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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1301_1400.s1390_four_divisors;

// #Medium #Array #Math #2022_03_17_Time_13_ms_(97.25%)_Space_42.3_MB_(95.60%)

/**
 * 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 <= 10⁴
 * *   1 <= nums[i] <= 10⁵
**/
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;
    }
}