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.

#2334

Hard

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LeetCode 题解工作台

元素值大于变化阈值的子数组

给你一个整数数组 nums 和一个整数 threshold 。找到长度为 k 的 nums 子数组，满足数组中每个元素都大于 threshold / k 。请你返回满足要求的任意子数组的大小。如果没有这样的子数组，返回 -1 。子数组是数组中一段连续非空的元素序列。示例 1：…

数组栈并查集单调栈

题目描述

给你一个整数数组 nums 和一个整数 threshold 。

找到长度为 k 的 nums 子数组，满足数组中每个元素都大于 threshold / k 。

请你返回满足要求的任意子数组的大小。如果没有这样的子数组，返回 -1 。

子数组 是数组中一段连续非空的元素序列。

示例 1：

输入：nums = [1,3,4,3,1], threshold = 6
输出：3
解释：子数组 [3,4,3] 大小为 3 ，每个元素都大于 6 / 3 = 2 。
注意这是唯一合法的子数组。

示例 2：

输入：nums = [6,5,6,5,8], threshold = 7
输出：1
解释：子数组 [8] 大小为 1 ，且 8 > 7 / 1 = 7 。所以返回 1 。
注意子数组 [6,5] 大小为 2 ，每个元素都大于 7 / 2 = 3.5 。
类似的，子数组 [6,5,6] ，[6,5,6,5] ，[6,5,6,5,8] 都是符合条件的子数组。
所以返回 2, 3, 4 和 5 都可以。

提示：

1 <= nums.length <= 10⁵
1 <= nums[i], threshold <= 10⁹

lightbulb

解题思路

方法一：并查集

考虑从大到小遍历数组 $nums$ 中的每个元素 $v$ ，用并查集来维护以 $v$ 作为子数组最小值的连通块。

遍历过程中：

$v$ 在数组 $nums$ 中的下标为 $i$ ，若下标 $i-1$ 对应的元素遍历过，可以将 $i-1$ 与 $i$ 进行合并，同理，若下标 $i+1$ 对应的元素也遍历过了，将 $i$ 与 $i+1$ 合并。合并过程中更新连通块的大小。

$v$ 作为当前连通块的最小值，当前连通块的大小为 $size[find(i)]$ ，若 $v>\frac{\textit{threshold}}{size[find(i)]}$ ，说明找到了满足条件的子数组，返回 $true$ 。

否则遍历结束，返回 $-1$ 。

时间复杂度 $O(nlogn)$ 。

相似题目：

1562. 查找大小为 M 的最新分组

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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

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

继续练习

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