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Formatted question description: https://leetcode.ca/all/1637.html

# 1637. Widest Vertical Area Between Two Points Containing No Points (Medium)

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

A vertical area is an area of fixed-width extending infinitely along the y-axis (i.e., infinite height). The widest vertical area is the one with the maximum width.

Note that points on the edge of a vertical area are not considered included in the area.

Example 1:

Input: points = [[8,7],[9,9],[7,4],[9,7]]
Output: 1
Explanation: Both the red and the blue area are optimal.


Example 2:

Input: points = [[3,1],[9,0],[1,0],[1,4],[5,3],[8,8]]
Output: 3


Constraints:

• n == points.length
• 2 <= n <= 105
• points[i].length == 2
• 0 <= xi, yi <= 109

Related Topics:
Sort

## Solution 1. Sort

Sort the points in ascending order of x values, then traverse to find the maximum distance of x values between adjacent points.

// OJ: https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int maxWidthOfVerticalArea(vector<vector<int>>& A) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[0]< b[0]; });
int ans = 0;
for (int i = 1; i < A.size(); ++i) ans = max(ans, A[i][0] - A[i - 1][0]);
return ans;
}
};

• class Solution {
public int maxWidthOfVerticalArea(int[][] points) {
Arrays.sort(points, new Comparator<int[]>() {
public int compare(int[] point1, int[] point2) {
if (point1[0] != point2[0])
return point1[0] - point2[0];
else
return point1[1] - point2[1];
}
});
int maxWidth = 0;
int length = points.length;
for (int i = 1; i < length; i++) {
int width = points[i][0] - points[i - 1][0];
maxWidth = Math.max(maxWidth, width);
}
return maxWidth;
}
}

############

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

• // OJ: https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int maxWidthOfVerticalArea(vector<vector<int>>& A) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[0]< b[0]; });
int ans = 0;
for (int i = 1; i < A.size(); ++i) ans = max(ans, A[i][0] - A[i - 1][0]);
return ans;
}
};

• class Solution:
def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:
points.sort()
return max(b[0] - a[0] for a, b in pairwise(points))

############

# 1637. Widest Vertical Area Between Two Points Containing No Points
# https://leetcode.com/problems/widest-vertical-area-between-two-points-containing-no-points/

class Solution:
def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:
if not points: return 0

points.sort(key = lambda x : (x[0], x[1]))

res = 0
prevX = points[0][0]
for x,y in points[1:]:
if x == prevX:
continue
else:
res = max(res, x - prevX)
prevX = x

return res

• 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
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• /**
* @param {number[][]} points
* @return {number}
*/
var maxWidthOfVerticalArea = function (points) {
points.sort((a, b) => a[0] - b[0]);
let ans = 0;
let px = points[0][0];
for (const [x, _] of points) {
ans = Math.max(ans, x - px);
px = x;
}
return ans;
};


• function maxWidthOfVerticalArea(points: number[][]): number {
const nums: number[] = points.map(point => point[0]);
const inf = 1 << 30;
const n = nums.length;
let mi = inf;
let mx = -inf;
for (const x of nums) {
mi = Math.min(mi, x);
mx = Math.max(mx, x);
}
const bucketSize = Math.max(1, Math.floor((mx - mi) / (n - 1)));
const bucketCount = Math.floor((mx - mi) / bucketSize) + 1;
const buckets = new Array(bucketCount).fill(0).map(() => [inf, -inf]);
for (const x of nums) {
const i = Math.floor((x - mi) / bucketSize);
buckets[i][0] = Math.min(buckets[i][0], x);
buckets[i][1] = Math.max(buckets[i][1], x);
}
let prev = inf;
let ans = 0;
for (const [left, right] of buckets) {
if (left > right) {
continue;
}
ans = Math.max(ans, left - prev);
prev = right;
}
return ans;
}