LeetCode in Kotlin

2183. Count Array Pairs Divisible by K

Hard

Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs (i, j) such that:

• 0 <= i < j <= n - 1 and
• nums[i] * nums[j] is divisible by k.

Example 1:

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

Output: 7

Explanation:

The 7 pairs of indices whose corresponding products are divisible by 2 are

(0, 1), (0, 3), (1, 2), (1, 3), (1, 4), (2, 3), and (3, 4).

Their products are 2, 4, 6, 8, 10, 12, and 20 respectively.

Other pairs such as (0, 2) and (2, 4) have products 3 and 15 respectively, which are not divisible by 2.

Example 2:

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

Output: 0

Explanation: There does not exist any pair of indices whose corresponding product is divisible by 5.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i], k <= 105

Solution

class Solution {
    fun countPairs(nums: IntArray, k: Int): Long {
        var count = 0L
        val map: MutableMap<Int, Long> = HashMap()
        for (num in nums) {
            val gd = gcd(num, k)
            val want = k / gd
            for ((key, value) in map) {
                if (key % want == 0) {
                    count += value
                }
            }
            map[gd] = map.getOrDefault(gd, 0L) + 1L
        }
        return count
    }

    private fun gcd(a: Int, b: Int): Int {
        if (a > b) {
            return gcd(b, a)
        }
        return if (a == 0) {
            b
        } else gcd(a, b % a)
    }
}

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