LeetCode in Kotlin
963. Minimum Area Rectangle II
Medium
You are given an array of points in the X-Y plane points where points[i] = [xi, yi].
Return the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the X and Y axes. If there is not any such rectangle, return 0.
Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input: points = [[1,2],[2,1],[1,0],[0,1]]
Output: 2.00000
Explanation: The minimum area rectangle occurs at [1,2],[2,1],[1,0],[0,1], with an area of 2.
Example 2:
Input: points = [[0,1],[2,1],[1,1],[1,0],[2,0]]
Output: 1.00000
Explanation: The minimum area rectangle occurs at [1,0],[1,1],[2,1],[2,0], with an area of 1.
Example 3:
Input: points = [[0,3],[1,2],[3,1],[1,3],[2,1]]
Output: 0
Explanation: There is no possible rectangle to form from these points.
Constraints:
1 <= points.length <= 50points[i].length == 20 <= xi, yi <= 4 * 104- All the given points are unique.
Solution
import kotlin.math.abs
class Solution {
fun minAreaFreeRect(points: Array<IntArray>): Double {
val map: MutableMap<Int, MutableSet<Int>> = HashMap()
var area: Double
for (point in points) {
map.putIfAbsent(point[0], HashSet())
map.getValue(point[0]).add(point[1])
}
var minArea = Double.MAX_VALUE
val n = points.size
for (i in 0 until n - 2) {
for (j in i + 1 until n - 1) {
val dx1 = points[j][0] - points[i][0]
val dy1 = points[j][1] - points[i][1]
// get the 3rd point
for (k in j + 1 until n) {
val dx2 = points[k][0] - points[i][0]
val dy2 = points[k][1] - points[i][1]
if (dx1 * dx2 + dy1 * dy2 != 0) {
continue
}
// find the 4th point
val x = dx1 + points[k][0]
val y = dy1 + points[k][1]
area = calculateArea(points, i, j, k)
if (area >= minArea) {
continue
}
// 4th point exists
if (map[x] != null && map.getValue(x).contains(y)) {
minArea = area
}
}
}
}
return if (minArea == Double.MAX_VALUE) 0.0 else minArea
}
private fun calculateArea(points: Array<IntArray>, i: Int, j: Int, k: Int): Double {
val first = points[i]
val second = points[j]
val third = points[k]
return abs(
first[0] * (second[1] - third[1]) + second[0] * (third[1] - first[1]) + third[0] * (first[1] - second[1])
).toDouble()
}
}