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.

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.

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.

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

Bats fly at night, 'cause they aren't we. And if we tried, we'd hit a tree -- Ogden Nash (or should've been).
Articles by Winston can be found here

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.