# 350.intersection-of-two-arrays-ii

## Statement

• Difficulty: Easy
• Tag: `数组` `哈希表` `双指针` `二分查找` `排序`

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

``````

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

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

• 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？

• Link: Intersection of Two Arrays II
• Difficulty: Easy
• Tag: `Array` `Hash Table` `Two Pointers` `Binary Search` `Sorting`

Given two integer arrays `nums1` and `nums2`, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.

Example 1:

``````Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
``````

Example 2:

``````Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.
``````

Constraints:

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

• What if the given array is already sorted? How would you optimize your algorithm?
• What if `nums1`'s size is small compared to `nums2`'s size? Which algorithm is better?
• What if elements of `nums2` are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

## Solution

``````from collections import Counter
from typing import List

class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
a = Counter(nums1)
b = Counter(nums2)
res = []
for k, v in a.items():
if k in b.keys():
res.extend([k] * min(v, b[k]))
return res
``````