
  • The more complex part is dealing with empty seats that lie between two occupied seats. The best seat will be in the middle of the empty gap, and the distance to the closest person will be half of the number of empty seats in that gap.

Approach: Linear Scan with Two-Pointer Technique

The optimal solution involves making a single pass through the array to determine the maximum distance to the closest person. We’ll use a two-pointer technique, where we keep track of the position of the last occupied seat. Here are the steps to solve the problem:

  1. Initial Setup:
  • We traverse the array from left to right.
  • We keep track of the position of the last occupied seat and calculate the distance for each empty seat.
  1. Handle Empty Seats at the Beginning:
  • If there are empty seats at the beginning of the row, the distance will be the position of the first occupied seat. For example, if seats = [0, 0, 1], the best seat is at index 0 with a distance of 2.
  1. Handle Empty Seats Between Two Occupied Seats:
  • For empty seats between two occupied seats, the best seat is in the middle of the gap. For example, if seats = [1, 0, 0, 1], the best seat is at index 2, with a distance of 1 (because the gap is two seats wide, and you want to sit halfway).
  1. Handle Empty Seats at the End:
  • If there are empty seats at the end of the row, the distance is equal to the number of seats from the last occupied seat to the end.

By systematically checking all these cases, we can determine the maximum possible distance.

Code Implementation

Here’s a Python implementation of the solution:

def maxDistToClosest(seats):
    n = len(seats)
    prev_occupied = -1  # Track the last occupied seat
    max_distance = 0  # Maximum distance to closest person

    for i in range(n):
        if seats[i] == 1:
            if prev_occupied == -1:
                # Handle empty seats at the beginning
                max_distance = i
            else:
                # Handle empty seats between two occupied seats
                max_distance = max(max_distance, (i - prev_occupied) // 2)
            prev_occupied = i

    # Handle empty seats at the end
    max_distance = max(max_distance, n - 1 - prev_occupied)

    return max_distance

Explanation of the Code:

  • Initial Setup: We initialize prev_occupied to -1, which helps us track the last seated person’s index. max_distance keeps track of the maximum distance encountered during the scan.
  • Traversing the Array:
  • As we iterate through the array, if we encounter an occupied seat (value 1), we compute the distance for the seats between this occupied seat and the previous one (prev_occupied).
  • If prev_occupied is still -1, it means we’re still processing the seats before the first occupied seat, so we calculate the distance from the first empty seat to this occupied seat.
  • Edge Cases: The code handles empty seats at the beginning and end of the row, which are critical edge cases for this problem.

Time Complexity Analysis

The time complexity of this solution is O(n), where n is the number of seats. This is because we only traverse the array once, performing constant-time operations for each element.

The space complexity is O(1) since we only use a few extra variables (prev_occupied and max_distance) to keep track of the current state during traversal.

Edge Cases and Challenges

  1. All Seats are Occupied:
  • If all seats are occupied, there’s no available seat for a person to sit, and the problem doesn’t explicitly handle this case. However, our implementation naturally returns a distance of 0 in such scenarios, which can be considered a valid result.
  1. Multiple Gaps:
  • In a row with multiple gaps between occupied seats, the algorithm will correctly calculate the middle point of each gap and return the maximum possible distance for the best seat.
  1. Single Occupied Seat at the Start or End:
  • If there’s only one occupied seat at either the start or the end of the row, the algorithm efficiently calculates the distance for the remaining empty seats.

Optimization Techniques

Our current solution works in linear time, which is optimal for most cases. However, there are certain situations where further optimizations could be applied based on specific constraints:

  1. Dynamic Seating Arrangement:
  • If seats are frequently added or removed (e.g., in a theater booking system), maintaining a list of occupied seats and recalculating the maximum distance dynamically could be more efficient.
  1. Multiple Queries:
  • If the problem involves multiple queries asking for the best seat repeatedly, a preprocessing step to store the distances between occupied seats could make querying faster.

The “Maximize Distance to Closest Person” problem presents a fascinating challenge that requires balancing edge cases and ensuring efficient traversal through the array. Our solution, using a simple linear scan and handling various seat configurations, ensures that we find the optimal seat in terms of maximizing the distance to the closest person.

The problem is commonly seen on platforms like LeetCode, and understanding it provides valuable insight into how to handle real-world scenarios of dynamic seating arrangements. The linear time complexity ensures that the algorithm scales well for large inputs, and edge cases such as empty seats at the beginning or end of the row are efficiently handled.

LeetCode Reference: Maximize Distance to Closest Person

This problem teaches essential concepts such as array traversal, handling edge cases, and optimizing for distance calculations—skills that are valuable across many coding challenges and

practical applications.

FAQs: Maximize Distance to Closest Person

1. What is the “Maximize Distance to Closest Person” problem?

The “Maximize Distance to Closest Person” problem involves finding the best empty seat in a row where some seats are occupied, such that the distance to the nearest occupied seat is maximized. The goal is to maximize your personal space by sitting as far away as possible from others.

2. Where can I find the “Maximize Distance to Closest Person” problem?

This problem is available on LeetCode, a popular online coding platform. You can find it under the problem name: Maximize Distance to Closest Person.

3. What approach is used to solve this problem efficiently?

The optimal solution uses a linear scan with a two-pointer technique. The array is traversed once to calculate the distance from empty seats to the closest occupied seat. This results in an efficient O(n) time complexity solution.

4. How does the algorithm handle seats at the beginning or end of the row?

The algorithm checks for empty seats at both the start and the end of the row. If there are empty seats before the first occupied seat or after the last occupied seat, the distance to the closest occupied seat is calculated based on the number of empty seats in those sections.

5. What happens if there are multiple gaps between occupied seats?

In cases where there are multiple gaps between occupied seats, the algorithm identifies the middle of each gap and calculates the distance to the closest person. The maximum possible distance from any gap is then chosen.

6. What is the time complexity of the solution?

The solution has a time complexity of O(n), where n is the length of the array. This means the array is traversed only once, making the solution efficient even for large input sizes.

7. What edge cases should I consider when solving this problem?

Edge cases include:

  • All seats are occupied.
  • All seats are empty (not explicitly covered in the problem but important to handle logically).
  • Empty seats only at the beginning or end of the row.
  • Large gaps between occupied seats.

8. What is the difference between optical and logical distance in this problem?

Optical distance refers to the physical seats in the array, while logical distance is the count of empty seats between two occupied seats. The solution focuses on logical distance to determine the best possible seat.

9. Can this algorithm be applied to dynamic seating arrangements?

Yes, but in a dynamic setup where people are constantly moving, you may need to update the distances more frequently. Optimizations such as maintaining a list of occupied seats or recalculating distances when seats change can be applied.

10. What is the significance of this problem in real-world scenarios?

The “Maximize Distance to Closest Person” problem simulates real-life scenarios like seating arrangements in theaters or conferences where people want to maximize personal space. The algorithmic approach teaches important concepts in efficient array traversal and distance calculation, applicable in other practical domains.

Also Read: Two Sum II – Input Array Is Sorted | Two Sum II – Input Array Is Sorted LeetCode Solution in Java

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