Problem description:

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

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:

For this question, we can easily use an unordered_map to count the number that appear in nums1. Then walk through the nums2, and store the same number to the result.

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class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> dict;
        vector<int> res;
        for(int i = 0; i < (int)nums1.size(); i++) 
            dict[nums1[i]]++;
        
        for(int i = 0; i < (int)nums2.size(); i++)
            if(dict.find(nums2[i]) != dict.end() && --dict[nums2[i]] >= 0) 
        
        res.push_back(nums2[i]);            
    return res;
    }
};

Follow up:

  • What if the given array is already sorted? How would you optimize your algorithm?
    If the array is already sorted, we can use binary search to speed up the process.

  • What if nums1’s size is small compared to nums2’s size? Which algorithm is better?
    It’s better to use the smaller array to construct the counter hash.

  • 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?
    Divide and conquer. Repeat the process frequently: Slice nums2 to fit into memory, process (calculate intersections), and write partial results to memory.

reference:
https://goo.gl/CqspgR
https://goo.gl/rewNgE