g1301_1400.s1362_closest_divisors.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1301_1400.s1362_closest_divisors;

// #Medium #Math #2022_03_21_Time_8_ms_(84.91%)_Space_39.9_MB_(79.25%)

/**
 * 1362 - Closest Divisors\.
 *
 * Medium
 *
 * Given an integer `num`, find the closest two integers in absolute difference whose product equals `num + 1` or `num + 2`.
 *
 * Return the two integers in any order.
 *
 * **Example 1:**
 *
 * **Input:** num = 8
 *
 * **Output:** [3,3]
 *
 * **Explanation:** For num + 1 = 9, the closest divisors are 3 & 3, for num + 2 = 10, the closest divisors are 2 & 5, hence 3 & 3 is chosen.
 *
 * **Example 2:**
 *
 * **Input:** num = 123
 *
 * **Output:** [5,25]
 *
 * **Example 3:**
 *
 * **Input:** num = 999
 *
 * **Output:** [40,25]
 *
 * **Constraints:**
 *
 * *   `1 <= num <= 10^9`
**/
public class Solution {
    public int[] closestDivisors(int num) {
        int sqrt1 = (int) Math.sqrt(num + 1.0);
        int sqrt2 = (int) Math.sqrt(num + 2.0);
        if (sqrt1 * sqrt1 == num + 1) {
            return new int[] {sqrt1, sqrt1};
        }
        if (sqrt2 * sqrt2 == num + 2) {
            return new int[] {sqrt2, sqrt2};
        }
        int[] ans1 = new int[2];
        for (int i = sqrt1; i >= 1; i--) {
            if ((num + 1) % i == 0) {
                ans1 = new int[] {i, (num + 1) / i};
                break;
            }
        }
        int[] ans2 = new int[2];
        for (int i = sqrt2; i >= 1; i--) {
            if ((num + 2) % i == 0) {
                ans2 = new int[] {i, (num + 2) / i};
                break;
            }
        }
        return Math.abs(ans2[0] - ans2[1]) < Math.abs(ans1[0] - ans1[1]) ? ans2 : ans1;
    }
}