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Tested Python solution for LeetCode 1726 with 36 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 1726 # by problem number
lcpy gen -s tuple_with_same_product # by problem name
nums of distinct positive integers, return the number of tuples (a, b, c, d) such that a * b = c * d where a, b, c, and d are elements of nums, and a != b != c != d.
Input: nums = [2,3,4,6]
Output: 8
Explanation: There are 8 valid tuples:
(2,6,3,4) , (2,6,4,3) , (6,2,3,4) , (6,2,4,3)
(3,4,2,6) , (4,3,2,6) , (3,4,6,2) , (4,3,6,2)
Input: nums = [1,2,4,5,10]
Output: 16
Explanation: There are 16 valid tuples:
(1,10,2,5) , (1,10,5,2) , (10,1,2,5) , (10,1,5,2)
(2,5,1,10) , (2,5,10,1) , (5,2,1,10) , (5,2,10,1)
(2,10,4,5) , (2,10,5,4) , (10,2,4,5) , (10,2,5,4)
(4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)
from collections import defaultdict
class Solution:
# Time: O(n^2)
# Space: O(n^2)
def tuple_same_product(self, nums: list[int]) -> int:
product_counts: dict[int, int] = defaultdict(int)
n = len(nums)
for i in range(n):
for j in range(i + 1, n):
product_counts[nums[i] * nums[j]] += 1
return sum(8 * c * (c - 1) // 2 for c in product_counts.values())
| Time | Space |
|---|---|
| O(n^2) | O(n^2) |