LeetCode-in-Java
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.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.TreeSet;
public class Solution {
public int validSubarraySize(int[] nums, int threshold) {
int n = nums.length;
int[] min = new int[n];
// base case
TreeSet<Integer> dead = new TreeSet<>(Arrays.asList(n, -1));
PriorityQueue<Integer> maxheap = new PriorityQueue<>(Comparator.comparingInt(o -> -min[o]));
for (int i = 0; i < n; i++) {
min[i] = threshold / nums[i] + 1;
if (min[i] > nums.length) {
// dead, this element should never appear in the answer
dead.add(i);
} else {
maxheap.offer(i);
}
}
while (!maxheap.isEmpty()) {
int 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;
}
}