# LeetCode 350. Intersection of Two Arrays II

## Description

https://leetcode.com/problems/intersection-of-two-arrays-ii/

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?

## Explanation

use a hashmap to check intersections

## Python Solution

``````class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
results = []

counter1 = {}
for num in nums1:
counter1[num] = counter1.get(num, 0) + 1

counter2 = {}
for num in nums2:
counter2[num] = counter2.get(num, 0) + 1

for key, value in counter1.items():
if key not in counter2:
continue
else:
results += [key] * min(value, counter2[key])

return results``````
• Time complexity: O(n+m), where n and m are the lengths of the arrays. We iterate through the first, and then through the second array; insert and lookup operations in the hash map take a constant time.
• Space complexity: O(min(n,m)). We use a hash map to store numbers (and their counts) from the smaller array.