notes/OJ notes/pages/Leetcode Intersection-of-Two-Arrays-II.md

Leetcode Intersection-of-Two-Arrays-II

2022-06-11

First revision 2022-06-27

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Data structures:

#DS #unordered_map

Algorithms:

#algorithm #two_pointers #sort

Difficulty:

#leetcode #coding_problem #difficulty_easy

Related topics:

Links:

• Link to problem
• Solution 2(two pointers method)
• Solution 3 and 4

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Problem

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.

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?

Examples

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

Thoughts

For the original problem, I thought up an O(m + n) algo, that uses C++'s cpp_std_unordered_map, and for the second one, I use double pointer method.

Tip

Because elements can be duplicated and we need to know how many, we should use unordered map to store the item's appereance times, (Maybe multiset can work too.)

[!tip] Use cpp_std_unordered_map for a O(1) hash table when not sorted #tip

[!tip] Use cpp_std_sort to sort anything. #tip

Solution

Unordered map way O(m + n)

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
      vector<int> answer;
      unordered_map<int, int> umap;
      for (int i = 0; i < nums1.size(); i++) {
        umap[nums1[i]]++;
      }

      for (int i = 0; i < nums2.size(); i++) {
        if (umap[nums2[i]] != 0) {
          answer.push_back(nums2[i]);
          umap[nums2[i]]--;
        }
      }
      return answer;
    }
};