Find Number of Ways Where Square Equals Product of Two Numbers in Python

Suppose we have two arrays nums1 and nums2, we have to find of triplets formed (type 1 and type 2) following these two rules −

• Triplet (i, j, k) if nums1[i]^2 = nums2[j] * nums2[k] where [0 <= i < size of nums1 and 0 <= j < k < size of nums2].
• Triplet (i, j, k) if nums2[i]^2 = nums1[j] * nums1[k] where [0 <= i < size of nums2 and 0 <= j < k < size of nums1].

So, if the input is like nums1 = [7,4] nums2 = [5,2,8,9], then the output will be 1 because there is a triplet of type 1, (1,1,2), nums1[1]^2 = nums2[1] * nums2[2] = (16 = 2*8).

To solve this, we will follow these steps:

• cnt1 := a map to hold each elements and its count of nums1
• cnt2 := a map to hold each elements and its count of nums2
• Define a function triplets() . This will take arr1, arr2
• ans := 0
• for each t, v in items() of arr1, do
  • k := arr2[t] if it is there, otherwise 0
  • tmp := k*(k - 1) / 2
  • sq := t * t
  • for each m in arr2, do
    • if m < t and sq mod m is same as 0, then
      • tmp := tmp + (arr2[m] if it is there, otherwise 0) * (arr2[quotient of sq/m] if it is there, otherwise 0)
• ans := ans + tmp * v
• return ans
• From the main method, do the following−
• return triplets(cnt1, cnt2) + triplets(cnt2, cnt1)

Example

Let us see the following implementation to get better understanding−

from collections import Counter
def solve(nums1, nums2):
   cnt1 = Counter(nums1)
   cnt2 = Counter(nums2)

   def triplets(arr1, arr2):
      ans = 0
      for t, v in arr1.items():
         k = arr2.get(t, 0)
         tmp = k * (k - 1) // 2
         sq = t * t
         for m in arr2:
            if m < t and sq % m == 0:
               tmp += arr2.get(m, 0) * arr2.get(sq // m, 0)
            ans += tmp * v
      return ans

   return triplets(cnt1, cnt2) + triplets(cnt2, cnt1)
nums1 = [7,4]
nums2 = [5,2,8,9]
print(solve(nums1, nums2))

Input

[7,4],[5,2,8,9]

Output

2