The technical question was: You're given an array of strings. Sort it so that the result returns an array of an array of strings sorted into its anagrams. (e.g. input: ["aa", "ad", "da"], output: [ ["aa"], ["ad", "da"] ]

Question

Sigiloso · Accepted Answer

def yp(words):
    words = set(words)      # to make searh O(1)
    out = []
    
    while words:            # cost O(N * O(1)) := O(N) 
        word = words.pop()        
        anagram = reversed(word)
                
        if anagram in words:
            words.discard(anagram)
            out.append([word, anagram])
        else:
            out.append([word])

    return sorted(out)      # O(C * N * log(N)) + O(N) := O(N * log(N))
    
if __name__ == "__main__":
    print( yp(["aa", "ad", "da"]) )

Sigiloso · Answer

from collections import defaultdict
def isAnagram(word, otherWord):
	if len(word) != len(otherWord):
		return False
	word_dict = defaultdict(int)
	otherWord_dict = defaultdict(int)
	for char in word:
		word_dict[char] += 1
	for char in otherWord:
		otherWord_dict[char] += 1
	for char, count in word_dict.items():
		if count != otherWord_dict[char]:
			return False
	return True

def main():
	lst = None
	with open('data.txt', 'r') as f:
		lst = f.readline().split()
	i = 0
	ans = []
	visited = set()
	for word in lst[i:]:
		if word not in visited:
			temp = [word]
			for otherWord in lst[i+1:]:
				if word != otherWord and otherWord not in visited and isAnagram(word, otherWord):
					temp.append(otherWord)
					visited.add(otherWord)

			visited.add(word)
			ans.append(temp)
			
	print ans

if __name__ == '__main__':
	main()

Anna · Answer

in ruby

def print_anagrams_group(arr)
  arr.group_by { |element| element.downcase.chars.sort }.values
end

print_anagrams_group(["aa", "ad", "da"]) #=> [["aa"], ["ad", "da"]]

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