Group Anagrams

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My Thoughts

What Went Well
I knew that all anagrams contain the same characters, so I implemented an approach using this attribute.
I sorted each word, and using the sorted characters of that word allowed me to have a key that would access all anagrams.

Data Structure Description

Hash Map (Dictionary) - A data structure that stores a collection of keys, each to their respective values. Each key may only appear once.

Solution Statistics

Time Spent Coding
5 minutes

Time Complexity
O(n * k) - The number of strings in strs is n, and the largest string in strs is of length k. Assuming the previous values of n and k, we must iterate through all strings in strs and each character in each of the strings in strs, resulting in the O(n * k) time complexity. Technically we are iterating through each stings characters twice, but since big O “ignores” constant multiples, the time complexity does not change.

Space Complexity
O(n) - We will never store more values or keys in the dictionary than elements in the strs list, resulting in the O(n) space complexity.

Runtime Beats
86.24% of other submissions

Memory Beats
97.57% of other sumbissions

Solution

class Solution(object):
    def groupAnagrams(self, strs):
        dic = {}

        for s in strs:
            cur = "".join(sorted(s))

            if cur in dic:
                dic[cur].append(s)
            else:
                dic[cur] = [s]
                
        return dic.values()