1577. 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:
- Type 1: Triplet (i, j, k) if
nums1[i]2 == nums2[j] * nums2[k]
where0 <= i < nums1.length
and0 <= j < k < nums2.length
. - Type 2: Triplet (i, j, k) if
nums2[i]2 == nums1[j] * nums1[k]
where0 <= i < nums2.length
and0 <= 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]. (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:
1 <= nums1.length, nums2.length <= 1000
1 <= nums1[i], nums2[i] <= 10^5
# @lc code=start
using LeetCode
function num_triplets(nums1::Vector{Int}, nums2::Vector{Int})
cnt1 = counter(nums1)
cnt2 = counter(nums2)
sqr_cnt1 = Dict(k ^ 2 => v for (k, v) in cnt1)
sqr_cnt2 = Dict(k ^ 2 => v for (k, v) in cnt2)
res = 0
for (k1, v1) in cnt1
for (k2, v2) in sqr_cnt2
if k1 ^ 2 == k2
res += (v1 - 1) * v1 * v2
else
res += cnt1[k2 ÷ k1] * v1 * v2
end
end
end
for (k1, v1) in cnt2
for (k2, v2) in sqr_cnt1
if k2 % k1 == 0
if k1 ^ 2 == k2
res += (v1 - 1) * v1 * v2
else
res += cnt2[k2 ÷ k1] * v1 * v2
end
end
end
end
res ÷ 2
end
# @lc code=end
num_triplets (generic function with 1 method)
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