# 350 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 `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?
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 `````` ``````class Solution: def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]: if len(nums1) > len(nums2): return self.intersect(nums2, nums1) m1 = {} for n in nums1: if n in m1: m1[n] += 1 else: m1[n] = 1 result = [] for n in nums2: if n in m1 and m1[n] > 0: result.append(n) m1[n] -= 1 return result``````