# Find two repeating elements in an Array

Raihan Jamal
Ranch Hand
Posts: 86
This code works fine. But it is O(n^2) complexity. Is there any other efficient way to do this. I want to do it in O(n) or O(log n)

Henry Wong
author
Marshal
Posts: 21490
84
Raihan Jamal wrote:This code works fine. But it is O(n^2) complexity. Is there any other efficient way to do this. I want to do it in O(n) or O(log n)

Keep in mind that a lower order doesn't mean that it is faster. It just means that it scales better.

So, if you want a lower order, you can actually move the data into a collection first. For example, if you simply add each element into a hashmap, but check if the element is already in the hashmap first (if the put() method returns null or not), then I can get it much lower.

Henry

Mohamed Sanaulla
Saloon Keeper
Posts: 3159
33
i like Henry's approach. you can populate the elements in the Hashmap- the key would be array element and the value will be number of times its repeating.
analysing the efficiency would be O(n) to loop through the list, O(1) to fetch element from Hashmap. so effectively it would be O(n). And then another loop to fetch the keys from the Hashmap.

Sagar Dabas
Ranch Hand
Posts: 47
Mohamed Sanaulla wrote:i like Henry's approach. you can populate the elements in the Hashmap- the key would be array element and the value will be number of times its repeating.
analysing the efficiency would be O(n) to loop through the list, O(1) to fetch element from Hashmap. so effectively it would be O(n). And then another loop to fetch the keys from the Hashmap.

Am i on the right track...??

Mohamed Sanaulla
Saloon Keeper
Posts: 3159
33
Did you try running the program? let us know the results and if it holds good for different inputs.

Sagar Dabas
Ranch Hand
Posts: 47
Mohamed Sanaulla wrote:Did you try running the program? let us know the results and if it holds good for different inputs.

Integer[] a = {1,2,3,3,0,1,5,1,1};
run:
1 repeats : 4
3 repeats : 2

Integer[] a = {1,2,3,4,0,1,5,2,1};
run:
1 repeats : 3
2 repeats : 2

Yes,it is working. Is it's efficiency O(n)??

Is containsKey(x) function not increasing the complexity??

Winston Gutkowski
Bartender
Posts: 10527
64
Sagar Dabas wrote:Yes,it is working. Is it's efficiency O(n)??

Is containsKey(x) function not increasing the complexity??

Yes, and so are the map.get() and map.put(), not to mention the final iteration to check occurrence values; but they aren't changing the behaviour of the algorithm with relation to time.

Big O notation is an indication of scalability, not of elapse time; so if a program has O(n) characteristics, if it takes time T to execute when n=1, it will take time 10T when n=10. But that says nothing about how big (or long) T actually is.

Winston

Aditya Jha
Ranch Hand
Posts: 227
If you're interested in finding repeated element(s) only (and not how many times they have repeated), you can do the following:
Just as a side note - at the end, the set will have only unique elements from the array.