LeetCode in Kotlin

447. Number of Boomerangs

Medium

You are given n points in the plane that are all distinct, where points[i] = [xi, yi]. A boomerang is a tuple of points (i, j, k) such that the distance between i and j equals the distance between i and k (the order of the tuple matters).

Return the number of boomerangs.

Example 1:

Input: points = [[0,0],[1,0],[2,0]]

Output: 2

Explanation: The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]].

Example 2:

Input: points = [[1,1],[2,2],[3,3]]

Output: 2

Example 3:

Input: points = [[1,1]]

Output: 0

Constraints:

• n == points.length
• 1 <= n <= 500
• points[i].length == 2
• -104 <= xi, yi <= 104
• All the points are unique.

Solution

class Solution {
    fun numberOfBoomerangs(points: Array<IntArray>): Int {
        val m: HashMap<Int, Int> = HashMap()
        var ans = 0
        for (i in points.indices) {
            for (j in points.indices) {
                if (i == j) {
                    continue
                }
                val dis = dist(points[i], points[j])
                val prev = m.getOrDefault(dis, 0)
                if (prev >= 1) {
                    ans += prev * 2
                }
                m[dis] = prev + 1
            }
            m.clear()
        }
        return ans
    }

    private fun dist(d1: IntArray, d2: IntArray): Int {
        return (d1[0] - d2[0]) * (d1[0] - d2[0]) + (d1[1] - d2[1]) * (d1[1] - d2[1])
    }
}

This site is open source. Improve this page.