g1501_1600.s1577_number_of_ways_where_square_of_number_is_equal_to_product_of_two_numbers

Class Solution

java.lang.Object

g1501_1600.s1577_number_of_ways_where_square_of_number_is_equal_to_product_of_two_numbers.Solution

public class Solution
extends Object

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

Medium

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

Constructor Summary

Constructors

Constructor and Description
Solution()

Method Summary

Modifier and Type	Method and Description
int	count(int[] a, int[] b)
int	numTriplets(int[] nums1, int[] nums2)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Detail

Solution

public Solution()

Method Detail

numTriplets

public int numTriplets(int[] nums1,
                       int[] nums2)

count

public int count(int[] a,
                 int[] b)