algoadvance

Leetcode 1637. Widest Vertical Area Between Two Points Containing No Points

Problem Statement

Given n points on a 2D plane where points[i] = [xi, yi], return the width of the widest vertical area between two points such that no points are inside the area.

A vertical area is an area of the plane where all points have different x-coordinates but have the same y-coordinate, and the width of a vertical area is the absolute difference between the x-coordinates of the two edges.

Clarifying Questions

  1. What is the range of the input array (points)?
    • The points array can have a size anywhere from 2 to 10^5 elements.
  2. What is the range of the x and y coordinates?
    • Both x and y coordinates are integers that can range from -10^9 to 10^9.
  3. Can there be duplicate points?
    • The problem implies unique points but does not explicitly prohibit duplicates. However, since vertical areas are based only on x-coordinates, duplicates in y_coordinates are not directly relevant.
  4. What should be returned if there are no vertical areas?
    • The problem guarantees at least two points, so there will always be a vertical area, and hence, this case does not need special handling.

Strategy

  1. Extract X-Coordinates: Focus on the x-coordinates since vertical areas are concerned with their differences.

  2. Sort X-Coordinates: Sorting the list of x-coordinates allows us to easily calculate the differences between consecutive x-coordinates.

  3. Compute Maximum Difference: Iterate through the sorted list of x-coordinates and compute the differences between consecutive elements, maintaining the maximum difference.

Code

Here’s the possible C++ implementation of the solution:

#include <vector>
#include <algorithm>

class Solution {
public:
    int maxWidthOfVerticalArea(std::vector<std::vector<int>>& points) {
        // Extract x-coordinates
        std::vector<int> xCoords;
        for (const auto& point : points) {
            xCoords.push_back(point[0]);
        }
        
        // Sort the x-coordinates
        std::sort(xCoords.begin(), xCoords.end());
        
        // Calculate the maximum difference between consecutive x-coordinates
        int maxWidth = 0;
        for (size_t i = 1; i < xCoords.size(); ++i) {
            maxWidth = std::max(maxWidth, xCoords[i] - xCoords[i - 1]);
        }
        
        return maxWidth;
    }
};

Time Complexity

Thus, the overall time complexity is dominated by the sorting step, which results in (O(n \log n)).

Cut your prep time in half and DOMINATE your interview with AlgoAdvance AI