No. 24 - Intersection of Sorted Arrays

No. 24 - Intersection of Sorted Arrays

 

Problem: Please implement a function which getsthe intersection of two sorted arrays. Assuming numbers in each array areunique.

 

For example, ifthe two sorted arrays as input are {1, 4, 7, 10, 13} and {1, 3, 5, 7, 9}, itreturns an intersection array with numbers {1, 7}.

 

Analysis: An intuitive solution for this problemis to check whether every number in the first array (denoted as array1) is in the second array (denoted as array2). If the length of array1 is m, and the length of array2 is n, its overall time complexity is O(m*n)based on linear search. We have two better solutions.

 

Solution 1: With O(m+n) Time

 

It is noticeablethat the two input arrays are sorted. Supposing a number number1 in array1equals to a number number2 in array2, the numbers after number1 in array1 should be greater than the numbersbefore number2 in array2. Therefore, it is not necessary tocompare the numbers after number1 in array1 with numbers before number2 in array2. It improves efficiency since manycomparisons are eliminated.

 

The sample codefor this solution is shown below:

 

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;

   }

}

 

Since it onlyrequires to scan two arrays once, its time complexity is O(m+n).

 

Solution 2: With O(nlogm) Time

 

As we know, abinary search algorithm requires O(logm) time to find a number in anarray with length m. Therefore, if we search each number of an array withlength n from an array with lengthm, itsoverall time complexity is O(nlogm). If m is much greater than n, O(nlogm) is actuallyless than O(m+n). Therefore, we can implement a new and bettersolution based on binary search in such a situation. 

 

For instance,the following same code is suitable when array1 is much longer than array2.

 

/* === Supposingarray1 is much longer than array2 === */

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);

   }

}

 

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