Widest Vertical Area Between Two Points Containing No Points solution and explanation

def maxWidthOfVerticalArea(points):
    x_coords = [point[0] for point in points]
    x_coords.sort()
    max_width = 0
    for i in range(1, len(x_coords)):
        max_width = max(max_width, x_coords[i] - x_coords[i - 1])
    return max_width

if __name__ == "__main__":
    points = [[3, 1], [9, 0], [1, 0], [1, 4], [5, 3], [8, 8]]
    print(maxWidthOfVerticalArea(points))

import java.util.Arrays;

public class Solution {
    public int maxWidthOfVerticalArea(int[][] points) {
        int[] xCoords = new int[points.length];
        for (int i = 0; i < points.length; i++) {
            xCoords[i] = points[i][0];
        }
        Arrays.sort(xCoords);
        int maxWidth = 0;
        for (int i = 1; i < xCoords.length; i++) {
            maxWidth = Math.max(maxWidth, xCoords[i] - xCoords[i - 1]);
        }
        return maxWidth;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        int[][] points = {{3, 1}, {9, 0}, {1, 0}, {1, 4}, {5, 3}, {8, 8}};
        System.out.println(solution.maxWidthOfVerticalArea(points));
    }
}

#include <iostream>
#include <vector>
#include <algorithm>

int maxWidthOfVerticalArea(std::vector<std::vector<int>>& points) {
    std::vector<int> xCoords;
    for (const auto& point : points) {
        xCoords.push_back(point[0]);
    }
    std::sort(xCoords.begin(), xCoords.end());
    int maxWidth = 0;
    for (size_t i = 1; i < xCoords.size(); ++i) {
        maxWidth = std::max(maxWidth, xCoords[i] - xCoords[i - 1]);
    }
    return maxWidth;
}

int main() {
    std::vector<std::vector<int>> points = {{3, 1}, {9, 0}, {1, 0}, {1, 4}, {5, 3}, {8, 8}};
    std::cout << maxWidthOfVerticalArea(points) << std::endl;
    return 0;
}

function maxWidthOfVerticalArea(points) {
    let xCoords = points.map(point => point[0]);
    xCoords.sort((a, b) => a - b);
    let maxWidth = 0;
    for (let i = 1; i < xCoords.length; i++) {
        maxWidth = Math.max(maxWidth, xCoords[i] - xCoords[i - 1]);
    }
    return maxWidth;
}

console.log(maxWidthOfVerticalArea([[3, 1], [9, 0], [1, 0], [1, 4], [5, 3], [8, 8]]));

#include <stdio.h>
#include <stdlib.h>

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

int maxWidthOfVerticalArea(int points[][2], int pointsSize) {
    int xCoords[pointsSize];
    for (int i = 0; i < pointsSize; i++) {
        xCoords[i] = points[i][0];
    }
    qsort(xCoords, pointsSize, sizeof(int), compare);
    int maxWidth = 0;
    for (int i = 1; i < pointsSize; i++) {
        int width = xCoords[i] - xCoords[i - 1];
        if (width > maxWidth) {
            maxWidth = width;
        }
    }
    return maxWidth;
}

int main() {
    int points[][2] = {{3, 1}, {9, 0}, {1, 0}, {1, 4}, {5, 3}, {8, 8}};
    int pointsSize = sizeof(points) / sizeof(points[0]);
    printf("%d\n", maxWidthOfVerticalArea(points, pointsSize));
    return 0;
}

using System;

class Solution {
    public int maxWidthOfVerticalArea(int[][] points) {
        int[] xCoords = new int[points.Length];
        for (int i = 0; i < points.Length; i++) {
            xCoords[i] = points[i][0];
        }
        Array.Sort(xCoords);
        int maxWidth = 0;
        for (int i = 1; i < xCoords.Length; i++) {
            maxWidth = Math.Max(maxWidth, xCoords[i] - xCoords[i - 1]);
        }
        return maxWidth;
    }

    static void Main() {
        Solution solution = new Solution();
        int[][] points = new int[][] { new int[] { 3, 1 }, new int[] { 9, 0 }, new int[] { 1, 0 }, new int[] { 1, 4 }, new int[] { 5, 3 }, new int[] { 8, 8 } };
        Console.WriteLine(solution.maxWidthOfVerticalArea(points));
    }
}

# Bucket approach is explained rather than implemented due to practical concerns with large domain ranges.

// Not implemented due to practical impracticalities with given constraints, useful as educational reflection.

// Similar educational example for C++ as in C.

// Logical exploration of specialized organized sorting for limited numerical spectra.

/* This approach is generally not needed for the given problem constraints but can be experimented with for understanding such tactics */

// Focus is on conceptual understanding instead of actual implementation for this context.

function maxWidthOfVerticalArea(points: number[][]): number {
    points.sort((a, b) => a[0] - b[0]);
    let ans = 0;
    for (let i = 1; i < points.length; ++i) {
        ans = Math.max(ans, points[i][0] - points[i - 1][0]);
    }
    return ans;
}

func maxWidthOfVerticalArea(points [][]int) (ans int) {
	sort.Slice(points, func(i, j int) bool { return points[i][0] < points[j][0] })
	for i, p := range points[1:] {
		ans = max(ans, p[0]-points[i][0])
	}
	return
}