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1577. Number of Ways Where Square of Number Is Equal to Product of Two Numbers

Description

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

  • Type 1: Triplet (i, j, k) if nums1[i]2 == nums2[j] * nums2[k] where 0 <= i < nums1.length and 0 <= j < k < nums2.length.
  • Type 2: Triplet (i, j, k) if nums2[i]2 == nums1[j] * nums1[k] where 0 <= i < nums2.length and 0 <= j < k < nums1.length.

 

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]. (42 = 2 * 8).

Example 2:

Input: nums1 = [1,1], nums2 = [1,1,1]
Output: 9
Explanation: All Triplets are valid, because 12 = 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].

 

Constraints:

  • 1 <= nums1.length, nums2.length <= 1000
  • 1 <= nums1[i], nums2[i] <= 105

Solutions

  • class Solution {
    public int numTriplets(int[] nums1, int[] nums2) {
        Map<Integer, Integer> cnt1 = new HashMap<>();
        Map<Integer, Integer> cnt2 = new HashMap<>();
        for (int v : nums1) {
            cnt1.put(v, cnt1.getOrDefault(v, 0) + 1);
        }
        for (int v : nums2) {
            cnt2.put(v, cnt2.getOrDefault(v, 0) + 1);
        }
        long ans = 0;
        for (var e1 : cnt1.entrySet()) {
            long a = e1.getKey(), x = e1.getValue();
            for (var e2 : cnt2.entrySet()) {
                long b = e2.getKey(), y = e2.getValue();
                if ((a * a) % b == 0) {
                    long c = a * a / b;
                    if (b == c) {
                        ans += x * y * (y - 1);
                    } else {
                        ans += x * y * cnt2.getOrDefault((int) c, 0);
                    }
                }
                if ((b * b) % a == 0) {
                    long c = b * b / a;
                    if (a == c) {
                        ans += x * (x - 1) * y;
                    } else {
                        ans += x * y * cnt1.getOrDefault((int) c, 0);
                    }
                }
            }
        }
        return (int) (ans >> 1);
    }
}

  • class Solution:
    def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
        cnt1 = Counter(nums1)
        cnt2 = Counter(nums2)
        ans = 0
        for a, x in cnt1.items():
            for b, y in cnt2.items():
                if a * a % b == 0:
                    c = a * a // b
                    if b == c:
                        ans += x * y * (y - 1)
                    else:
                        ans += x * y * cnt2[c]
                if b * b % a == 0:
                    c = b * b // a
                    if a == c:
                        ans += x * (x - 1) * y
                    else:
                        ans += x * y * cnt1[c]
        return ans >> 1


  • func numTriplets(nums1 []int, nums2 []int) (ans int) {
	cnt1 := map[int]int{}
	cnt2 := map[int]int{}
	for _, v := range nums1 {
		cnt1[v]++
	}
	for _, v := range nums2 {
		cnt2[v]++
	}
	for a, x := range cnt1 {
		for b, y := range cnt2 {
			if a*a%b == 0 {
				c := a * a / b
				if b == c {
					ans += x * y * (y - 1)
				} else {
					ans += x * y * cnt2[c]
				}
			}
			if b*b%a == 0 {
				c := b * b / a
				if a == c {
					ans += x * (x - 1) * y
				} else {
					ans += x * y * cnt1[c]
				}
			}
		}
	}
	ans /= 2
	return
}

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