Description
Given an array 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.
Example 1:
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)
Example 2:
Input: nums = [1,2,4,5,10]
Output: 16
Explanation: There are 16 valids 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) , (10,2,4,5)
(4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)
Example 3:
Input: nums = [2,3,4,6,8,12]
Output: 40
Example 4:
Input: nums = [2,3,5,7]
Output: 0
Constraints:
- 1 <= nums.length <= 1000
- 1 <= nums[i] <= 104
- All elements in nums are distinct.
分析
题目的意思是：这道题要求出有相同乘积的四个组合数。我看了一下思路，建立一个字典，对两两乘积进行统计，然后计算频率大于1的组合数就行了，公式维：
n*(n-1)/2*8
n为统计的频率.
因为每两对相同乘积的数，有8种不同的组合，所以排列组合一下为C(n,2)*8
代码
class Solution:
def tupleSameProduct(self, nums: List[int]) -> int:
n=len(nums)
d=defaultdict(int)
for i in range(n):
for j in range(i+1,n):
key=nums[i]*nums[j]
d[key]+=1
res=0
# print(d)
for k,v in d.items():
if(v>1):
res+=v*(v-1)/2*8
return int(res)