How does the intersection of two arrays work in this problem?

The problem asks you to find common elements between two arrays, with the constraint that each element in the result must appear as many times as it appears in both arrays.

What is the optimal approach for solving the Intersection of Two Arrays II?

The most optimal solution uses a hash table to store counts of elements from one array and then checks for these elements in the second array, ensuring efficient time and space usage.

What are some common mistakes when solving the Intersection of Two Arrays II?

Common mistakes include not handling duplicates properly, inefficient algorithms that lead to time complexity issues, and using too much memory by not optimizing the space usage.

How does sorting help in solving the Intersection of Two Arrays II?

Sorting allows you to apply the two-pointer technique, which is an efficient way to find common elements in sorted arrays without using extra space for hash tables.

Is the Intersection of Two Arrays II problem easy or hard?

This problem is considered easy as it can be solved with multiple approaches, but it requires careful consideration of time and space complexity to handle larger inputs.

#350

Easy

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

两个数组的交集 II

给你两个整数数组 nums1 和 nums2 ，请你以数组形式返回两数组的交集。返回结果中每个元素出现的次数，应与元素在两个数组中都出现的次数一致（如果出现次数不一致，则考虑取较小值）。可以不考虑输出结果的顺序。示例 1：输入： nums1 = [1,2,2,1], nums2 = [2,2] …

数组哈希表双指针二分查找排序

题目描述

给你两个整数数组 nums1 和 nums2 ，请你以数组形式返回两数组的交集。返回结果中每个元素出现的次数，应与元素在两个数组中都出现的次数一致（如果出现次数不一致，则考虑取较小值）。可以不考虑输出结果的顺序。

示例 1：

输入：nums1 = [1,2,2,1], nums2 = [2,2]
输出：[2,2]

示例 2:

输入：nums1 = [4,9,5], nums2 = [9,4,9,8,4]
输出：[4,9]

提示：

1 <= nums1.length, nums2.length <= 1000
0 <= nums1[i], nums2[i] <= 1000

进阶：

如果给定的数组已经排好序呢？你将如何优化你的算法？
如果 nums1 的大小比 nums2 小，哪种方法更优？
如果 nums2 的元素存储在磁盘上，内存是有限的，并且你不能一次加载所有的元素到内存中，你该怎么办？

lightbulb

解题思路

方法一：哈希表

我们可以用一个哈希表 $\textit{cnt}$ 统计数组 $\textit{nums1}$ 中每个元素出现的次数，然后遍历数组 $\textit{nums2}$ ，如果元素 $x$ 在 $\textit{cnt}$ 中，并且 $x$ 的出现次数大于 $0$ ，那么将 $x$ 加入答案，然后将 $x$ 的出现次数减一。

遍历结束后，返回答案数组即可。

时间复杂度 $O(m + n)$ ，空间复杂度 $O(m)$ 。其中 $m$ 和 $n$ 分别是数组 $\textit{nums1}$ 和 $\textit{nums2}$ 的长度。

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class Solution:
    def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
        cnt = Counter(nums1)
        ans = []
        for x in nums2:
            if cnt[x]:
                ans.append(x)
                cnt[x] -= 1
        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
Ability to implement hash table for counting frequencies efficiently.
question_mark
Demonstrates understanding of time-space tradeoffs with different approaches.
question_mark
Clear grasp of problem constraints and optimizations for large inputs.

warning

常见陷阱

外企场景

error
Not handling duplicates correctly, which can lead to incorrect results.
error
Using inefficient algorithms that result in high time complexity, especially with large input sizes.
error
Failure to optimize space usage, leading to unnecessary extra memory consumption.

swap_horiz

进阶变体

外企场景

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Intersection of three arrays with similar constraints.
arrow_right_alt
Finding the union of two arrays with the same frequency consideration.
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Intersection problem in an unsorted list with missing elements.

help

常见问题

外企场景

继续练习

#349 两个数组的交集 #532 数组中的 k-diff 数对 #1346 检查整数及其两倍数是否存在