Coding Interview Questions: No. 24 - Intersection of Sorted Arrays


Problem: Please implement a function which gets the intersection of two sorted arrays. Assuming numbers in each array are unique.
Solution 1: With O(m+n) Time
void GetIntersection_solution1(const vector<int>& array1,
                     const vector<int>& array2,
                     vector<int>& intersection)
{
    vector<int>::const_iterator iter1 = array1.begin();
    vector<int>::const_iterator iter2 = array2.begin();

    intersection.clear();

    while(iter1 != array1.end() && iter2 != array2.end())
    {
        if(*iter1 == *iter2)
        {
            intersection.push_back(*iter1);
            ++ iter1;
            ++ iter2;
        }
        else if(*iter1 < *iter2)
            ++ iter1;
        else
            ++ iter2;
    }
}
Solution 2: With O(nlogm) Time
Therefore, if we search each number of an array with length n from an array with lengthm, its overall time complexity is O(nlogm). If m is much greater than n, O(nlogm) is actually less than O(m+n).
==One small improvement is to track the last return index when do binary search, and ignore the smaller part in next binary search.

void GetIntersection_solution2(const vector<int>& array1,
                     const vector<int>& array2,
                     vector<int>& intersection)
{
    intersection.clear();
   
    vector<int>::const_iterator iter1 = array1.begin();
    while(iter1 != array1.end())
    {
        if(binary_search(array2.begin(), array2.end(), *iter1))
            intersection.push_back(*iter1);
    }
}
The second solution works nicely even if the shorter array (here length n) is unsorted. 
Read full article from Coding Interview Questions: No. 24 - Intersection of Sorted Arrays

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