Naive Python (30%) with explanation

  • 0
    class Solution(object):
        def minAbbreviation(self, target, dictionary):
            :type target: str
            :type dictionary: List[str]
            :rtype: str
            def checkWord(word, dictionary, l):
                # return if a word in valid according to the dictionary
                for check_word in dictionary:
                    if l!=len(check_word):
                    for element in abbr_word:
                        if type(element)==int: # a number indicates some str, just skip.
                            i += element
                            if element != check_word[i]:
                                break # not match, continue to the next check_word
                            i += 1
                    if i==l:
                        return False # match! not valid!
                return True # valid
            # 0. filter the dictionary
            l = len(target)
            dictionary = [d for d in dictionary if len(d)==l]
            if not dictionary: return str(l)
            # 1. list all the abbreviations
            abbr = [] # store all the abbr.
            for i in range(2**l):
                # bin(i) denotes an abbr.
                t = bin(i)[2:].zfill(l) # abbr in string format
                j = 0
                abbr_word = [] # a formatted abbr.
                while j<l:
                    if t[j]=='1': # abbr starts here at j
                        k = j+1
                        while k<l and t[k]=='1':
                            k += 1
                        n = k-j # length of abbreviated chunk of string
                        j = k
                        j += 1
            # 2. sort the list
            abbr = sorted(abbr, key=len)
            # 3. iterate through the elements of the list and check, if valid abbr. found, return instantly.
            for abbr_word in abbr:
                if checkWord(abbr_word, dictionary, l):
                    return ''.join(map(lambda x: str(x), abbr_word))

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