2334. Subarray With Elements Greater Than Varying Threshold

Problem Link

Description

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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 <= 105
• 1 <= nums[i], threshold <= 109

Solution

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Python3

class Solution:
    def validSubarraySize(self, nums: List[int], threshold: int) -> int:
        N = len(nums)
        prevSmaller = [-1] * N
        nextSmaller = [N] * N
        res = -1
 
        # construct nextSmaller array
        stack = []
        for i, x in enumerate(nums):
            while stack and x < nums[stack[-1]]:
                nextSmaller[stack.pop()] = i
            
            stack.append(i)
        
        # construct prevSmaller array
        stack = []
        for i in range(N - 1, -1, -1):
            while stack and nums[i] < nums[stack[-1]]:
                prevSmaller[stack.pop()] = i
            
            stack.append(i)
 
        for i, x in enumerate(nums):
            # fix x as the smallest element in the subarray
            length = nextSmaller[i] - prevSmaller[i] - 1
            if x > threshold // length:
                res = max(res, length)
 
        return res