Shortest Unique prefix for every word

class Solution:
    def findPrefixes(self, arr, N):
    '''
    while adding each character to trie
    If character is not present then create a new dictionary for that character with also an item (freq:1) and get into it.

    If character is already prsent in the current dictionary
    then increase the count of that branch i.e. freq which represents the no of words passing through
    that character.

    '''
        def add(word):
            cur=root
            for w in word:
                if w in cur:
                    cur[w]['freq']+=1
                else:
                    cur[w]={'freq':1}
                cur=cur[w]
            cur['*']=True
        '''
        Now traverse each character of word through trie.
        First add character to ans and check whether the frequency of that character==1 .If so return ans
        else Traverse
        '''
        def find(word):
            ans=''
            cur=root
            for w in word:
                ans=ans+w
                if cur[w]['freq']==1:
                    return ans
                cur=cur[w]
        ans=[]
        root={}
        for word in arr:
            add(word)
        for word in arr:
            ans.append(find(word))
        return ans