What is the best approach for solving the Intersection of Multiple Arrays problem?

The most efficient solution uses array scanning along with a hash table to track the frequency of each integer. Elements with a frequency equal to the number of subarrays are common.

Can the solution for the Intersection of Multiple Arrays problem be optimized further?

Yes, sorting the subarrays and using set intersections can sometimes simplify the solution. Hash maps generally offer better time complexity for large inputs.

What is the time complexity of solving the Intersection of Multiple Arrays problem?

The time complexity is O(n * m) for the array scanning with hash map approach, where n is the number of subarrays and m is the average length of the subarrays.

How do hash tables help in solving the Intersection of Multiple Arrays problem?

Hash tables allow you to efficiently count the occurrences of each integer across subarrays, helping to identify common integers quickly.

What if there are no common elements in the Intersection of Multiple Arrays problem?

In that case, the correct output would be an empty list, as there are no integers appearing in every subarray.

#2248

Easy

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

多个数组求交集

给你一个二维整数数组 nums ，其中 nums[i] 是由不同正整数组成的一个非空数组，按升序排列返回一个数组，数组中的每个元素在 nums 所有数组中都出现过。示例 1：输入： nums = [[ 3 ,1,2, 4 ,5],[1,2, 3 , 4 ],[ 3 , 4 ,5,6]]…

数组哈希表排序计数

题目描述

给你一个二维整数数组 nums ，其中 nums[i] 是由不同正整数组成的一个非空数组，按 升序排列 返回一个数组，数组中的每个元素在 nums 所有数组 中都出现过。

示例 1：

输入：nums = [[3,1,2,4,5],[1,2,3,4],[3,4,5,6]]
输出：[3,4]
解释：
nums[0] = [3,1,2,4,5]，nums[1] = [1,2,3,4]，nums[2] = [3,4,5,6]，在 nums 中每个数组中都出现的数字是 3 和 4 ，所以返回 [3,4] 。

示例 2：

输入：nums = [[1,2,3],[4,5,6]]
输出：[]
解释：
不存在同时出现在 nums[0] 和 nums[1] 的整数，所以返回一个空列表 [] 。

提示：

1 <= nums.length <= 1000
1 <= sum(nums[i].length) <= 1000
1 <= nums[i][j] <= 1000
nums[i] 中的所有值 互不相同

lightbulb

解题思路

方法一：计数

遍历数组 nums，对于每个数组 arr，统计数组 arr 中每个数字出现的次数，然后遍历计数数组，统计出现次数等于数组 nums 的长度的数字，即为答案。

时间复杂度 $O(N)$ ，空间复杂度 $O(1000)$ 。其中 $N$ 为数组 nums 中数字的总数。

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class Solution:
    def intersection(self, nums: List[List[int]]) -> List[int]:
        cnt = [0] * 1001
        for arr in nums:
            for x in arr:
                cnt[x] += 1
        return [x for x, v in enumerate(cnt) if v == len(nums)]

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
Strong candidates will use a hash table to efficiently count occurrences and check common elements.
question_mark
Look for solutions that optimize the time complexity using sorting or direct set operations.
question_mark
Candidates should demonstrate an understanding of array scanning and hash table usage.

warning

常见陷阱

外企场景

error
Failing to account for the case where no common elements exist, leading to incorrect results.
error
Not efficiently handling large inputs, causing time or space complexity issues.
error
Misusing sorting or set operations that increase the problem's complexity unnecessarily.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Change the problem to find common elements that appear in at least half the subarrays.
arrow_right_alt
Alter the problem to return only those common elements that appear more than once in each subarray.
arrow_right_alt
Modify the problem to require returning the common elements in sorted order.

help

常见问题

外企场景

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