LeetCode in Kotlin
2334. Subarray With Elements Greater Than Varying Threshold
Hard
You are given an integer array nums and an integer threshold.
Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k.
Return the size of any such subarray. If there is no such subarray, return -1.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,3,4,3,1], threshold = 6
Output: 3
Explanation: The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2.
Note that this is the only valid subarray.
Example 2:
Input: nums = [6,5,6,5,8], threshold = 7
Output: 1
Explanation: The subarray [8] has a size of 1, and 8 > 7 / 1 = 7. So 1 is returned.
Note that the subarray [6,5] has a size of 2, and every element is greater than 7 / 2 = 3.5.
Similarly, the subarrays [6,5,6], [6,5,6,5], [6,5,6,5,8] also satisfy the given conditions.
Therefore, 2, 3, 4, or 5 may also be returned.
Constraints:
1 <= nums.length <= 1051 <= nums[i], threshold <= 109
Solution
import java.util.PriorityQueue
import java.util.TreeSet
class Solution {
fun validSubarraySize(nums: IntArray, threshold: Int): Int {
val n = nums.size
val min = IntArray(n)
// base case
val dead = TreeSet(listOf(n, -1))
val maxheap = PriorityQueue(compareBy { o: Int -> -min[o] })
for (i in 0 until n) {
min[i] = threshold / nums[i] + 1
if (min[i] > nums.size) {
// dead, this element should never appear in the answer
dead.add(i)
} else {
maxheap.offer(i)
}
}
while (maxheap.isNotEmpty()) {
val cur = maxheap.poll()
if (dead.higher(cur)!! - dead.lower(cur)!! - 1 < min[cur]) {
// widest open range < minimum required length, this index is also bad.
dead.add(cur)
} else {
// otherwise we've found it!
return min[cur]
}
}
return -1
}
}