group anagrams

Author: always-maap

49.GroupAnagrams.py

@@ -0,0 +1,63 @@
+"""
+49. Group Anagrams
+Medium
+Array | Hash Table | String | Sorting
+---
+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.
+"""
+
+import collections
+from typing import List
+
+
+# O(m log n) time | O(m) space
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        dict = {}
+
+        for i in range(len(strs)):
+            curr = strs[i]
+            sorted_s = "".join(sorted(curr))
+
+            if dict.get(sorted_s):
+                dict[sorted_s].append(curr)
+            else:
+                dict[sorted_s] = [curr]
+
+        return dict.values()
+
+
+# O(m log n) time | O(m) space
+class Solution2:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        dict = collections.defaultdict(list)
+
+        for s in strs:
+            count = [0] * 26
+
+            for c in s:
+                count[ord(c) - ord("a")] += 1
+
+            dict[tuple(count)].append(s)
+
+        return dict.values()