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