Description
Given an integer array nums, return the sum of divisors of the integers in that array that have exactly four divisors. If there is no such integer in the array, return 0.
Example 1:
Input: nums = [21,4,7] Output: 32 Explanation: 21 has 4 divisors: 1, 3, 7, 21 4 has 3 divisors: 1, 2, 4 7 has 2 divisors: 1, 7 The answer is the sum of divisors of 21 only.
Example 2:
Input: nums = [21,21] Output: 64
Example 3:
Input: nums = [1,2,3,4,5] Output: 0
Constraints:
1 <= nums.length <= 1041 <= nums[i] <= 105
Solution
Python3
class Solution:
def sumFourDivisors(self, nums: List[int]) -> int:
res = 0
for num in nums:
s = set()
for i in range(1, floor(sqrt(num)) + 1):
if num % i == 0:
s.add(num//i)
s.add(i)
if len(s) > 4:
break
if len(s) == 4:
res += sum(s)
return res