Question

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?

Difficulty:Easy

Category:Hash-Table, Two-Points, Binary-Search, Sort

Solution

Solution 1: HashTable

Time Complexity: O(n) Space Complexity: O(n)

class Solution {
 public:
  vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
    vector<int> ans;
    unordered_map<int, int> cnts;
    for (int& num : nums1) ++cnts[num];
    for (int& num : nums2) {
      if (cnts[num]) {
        ans.emplace_back(num);
        --cnts[num];
      }
    }
    return ans;
  }
};

Solution 2: Sorting

Time Complexity: O(n log n) Space Complexity: O(n)

class Solution {
 public:
  vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
    vector<int> res;
    int i = 0, j = 0;
    sort(nums1.begin(), nums1.end());
    sort(nums2.begin(), nums2.end());
    while (i < nums1.size() && j < nums2.size()) {
      if (nums1[i] == nums2[j]) {
        res.push_back(nums1[i]);
        ++i;
        ++j;
      } else if (nums1[i] > nums2[j]) {
        ++j;
      } else {
        ++i;
      }
    }
    return res;
  }
};