Problem Description

You are given n points on a 2D plane, where each point is represented as points[i] = [xi, yi] with xi being the x-coordinate and yi being the y-coordinate.

The task is to find the widest vertical area between two points such that no other points exist inside this area.

A vertical area is defined as:

• A region with fixed width that extends infinitely in both the positive and negative y-directions (infinite height)
• Bounded by two vertical lines parallel to the y-axis
• The width is the horizontal distance between these two vertical lines

Key constraints to note:

• Points that lie exactly on the boundary (edge) of the vertical area are not considered to be inside the area
• You need to find the maximum possible width among all valid vertical areas

In simpler terms, imagine drawing vertical lines through some of the given points. You need to find the largest horizontal gap between any two consecutive vertical lines where no points exist strictly between them (points on the lines themselves are allowed).

For example, if you have points at x-coordinates [1, 3, 8, 10], the gaps between consecutive x-values are 2, 5, and 2. The widest vertical area would have width 5 (between x = 3 and x = 8).

Intuition

When looking for the widest vertical area with no points inside, we need to think about what creates these "empty" vertical areas. Since a vertical area extends infinitely along the y-axis, the y-coordinates of the points don't actually matter - only the x-coordinates determine where the vertical boundaries can be placed.

The key insight is that the widest empty vertical area must be bounded by two points from our dataset. Why? Because if we try to place a vertical boundary at any x-coordinate that doesn't contain a point, we could always move it closer to the nearest point to make the area wider without including any new points inside.

This means we're essentially looking for the largest gap between x-coordinates of our points. To find this systematically:

1. We only care about the x-coordinates of all points
2. The potential vertical areas are the gaps between consecutive x-values when arranged in order
3. We need to find the maximum among all these gaps

By sorting the points by their x-coordinates, we line them up from left to right on the plane. The width of each possible vertical area is simply the difference between consecutive x-values in this sorted order. For instance, if after sorting we have x-coordinates [1, 3, 8, 10], the possible widths are 3-1=2, 8-3=5, and 10-8=2.

The solution becomes straightforward: sort the points by x-coordinate, calculate all gaps between consecutive points, and return the maximum gap. This is why the code sorts the points and then uses pairwise to look at each consecutive pair (a, b), calculating b[0] - a[0] (the difference in x-coordinates) and finding the maximum of all such differences.

Learn more about Sorting patterns.

Solution Approach

The implementation follows a straightforward approach based on our intuition:

Step 1: Sort the points

points.sort()

• We sort the entire points array using Python's built-in sort() method
• By default, Python sorts lists of lists by the first element (x-coordinate), then by the second element (y-coordinate) if needed
• Since we only care about x-coordinates for finding vertical gaps, this default behavior works perfectly
• Time complexity: O(n log n) where n is the number of points

Step 2: Find consecutive differences

pairwise(points)

• The pairwise function (from Python's itertools) creates an iterator of consecutive pairs
• For a sorted list like [[1,2], [3,4], [8,1], [10,5]], it generates pairs: ([1,2], [3,4]), ([3,4], [8,1]), ([8,1], [10,5])
• This allows us to examine each adjacent pair of points efficiently

Step 3: Calculate maximum gap

max(b[0] - a[0] for a, b in pairwise(points))

• For each consecutive pair (a, b), we calculate the horizontal distance: b[0] - a[0]
• a[0] represents the x-coordinate of the left point
• b[0] represents the x-coordinate of the right point
• The generator expression computes all these differences
• The max() function returns the largest difference, which is our answer

Complete Solution Flow:

1. Input: points = [[8,7],[9,9],[7,4],[9,7]]
2. After sorting: [[7,4],[8,7],[9,9],[9,7]]
3. Consecutive differences: 8-7=1, 9-8=1, 9-9=0
4. Maximum difference: 1

Time Complexity: O(n log n) - dominated by the sorting step Space Complexity: O(1) - if we don't count the space used by sorting (in-place sort), otherwise O(n) for the sorting algorithm's temporary space

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example Input: points = [[3,2], [9,5], [1,7], [3,8], [7,1]]

Step 1: Extract and visualize x-coordinates Looking at our points, the x-coordinates are: 3, 9, 1, 3, 7 (The y-coordinates don't matter for finding vertical areas)

Step 2: Sort the points by x-coordinate After sorting: [[1,7], [3,2], [3,8], [7,1], [9,5]] The sorted x-coordinates are: 1, 3, 3, 7, 9

Step 3: Find consecutive gaps Now we examine each pair of consecutive x-values:

