LeetCode in Kotlin
2523. Closest Prime Numbers in Range
Medium
Given two positive integers left and right, find the two integers num1 and num2 such that:
left <= nums1 < nums2 <= right.nums1andnums2are both prime numbers.nums2 - nums1is the minimum amongst all other pairs satisfying the above conditions.
Return the positive integer array ans = [nums1, nums2]. If there are multiple pairs satisfying these conditions, return the one with the minimum nums1 value or [-1, -1] if such numbers do not exist.
A number greater than 1 is called prime if it is only divisible by 1 and itself.
Example 1:
Input: left = 10, right = 19
Output: [11,13]
Explanation: The prime numbers between 10 and 19 are 11, 13, 17, and 19.
The closest gap between any pair is 2, which can be achieved by [11,13] or [17,19].
Since 11 is smaller than 17, we return the first pair.
Example 2:
Input: left = 4, right = 6
Output: [-1,-1]
Explanation: There exists only one prime number in the given range, so the conditions cannot be satisfied.
Constraints:
1 <= left <= right <= 106
Solution
class Solution {
fun closestPrimes(left: Int, right: Int): IntArray {
var diff = -1
var x = -1
var y = -1
var prev = -1
for (i in left..right) {
val isPrime = isPrime(i)
if (isPrime) {
if (prev != -1) {
val d = i - prev
if (diff == -1 || d < diff) {
diff = d
x = prev
y = i
if (diff <= 2) {
return intArrayOf(x, y)
}
}
}
prev = i
}
}
return intArrayOf(x, y)
}
private fun isPrime(x: Int): Boolean {
if (x == 1) {
return false
}
val sqrt = Math.sqrt(x.toDouble())
var i = 2
while (i <= sqrt) {
if (x % i == 0) {
return false
}
i++
}
return true
}
}