25 lines
744 B
Python
25 lines
744 B
Python
# Time: O(N · M)
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# Space: O(N)
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from collections import Counter, defaultdict
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class Solution:
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def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
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# If key doesn't exist, creates empty list using `list`
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# factory/ctor function
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groups = defaultdict(list)
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for s in strs:
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# Keep counter for num of times each letter appeared
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counts = [0] * 26
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for c in s:
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i = ord(c) - ord('a')
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counts[i] += 1
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# Lists aren't hashable, so we need to convert the counts
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# to tuple so that we can key them
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groups[tuple(counts)].append(s)
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return list(groups.values())
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