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] <= 105class Solution {
public int sumFourDivisors(int[] nums) {
int ans = 0;
for (int i : nums)
ans += validDivSum(i);
return ans;
}
private int validDivSum(int n) {
Set<Integer> set = new HashSet<>();
for (int i = 1; i * i <= n; i++) {
if (n % i == 0) {
set.add(i);
set.add(n / i);
}
}
if (set.size() == 4) {
int sum = 0;
for (int i : set)
sum += i;
return sum;
}
return 0;
}
}class Solution {
public int sumFourDivisors(int[] nums) {
int ans = 0;
for (int i : nums)
ans += validDivSum(i);
return ans;
}
private int validDivSum(int n) {
int sum = 0, count = 0, i = 1;
for (; i * i < n; i++) {
if (n % i != 0)
continue;
sum += i + n / i;
if ((count += 2) > 4)
return 0;
}
if (i * i == n) {
sum += i;
count++;
}
return count == 4 ? sum : 0;
}
}