面试时问的时间复杂度问题

Question

int[] a = {3,5,4,3,2,2,1};

        System.out.println("Duplicate elements are: ");

        for(int i=0; i<a.length; i++) {
            for(int j = i+1; j<a.length; j++) {
                if(a[i]==a[j]  && i!=j){
                    System.out.println(a[j] + " ");
                }
            }
        }

This codes time complexity is O(n^2), interviewer asked me to change it to O(n). How can we do that?

Answer 1

This is one solution which would meet the O(n) requirement, that would be to use a HashSet - A HashSet will not accept a duplicate value, so one loop is all you need, that is if the requirements allow for use of a declared HashSet :

int[] a = {3,5,4,3,2,2,1};
// Create a set 
Set<Integer> set = new HashSet<Integer>();
//Create a set for storing duplicates
Set<Integer> setDupes = new HashSet<Integer>();

System.out.println("Duplicate elements are: ");

//loop the array once 
for (int num : a) 
{ 
     if (set.add(num) == false) 
     { 
        // your duplicate element
        if(setDupes.add(num) == true) 
           System.out.println(num + " ");
     } 
}

Answer 2

It depends on what you want to do inside the loop. If they are not dependent then you can merge both of them in one

Answer 3

When the loops are not dependent you can re-write the loop as:

    for (int i = 0, j=0; i < n && j<n; i++;j++) {
        //do something in constant time
    }