LeetCode-in-Java
1390. Four Divisors
Medium
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
public class Solution {
public int sumFourDivisors(int[] nums) {
int sum = 0;
for (int num : nums) {
int sqrt = (int) Math.sqrt(num);
if (sqrt * sqrt == num) {
continue;
}
int tmpSum = num + 1;
int count = 0;
for (int i = 2; i <= sqrt; i++) {
if (num % i == 0) {
count++;
tmpSum += (i + num / i);
}
if (count > 1) {
break;
}
}
if (count == 1) {
sum += tmpSum;
}
}
return sum;
}
}