# 350. Intersection of Two Arrays II
Given two arrays, write a function to compute their intersection.
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]
Note:
Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.
Follow up:
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
Approach 1: sort, then use two pointers.
Approach 2: store frequencies in hashtables, then compare.
# Code (Python)
Approach 1:
class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
# sort first
nums1.sort()
nums2.sort()
p1, p2 = 0, 0
intersection = []
while p1 < len(nums1) and p2 < len(nums2):
if nums1[p1] == nums2[p2]:
intersection.append(nums1[p1])
p1 += 1
p2 += 1
elif nums1[p1] < nums2[p2]:
p1 += 1
else:
p2 += 1
return intersection
Approach 2:
import collections
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
# use counter
intersection = []
if len(nums1) > len(nums2):
nums1, nums2 = nums2, nums1
counts1 = collections.Counter(nums1)
counts2 = collections.Counter(nums2)
for num, count in counts1.items():
if num in counts2:
intersection.extend([num for _ in range(min(counts1[num], counts2[num]))])
return intersection
# Code (C++)
Approach 1:
Approach 2: