Number of Ways Where Square of Number Is Equal to Product of Two Numbers

Given two arrays of integers nums1 and nums2, return the number of triplets formed (type 1 and type 2) under the following rules:

 

Example 1:

Input: nums1 = [7,4], nums2 = [5,2,8,9]
Output: 1
Explanation: Type 1: (1,1,2), nums1[1]^2 = nums2[1] * nums2[2]. (4^2 = 2 * 8).
Example 2:

Input: nums1 = [1,1], nums2 = [1,1,1]
Output: 9
Explanation: All Triplets are valid, because 1^2 = 1 * 1.
Type 1: (0,0,1), (0,0,2), (0,1,2), (1,0,1), (1,0,2), (1,1,2).  nums1[i]^2 = nums2[j] * nums2[k].
Type 2: (0,0,1), (1,0,1), (2,0,1). nums2[i]^2 = nums1[j] * nums1[k].
Example 3:

Input: nums1 = [7,7,8,3], nums2 = [1,2,9,7]
Output: 2
Explanation: There are 2 valid triplets.
Type 1: (3,0,2).  nums1[3]^2 = nums2[0] * nums2[2].
Type 2: (3,0,1).  nums2[3]^2 = nums1[0] * nums1[1].
Example 4:

Input: nums1 = [4,7,9,11,23], nums2 = [3,5,1024,12,18]
Output: 0
Explanation: There are no valid triplets.
 

Constraints:


Solution:

class Solution {
    public int numTriplets(int[] nums1, int[] nums2) {
        int res = 0;
        for (int val : nums1) {
            long v = (long) val;
            res += twoSum(v * v, nums2);
        }
        for (int val : nums2) {
            long v = (long) val;
            res += twoSum(v * v, nums1);
        }
        return res;
    }
    
    private int twoSum(long target, int[] arr) {
        Map<Long, Integer> map = new HashMap();
        int count = 0;
        for (int i = 0; i < arr.length; i ++) {
            long curr = (long) arr[i];
            long d = target / curr;
            if (d * curr != target) continue;
            if (map.containsKey(d)) {
                count += map.get(d);
            }
            map.put(curr, map.getOrDefault(curr, 0) + 1);
        }
        return count;
    }
}