What is the primary pattern used to solve the Number of Subarrays with Bounded Maximum problem?

The primary pattern is two-pointer scanning with invariant tracking, where the two pointers help manage subarray bounds and ensure the maximum remains within the specified range.

How does the two-pointer technique help solve this problem efficiently?

The two-pointer technique allows for efficient sliding window management, where the pointers expand and contract based on the maximum element in the current window, avoiding redundant calculations.

Can you describe the time and space complexity of this solution?

The time complexity is O(n) because each element is processed at most twice, and the space complexity is O(1) since only a few variables are used to track the current window state.

What are some common mistakes in solving this problem?

Common mistakes include failing to properly track subarray bounds, recalculating the maximum for each subarray, and not using a sliding window approach effectively.

How can GhostInterview assist in improving my solution to this problem?

GhostInterview provides instant feedback, helping refine your understanding of the two-pointer technique, pointing out inefficiencies, and ensuring optimal subarray calculations.

#795

Medium

auto_awesome双·指针·invariant

LeetCode 题解工作台

区间子数组个数

给你一个整数数组 nums 和两个整数： left 及 right 。找出 nums 中连续、非空且其中最大元素在范围 [left, right] 内的子数组，并返回满足条件的子数组的个数。生成的测试用例保证结果符合 32-bit 整数范围。示例 1：输入： nums = [2,1,4,3],…

数组双指针

题目描述

给你一个整数数组 nums 和两个整数：left 及 right 。找出 nums 中连续、非空且其中最大元素在范围 [left, right] 内的子数组，并返回满足条件的子数组的个数。

生成的测试用例保证结果符合 32-bit 整数范围。

示例 1：

输入：nums = [2,1,4,3], left = 2, right = 3
输出：3
解释：满足条件的三个子数组：[2], [2, 1], [3]

示例 2：

输入：nums = [2,9,2,5,6], left = 2, right = 8
输出：7

提示：

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹
0 <= left <= right <= 10⁹

lightbulb

解题思路

方法一：区间计数

题目要我们统计数组 nums 中，最大值在区间 $[left, right]$ 范围内的子数组个数。

对于区间 $[left,..right]$ 问题，我们可以考虑将其转换为 $[0,..right]$ 然后再减去 $[0,..left-1]$ 的问题。也就是说，所有最大元素不超过 $right$ 的子数组个数，减去所有最大元素不超过 $left-1$ 的子数组个数，剩下的就是最大元素在区间 $[left,..right]$ 范围内的子数组个数，即题目要求的结果。

ans = \sum_{i=0}^{right} ans_i - \sum_{i=0}^{left-1} ans_i

对于本题，我们设计一个函数 $f(x)$ ，表示数组 nums 中，最大值不超过 $x$ 的子数组个数。那么答案为 $f(right) - f(left-1)$ 。函数 $f(x)$ 的执行逻辑如下：

用变量 $cnt$ 记录最大值不超过 $x$ 的子数组的个数，用 $t$ 记录当前子数组的长度。
遍历数组 nums，对于每个元素 $nums[i]$ ，如果 $nums[i] \leq x$ ，则当前子数组的长度加一，即 $t=t+1$ ，否则当前子数组的长度重置为 0，即 $t=0$ 。然后将当前子数组的长度加到 $cnt$ 中，即 $cnt = cnt + t$ 。
遍历结束，将 $cnt$ 返回即可。

时间复杂度 $O(n)$ ，空间复杂度 $O(1)$ 。其中 $n$ 为数组 nums 的长度。

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class Solution:
    def numSubarrayBoundedMax(self, nums: List[int], left: int, right: int) -> int:
        def f(x):
            cnt = t = 0
            for v in nums:
                t = 0 if v > x else t + 1
                cnt += t
            return cnt

        return f(right) - f(left - 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 demonstrates proficiency with sliding window and two-pointer techniques.
question_mark
Candidate can clearly explain how the two-pointer technique avoids redundant calculations.
question_mark
Candidate optimizes their approach to avoid recalculating the maximum for each subarray.

warning

常见陷阱

外企场景

error
Failing to properly track subarray bounds and missing valid subarrays.
error
Over-complicating the solution with nested loops or redundant calculations.
error
Not efficiently managing the sliding window and recalculating the maximum for each subarray.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the subarrays can contain more than one element with the same maximum value?
arrow_right_alt
What if there were additional constraints on subarray lengths?
arrow_right_alt
How would the solution change if the input array was sorted?

help

常见问题

外企场景

继续练习

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