You can also sort the array, then traverse it with a (for i = 0; i < array.length; i +=2). If array[index] == array[index + 1] then it's a duplicate.

first loop: just use one hashset of integers and put all the numbers in the array into the hashset but before you put them in the hashset, remove the number if it already exists in the hashset 2nd loop: then loop through the hash set again with a 2nd for loop and check if the set contains the ith element in the array, the set should only have one element in it and that is the number that appeared once. one hashset only, and two for loops, O(2n) (assuming insertion/deletion O(1))

Since its from infinity to infinity, assuming that we have a large heap. 1. Create a "Set"; call it a "unique" set. 2. As we go though a number, check the "Set" to see if it already exists in the set. a. if the number is not in the set, simply put it in the set. (as this is a candiate for unique number) b. if the number is already present in the set, remove the number from the set as it is dupe. With this algoritm you should have "unique" number at any given time. Another way is to populate a map, with the number as key and "count" as value.

That's a dumb question. You can't add up or sort an infinite array.