What is the sliding window technique in the Maximum Subarray With Equal Products problem?

The sliding window technique involves maintaining a subarray and updating its properties, such as product, LCM, and GCD, as the window slides through the array.

How do you efficiently update the product, LCM, and GCD in this problem?

You update the product, LCM, and GCD as elements enter and exit the window, instead of recalculating them for every subarray, which improves efficiency.

What should I do if there are no valid subarrays in the input array?

In such cases, you should return 0, indicating that no subarray meets the product equivalent condition.

How can GhostInterview assist with this problem?

GhostInterview guides you through the sliding window approach and helps you understand how to manage running states, making the solution more efficient.

What are some common mistakes in solving the Maximum Subarray With Equal Products problem?

Common mistakes include recalculating values from scratch instead of using running state updates and neglecting to handle edge cases like small arrays.

#3411

Easy

auto_awesome滑动窗口（状态滚动更新）

LeetCode 题解工作台

最长乘积等价子数组

给你一个由正整数组成的数组 nums 。如果一个数组 arr 满足 prod(arr) == lcm(arr) * gcd(arr) ，则称其为乘积等价数组，其中： prod(arr) 表示 arr 中所有元素的乘积。 gcd(arr) 表示 arr 中所有元素的最大公因数 ( GCD )…

数组数学滑动窗口枚举数论

题目描述

给你一个由 正整数 组成的数组 nums。

如果一个数组 arr 满足 prod(arr) == lcm(arr) * gcd(arr)，则称其为 乘积等价数组 ，其中：

prod(arr) 表示 arr 中所有元素的乘积。
gcd(arr) 表示 arr 中所有元素的最大公因数 (GCD)。
lcm(arr) 表示 arr 中所有元素的最小公倍数 (LCM)。

返回数组 nums 的最长 乘积等价子数组 的长度。

示例 1：

输入： nums = [1,2,1,2,1,1,1]

输出： 5

解释：

最长的乘积等价子数组是 [1, 2, 1, 1, 1]，其中 prod([1, 2, 1, 1, 1]) = 2， gcd([1, 2, 1, 1, 1]) = 1，以及 lcm([1, 2, 1, 1, 1]) = 2。

示例 2：

输入： nums = [2,3,4,5,6]

输出： 3

解释：

最长的乘积等价子数组是 [3, 4, 5]。

示例 3：

输入： nums = [1,2,3,1,4,5,1]

输出： 5

提示：

2 <= nums.length <= 100
1 <= nums[i] <= 10

lightbulb

解题思路

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class Solution:
    def maxLength(self, nums: List[int]) -> int:
        n = len(nums)
        ans = 0
        max_p = lcm(*nums) * max(nums)
        for i in range(n):
            p, g, l = 1, 0, 1
            for j in range(i, n):
                p *= nums[j]
                g = gcd(g, nums[j])
                l = lcm(l, nums[j])
                if p == g * l:
                    ans = max(ans, j - i + 1)
                if p > max_p:
                    break
        return ans

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
Check if the candidate can efficiently handle the updates for product, LCM, and GCD as the window slides.
question_mark
Look for understanding of the sliding window technique and its application to this specific problem.
question_mark
Assess the candidate's ability to handle edge cases effectively while maintaining performance.

warning

常见陷阱

外企场景

error
Recalculating product, LCM, and GCD for every subarray instead of using a sliding window with running state.
error
Failing to handle edge cases such as very small arrays or arrays without any valid subarrays.
error
Not considering how to manage the state efficiently when the window expands or contracts.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider a variant where the numbers in the array are larger, requiring more efficient GCD and LCM calculations.
arrow_right_alt
Change the problem to find the longest subarray where the product is exactly equal to a given constant.
arrow_right_alt
Extend the problem to work with arrays containing negative numbers, adding complexity to the LCM and GCD calculations.

help

常见问题

外企场景

继续练习

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