#2334

Hard

auto_awesomeStack-based state management

LeetCode Problem Workspace

Subarray With Elements Greater Than Varying Threshold

Find the size of a subarray with all elements greater than threshold divided by length using stack-based state management.

array stack union find monotonic stack

Problem Statement

You are given an integer array nums and an integer threshold. The task is to find a subarray in nums of length k such that every element in the subarray is greater than threshold / k.

Return the size of any such subarray. If no such subarray exists, return -1. The problem requires handling this efficiently, using stack-based state management to check the conditions for valid subarrays.

Examples

Example 1

Input: nums = [1,3,4,3,1], threshold = 6

Output: 3

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

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 Approach

Stack-Based State Management

To solve this problem, maintain a stack to track the current subarray and its properties. By iterating through the array, for each subarray of length k, check if the minimum element is greater than threshold / k. If the condition is met, return the length of the subarray.

Sliding Window Technique

Another approach involves using a sliding window over the array. Maintain the window size while verifying the condition for every element inside the window. This method helps reduce redundant calculations and ensures the algorithm runs efficiently.

Binary Search for Optimization

For optimization, binary search can be employed on the subarray length. For each length, use a stack or sliding window approach to verify if a valid subarray exists, narrowing down the possibilities until the correct answer is found.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the stack-based approach depends on the final approach used but is typically O(n) where n is the length of the array. Space complexity will also vary, typically O(k) for maintaining the stack or window.

What Interviewers Usually Probe

Candidate effectively utilizes stack-based state management to track valid subarrays.
Candidate suggests a sliding window to avoid redundant checks and improve efficiency.
Candidate demonstrates an understanding of binary search for optimization of subarray length checks.

Common Pitfalls or Variants

Common pitfalls

Incorrectly managing the sliding window or stack, leading to incorrect subarray evaluations.
Failing to account for edge cases like the smallest or largest subarray sizes.
Overlooking the need for an efficient solution, potentially resulting in a time complexity of O(n^2).

Follow-up variants

Changing the condition for valid subarrays (e.g., using a different mathematical relationship for the elements).
Restricting the array to smaller or larger sizes, affecting the stack or sliding window management.
Including additional constraints like negative numbers or specific ranges for elements.

FAQ

What is the stack-based state management approach for this problem?

In this problem, stack-based state management helps track the elements in the subarray, ensuring that the minimum element in the subarray is greater than threshold / k. The stack stores elements while maintaining their valid subarray status.

How do I optimize the solution for this problem?

You can optimize the solution by using binary search on subarray lengths and combining it with stack-based or sliding window checks for valid subarrays, reducing redundant computations.

What is the significance of `threshold / k` in this problem?

The condition that each element in the subarray must be greater than threshold / k ensures that the subarray's elements meet a specific threshold condition based on its length. This relationship is key to validating potential subarrays.

Can a sliding window approach be used here?

Yes, a sliding window approach can be used to maintain and check the subarray's validity while iterating through the array, ensuring each element satisfies the required condition.

What are some common pitfalls to avoid when solving this problem?

Some common pitfalls include incorrectly managing the sliding window or stack, failing to handle edge cases like empty subarrays, and not optimizing the solution to avoid excessive time complexity.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def validSubarraySize(self, nums: List[int], threshold: int) -> int:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        def merge(a, b):
            pa, pb = find(a), find(b)
            if pa == pb:
                return
            p[pa] = pb
            size[pb] += size[pa]

        n = len(nums)
        p = list(range(n))
        size = [1] * n
        arr = sorted(zip(nums, range(n)), reverse=True)
        vis = [False] * n
        for v, i in arr:
            if i and vis[i - 1]:
                merge(i, i - 1)
            if i < n - 1 and vis[i + 1]:
                merge(i, i + 1)
            if v > threshold // size[find(i)]:
                return size[find(i)]
            vis[i] = True
        return -1

Solution 2