LeetCode in Kotlin

2857. Count Pairs of Points With Distance k

Medium

You are given a 2D integer array coordinates and an integer k, where coordinates[i] = [xi, yi] are the coordinates of the ith point in a 2D plane.

We define the distance between two points (x1, y1) and (x2, y2) as (x1 XOR x2) + (y1 XOR y2) where XOR is the bitwise XOR operation.

Return the number of pairs (i, j) such that i < j and the distance between points i and j is equal to k.

Example 1:

Input: coordinates = [[1,2],[4,2],[1,3],[5,2]], k = 5

Output: 2

Explanation: We can choose the following pairs:

• (0,1): Because we have (1 XOR 4) + (2 XOR 2) = 5.
• (2,3): Because we have (1 XOR 5) + (3 XOR 2) = 5.

Example 2:

Input: coordinates = [[1,3],[1,3],[1,3],[1,3],[1,3]], k = 0

Output: 10

Explanation: Any two chosen pairs will have a distance of 0. There are 10 ways to choose two pairs.

Constraints:

• 2 <= coordinates.length <= 50000
• 0 <= xi, yi <= 106
• 0 <= k <= 100

Solution

class Solution {
    fun countPairs(coordinates: List<List<Int>>, k: Int): Int {
        var ans = 0
        val map: MutableMap<Long, Int> = HashMap()
        for (p in coordinates) {
            val p0 = p[0]
            val p1 = p[1]
            for (i in 0..k) {
                val x1 = i xor p0
                val y1 = (k - i) xor p1
                val key2 = hash(x1, y1)
                if (map.containsKey(key2)) {
                    ans += map[key2]!!
                }
            }
            val key = hash(p0, p1)
            map[key] = map.getOrDefault(key, 0) + 1
        }
        return ans
    }

    private fun hash(x1: Int, y1: Int): Long {
        val r = 1e8.toLong()
        return x1 * r + y1
    }
}

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