Description#
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:
![](https://spcdn.pages.dev/leetcode/problems/1637.Widest%20Vertical%20Area%20Between%20Two%20Points%20Containing%20No%20Points/images/points3.png)
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
Solutions#
Solution 1#
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| class Solution:
def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:
points.sort()
return max(b[0] - a[0] for a, b in pairwise(points))
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| 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;
}
}
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| class Solution {
public:
int maxWidthOfVerticalArea(vector<vector<int>>& points) {
sort(points.begin(), points.end());
int ans = 0;
for (int i = 0; i < points.size() - 1; ++i) {
ans = max(ans, points[i + 1][0] - points[i][0]);
}
return ans;
}
};
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| 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
}
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| 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;
}
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| /**
* @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;
};
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Solution 2#
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| class Solution:
def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:
nums = [x for x, _ in points]
n = len(nums)
mi, mx = min(nums), max(nums)
bucket_size = max(1, (mx - mi) // (n - 1))
bucket_count = (mx - mi) // bucket_size + 1
buckets = [[inf, -inf] for _ in range(bucket_count)]
for x in nums:
i = (x - mi) // bucket_size
buckets[i][0] = min(buckets[i][0], x)
buckets[i][1] = max(buckets[i][1], x)
ans = 0
prev = inf
for curmin, curmax in buckets:
if curmin > curmax:
continue
ans = max(ans, curmin - prev)
prev = curmax
return ans
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| class Solution {
public int maxWidthOfVerticalArea(int[][] points) {
int n = points.length;
int[] nums = new int[n];
for (int i = 0; i < n; ++i) {
nums[i] = points[i][0];
}
final int inf = 1 << 30;
int mi = inf, mx = -inf;
for (int v : nums) {
mi = Math.min(mi, v);
mx = Math.max(mx, v);
}
int bucketSize = Math.max(1, (mx - mi) / (n - 1));
int bucketCount = (mx - mi) / bucketSize + 1;
int[][] buckets = new int[bucketCount][2];
for (var bucket : buckets) {
bucket[0] = inf;
bucket[1] = -inf;
}
for (int v : nums) {
int i = (v - mi) / bucketSize;
buckets[i][0] = Math.min(buckets[i][0], v);
buckets[i][1] = Math.max(buckets[i][1], v);
}
int prev = inf;
int ans = 0;
for (var bucket : buckets) {
if (bucket[0] > bucket[1]) {
continue;
}
ans = Math.max(ans, bucket[0] - prev);
prev = bucket[1];
}
return ans;
}
}
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| class Solution {
public:
int maxWidthOfVerticalArea(vector<vector<int>>& points) {
int n = points.size();
vector<int> nums;
for (auto& p : points) {
nums.push_back(p[0]);
}
const int inf = 1 << 30;
int mi = inf, mx = -inf;
for (int v : nums) {
mi = min(mi, v);
mx = max(mx, v);
}
int bucketSize = max(1, (mx - mi) / (n - 1));
int bucketCount = (mx - mi) / bucketSize + 1;
vector<pair<int, int>> buckets(bucketCount, {inf, -inf});
for (int v : nums) {
int i = (v - mi) / bucketSize;
buckets[i].first = min(buckets[i].first, v);
buckets[i].second = max(buckets[i].second, v);
}
int ans = 0;
int prev = inf;
for (auto [curmin, curmax] : buckets) {
if (curmin > curmax) continue;
ans = max(ans, curmin - prev);
prev = curmax;
}
return ans;
}
};
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| func maxWidthOfVerticalArea(points [][]int) (ans int) {
n := len(points)
nums := make([]int, 0, n)
for _, p := range points {
nums = append(nums, p[0])
}
const inf = 1 << 30
mi, mx := inf, -inf
for _, v := range nums {
mi = min(mi, v)
mx = max(mx, v)
}
bucketSize := max(1, (mx-mi)/(n-1))
bucketCount := (mx-mi)/bucketSize + 1
buckets := make([][]int, bucketCount)
for i := range buckets {
buckets[i] = []int{inf, -inf}
}
for _, v := range nums {
i := (v - mi) / bucketSize
buckets[i][0] = min(buckets[i][0], v)
buckets[i][1] = max(buckets[i][1], v)
}
prev := inf
for _, bucket := range buckets {
if bucket[0] > bucket[1] {
continue
}
ans = max(ans, bucket[0]-prev)
prev = bucket[1]
}
return ans
}
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| 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;
}
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| /**
* @param {number[][]} points
* @return {number}
*/
var maxWidthOfVerticalArea = function (points) {
const nums = 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;
};
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