LeetCode in Kotlin
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
Solution
class Solution {
fun closestDivisors(num: Int): IntArray {
val sqrt1 = Math.sqrt(num + 1.0).toInt()
val sqrt2 = Math.sqrt(num + 2.0).toInt()
if (sqrt1 * sqrt1 == num + 1) {
return intArrayOf(sqrt1, sqrt1)
}
if (sqrt2 * sqrt2 == num + 2) {
return intArrayOf(sqrt2, sqrt2)
}
var ans1 = IntArray(2)
for (i in sqrt1 downTo 1) {
if ((num + 1) % i == 0) {
ans1 = intArrayOf(i, (num + 1) / i)
break
}
}
var ans2 = IntArray(2)
for (i in sqrt2 downTo 1) {
if ((num + 2) % i == 0) {
ans2 = intArrayOf(i, (num + 2) / i)
break
}
}
return if (Math.abs(ans2[0] - ans2[1]) < Math.abs(ans1[0] - ans1[1])) ans2 else ans1
}
}