# Find smallest subset prefixes

Here is my code for this problem. Any bugs, performance in terms of algorithm time complexity, code style advice are appreciated.

The areas which I want to improve, but do not know how to improve are:

1. I sort the strings by length and I'm not sure if it's a smarter idea not using sorting at all.
2. I used a few set operations, which looks a bit ugly. I want to learn smarter ideas.

Problem:

Given a set of strings, return the smallest subset that contains the longest possible prefixes for every string.

If the list is ['foo', 'foog', 'food', 'asdf'] return ['foo', 'asdf']

The return is foo since foo is prefix for foo (itself), prefix for foog and prefix for food (in other words, foo could "represent" longer string like foog and food). Output also contains asdf because it is not prefix for any other words in the input list, so output itself.

The empty set is not a correct answer because it does not contain the longest possible prefixes.

Source code:

from collections import defaultdict
class TrieTreeNode:
def __init__(self):
self.children = defaultdict(TrieTreeNode)
self.isEnd = False
self.count = 0
def insert(self, word, result_set):
node = self
for i,w in enumerate(word):
node = node.children[w]
if node.isEnd:
node.isEnd = True

if __name__ == "__main__":
words = ['foo', 'foog', 'food', 'asdf']
words_length = []
sorted_words = []
for w in words:
words_length.append(len(w))
for t in sorted(zip(words_length, words)):
sorted_words.append(t[1])
result_set = defaultdict(set)
root = TrieTreeNode()
for w in sorted_words:
root.insert(w, result_set)
match_set = set()
for k,v in result_set.items():
for i in v:
unmatch_set = set(words) - match_set
print unmatch_set | set(result_set.keys())

• I am afraid the problem is not well defined. An empty set is obviously a solution.
– vnp
Commented Dec 21, 2016 at 6:45
• Why do you have 'asdf' in result if it's not a prefix for first 3 strings?
– Alex
Commented Dec 21, 2016 at 10:59
• The problem you've solved needs a better description.
– Mast
Commented Dec 21, 2016 at 16:32
• @LinMa isn't the shortest prefix between "Foo" and "Bar" nothing? As it'd be ("" + "Foo", "" + "Bar"). If however it were "Foo" and "Food" then it would be ("Foo" + "", "Foo" + "d"). Is your code broken, or did it pass the online test? Commented Dec 22, 2016 at 0:10
• I've edited the question to be more specific about what I believe is the purpose of the program. Commented Dec 22, 2016 at 5:37

The best solution I have for this. If my understanding is good, this is a fairly simple to solve and you do not need anything else than a set.

There might be a better way than rebuilding entirely the set.

words = ['foo', 'foog', 'food', 'asdf']
prefixes = set()

for w in words:
if any(p for p in prefixes if w.startswith(p)):
# A prefix exists in the prefixes set.
continue

# Re-build the set with all elements not matching the set.
prefixes = set(p for p in prefixes if not p.startswith(w))

print(prefixes)

• Thanks FunkySayu, I think you compare each string by one by one and time complexity is O(n^2), do you think using Trie tree or any other solutions help on improving time complexity? Commented Dec 23, 2016 at 23:43
• Hi FunkySayu, I have written a more elegant version of code, your advice is highly appreciated => codereview.stackexchange.com/questions/150911/… Commented Dec 27, 2016 at 3:37

If you want to keep the Trie, then you should simplify your code. You can get the code to work with insert, without the if or result_set. You also don't need the self.count, and so I'd change your code to:

class TrieTreeNode:
def __init__(self):
self.children = defaultdict(TrieTreeNode)
self.is_end = False

def insert(self, word):
node = self
for char in word:
node = node.children[char]
node.isEnd = True

if __name__ == "__main__":
words = ['foo', 'foog', 'food', 'asdf']
root = TrieTreeNode()
for word in words:
root.insert(word)


After this you just need a loop that will tell you the roots. The children from root all have one child, otherwise that's the end of the word, or if it reaches the end of a word. And so you should be able to get:

output = []
for char, child in root.children.items():
word = [char]
while not child.is_end and len(child.children) == 1:
children = child.children
letter = children.keys()[0]
word.append(letter)
child = children[letter]
output.append(''.join(word))
print output

• Thanks Peilonrayz, love your implementation and vote up, but do not know why you need this condition len(child.children) == 1? Commented Dec 27, 2016 at 3:09
• Hi Peilonrayz, I tested and it seems len(child.children) == 1 is not needed, for example, if input is abc and abd, we need both as output, ab is common prefix, but not a complete word. Please feel free to correct me if I am wrong. Commented Dec 27, 2016 at 3:26
• Hi Peilonrayz, I have written a more elegant version of code, your advice is highly appreciated => codereview.stackexchange.com/questions/150911/… Commented Dec 27, 2016 at 3:37
• @LinMa Removing just the and len(child.children) == 1 causes it to output abc. Please recheck your code. As for your last comment, please don't do that, I'm under no obligation to help you. But learn SRP. Commented Dec 27, 2016 at 4:09