add group-anagrams

0049_group-anagrams/README.md

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+Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.
+
+An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
+
+**Example 1:**
+
+    Input: strs = ["eat","tea","tan","ate","nat","bat"]
+    Output: [["bat"],["nat","tan"],["ate","eat","tea"]]
+    
+
+**Example 2:**
+
+    Input: strs = [""]
+    Output: [[""]]
+    
+
+**Example 3:**
+
+    Input: strs = ["a"]
+    Output: [["a"]]
+    
+
+**Constraints:**
+
+*   `1 <= strs.length <= 104`
+*   `0 <= strs[i].length <= 100`
+*   `strs[i]` consists of lowercase English letters.

0049_group-anagrams/python3/naive.py

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+# Time: O(N * MlogM) where N is len(strs) and M is num of chars in a string
+# Space: O(N)
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        groups = {}
+        
+        for s in strs:
+            s_sorted = ''.join(sorted(s))
+            
+            groups[s_sorted] = groups.get(s_sorted, []) + [s]
+        
+        return list(groups.values())

0049_group-anagrams/python3/solution.py

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