Remove duplicate characters in a given string keeping only the first occurrences. For example, if the input is ‘tree traversal’ the output will be ‘tre avsl’.
We need a data structure to keep track of the characters we have seen so far, which can perform efficient find operation. If the input is guaranteed to be in standard ASCII form, we can just create a boolean array of size 128 and perform lookups by accessing the index of the character’s ASCII value in constant time. But if the string is Unicode then we would need a much larger array of size more than 100K, which will be a waste since most of it would generally be unused.
Set data structure perfectly suits our purpose. It stores keys and provides constant time search for key existence. So, we’ll loop over the characters of the string, and at each iteration we’ll check whether we have seen the current character before by searching the set. If it’s in the set then it means we’ve seen it before, so we ignore it. Otherwise, we include it in the result and add it to the set to keep track for future reference. The code is easier to understand:
def removeDuplicates(string): result= seen=set() for char in string: if char not in seen: seen.add(char) result.append(char) return ''.join(result)
The time complexity of the algorithm is O(N) where N is the number of characters in the input string, because set supports O(1) insert and find. This is an optimal solution to one of the most common string interview questions.
Note: Complexity of insert and find for set is language and implementation dependent. Average case is mostly constant, but worst case can be logarithmic or even linear. But in an interview setting I think it’s safe to assume constant time insert and find for set.