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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?

1 Answer

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by Goeduhub's Expert (2.3k points)
 
Best answer

Simple count all the occurance of each number for nums 1 and then compare its freq from nums 2 to get final ans.

class Solution:

    def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:

        dict1 = dict()

        for i in nums1:

            if i not in dict1:

                dict1[i] = 1

            else:

                dict1[i] += 1

        ret = []

        for i in nums2:

            if i in dict1 and dict1[i]>0:

                ret.append(i)

                dict1[i] -= 1

        return ret

eg : ans = Solution()

nums1 = [1,2,2,1], nums2 = [2,2]

ans.intersect(nums1, nums2)

Output -

[2,2]

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