Question

i have seen a few days ago such problem

there is given two array find      elements which are common   of these array

one of the solution was sort big array and then use binary search algorithm
and also there is another algorithm- brute-force algorithm

 for (int i=0;i<array1.length;i++){
  for (int j=0;j<array2.length;j++){
 if (array1[i]==array2[j]){
//code here
}
}

it’s complexity is O(array1.lengtharray2.length);
and i am interested the first method’s complexity is also same yes?
because we should sort array first and then use search method
binary search algorithm’s complexity is log_2(n) so it means that total time will be
array.lengthlog_2(n) and about sort?
please explain me which is better

Answer 1

An O(M log N) solution

Let the length of arr1 be O(M), and the length of arr2 be O(N). The sort/binary search algorithm is O(M log N).

The pseudocode is as follows:

SORT(arr2)   # N log N

FOR EACH element x OF arr1             # M
   IF binarySearch(x, arr2) is FOUND   # log N
       DECLARE DUP x

O(M log N) is vastly better than O(MN).

---

A linear-time solution

There’s also a third way which is O(M+N), using a set that has a O(1) insertion and test. A hash-based set meets this expectation.

The pseudocode is as follows:

INIT arr1set AS emptySet

FOR EACH element x OF arr1    # M
   INSERT x INTO arr1set      # 1

FOR EACH element x OF arr2    # N
   IF arr1set CONTAINS x      # 1
      DECLARE DUP x