Why not sorting the data first? If it is sorted, you can easily do a dichotomic search for the second number and get O(n(log(n)). Because sorting can be done in the same complexity, you get something better than the O(n2) algorithm above, with a space complexity of O(1)
You could also sort using a binary tree, then it becomes trivial to search for the second number in the tree while keeping the same complexity (with a space complexity of O(n) though)
If you use the map solution of Andrei, you don't know how much space the map will take (it depends on the implementation) but with a time complexity of O(n). If you use the set, you will get the same thing as my first solution.
We're a place where coders share, stay up-to-date and grow their careers.
We strive for transparency and don't collect excess data.