• Gap between x=1 and x=3: 3 - 1 = 2
• Gap between x=3 and x=3: 3 - 3 = 0 (duplicate x-values)
• Gap between x=3 and x=7: 7 - 3 = 4
• Gap between x=7 and x=9: 9 - 7 = 2

Step 4: Find the maximum gap The gaps are: [2, 0, 4, 2] Maximum gap = 4

Visual representation:

x:  1   3   7   9
    |   |   |   |
    <-2-> 0 <-4-> <-2->
            ^^^^
         widest area

The widest vertical area is between x=3 and x=7 with width 4. This area contains no points inside it (the points at x=3 and x=7 are on the boundaries, which is allowed).

Code execution trace:

points = [[3,2], [9,5], [1,7], [3,8], [7,1]]
points.sort()  # [[1,7], [3,2], [3,8], [7,1], [9,5]]

# pairwise generates: ([1,7],[3,2]), ([3,2],[3,8]), ([3,8],[7,1]), ([7,1],[9,5])
# Differences: 3-1=2, 3-3=0, 7-3=4, 9-7=2
# max(2, 0, 4, 2) = 4

The answer is 4.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n log n)

The time complexity is dominated by the sorting operation points.sort(), which takes O(n log n) time where n is the number of points. After sorting, the code iterates through consecutive pairs of points using pairwise(), which takes O(n) time to generate pairs and compute differences. The max() function also runs in O(n) time. Therefore, the overall time complexity is O(n log n) + O(n) = O(n log n).

Space Complexity: O(n)

The space complexity depends on the sorting algorithm used by Python's sort() method. Python uses Timsort, which has a worst-case space complexity of O(n). Additionally, pairwise() creates an iterator that generates pairs on-the-fly, but the generator expression (b[0] - a[0] for a, b in pairwise(points)) is consumed by max(), which only needs O(1) additional space to track the maximum value. Therefore, the overall space complexity is O(n).

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

Common Pitfalls

1. Forgetting to Handle Duplicate X-Coordinates

One of the most common mistakes is not realizing that multiple points can share the same x-coordinate. When this happens, the width between these points is 0, which might be overlooked.

Example of the pitfall:

points = [[1,2], [3,4], [3,8], [5,6]]
# After sorting: [[1,2], [3,4], [3,8], [5,6]]
# Gap between [3,4] and [3,8] is 0 (same x-coordinate)

Why this matters: If you're not careful with the implementation, you might accidentally skip these zero-width gaps or handle them incorrectly. While the current solution handles this correctly (since 3 - 3 = 0 is a valid calculation), developers might be tempted to "optimize" by skipping duplicate x-coordinates, which could lead to incorrect indexing.

2. Attempting to Use Only X-Coordinates (Set-Based Approach)

A tempting optimization is to extract only the unique x-coordinates using a set, then find the maximum gap:

Incorrect approach:

def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:
    # WRONG: This loses duplicate x-coordinates
    x_coords = sorted(set(point[0] for point in points))
    return max(x_coords[i+1] - x_coords[i] for i in range(len(x_coords)-1))

The problem: While this seems more efficient, it has a critical edge case. When there's only one unique x-coordinate (all points lie on the same vertical line), the function will crash because there are no consecutive pairs to compare.

Correct handling:

def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:
    x_coords = sorted(set(point[0] for point in points))
  
    # Handle edge case: all points on same vertical line
    if len(x_coords) == 1:
        return 0
  
    return max(x_coords[i+1] - x_coords[i] for i in range(len(x_coords)-1))

3. Not Handling Edge Cases Properly

Several edge cases can trip up implementations:

Edge Case 1: Only two points

points = [[1,1], [3,3]]
# Should return 2, not crash or return 0

Edge Case 2: All points on the same vertical line

points = [[5,1], [5,2], [5,3]]
# Should return 0, as there's no gap between different x-coordinates

Robust solution that handles all cases:

class Solution:
    def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:
        # Extract and sort unique x-coordinates
        x_coords = sorted(set(point[0] for point in points))
      
        # If all points are on the same vertical line
        if len(x_coords) <= 1:
            return 0
      
        # Find maximum gap between consecutive x-coordinates
        max_width = 0
        for i in range(1, len(x_coords)):
            width = x_coords[i] - x_coords[i-1]
            max_width = max(max_width, width)
      
        return max_width

This approach is actually more efficient as it eliminates duplicate x-coordinates, reducing the number of comparisons needed while properly handling all edge cases.