Update Logic

Author: Haleshot

Neetcode/Neetcode_150/Group_Anagrams/Group_Anagrams.py

@@ -1,16 +1,29 @@
 # Neetcode
 class Solution:
     def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
-        results = []
-        for i in strs:
-            for j in range(strs.index(i) + 1, len(strs)):
-                if sorted(i) == sorted(strs[j]):
-                    results.append([i, strs[j]])
-        for i in results:
-            for j in range(len(i) - 1):
-                if i[j] == i[j + 1]:
-                    print(i[j])
-        return results
+
+        # # Step 1
+        # anagrams = defaultdict(list)  # Initialize a dictionary to hold lists of anagrams
+        
+        # # Iterate over each word in the list
+        # for word in strs:
+        #     # Sort the characters of the word to use as the key
+        #     sorted_word = ''.join(sorted(word))
+        #     # Group anagrams together using the sorted word as the key
+        #     anagrams[sorted_word].append(word)
+        
+        # # Convert the dictionary values (which are lists of anagrams) to a list of lists
+        # return list(anagrams.values())
+
+        # # Step 2:
+        ans = collections.defaultdict(list)
+
+        for s in strs:
+            count = [0] * 26
+            for c in s:
+                count[ord(c) - ord("a")] += 1
+            ans[tuple(count)].append(s)
+        return ans.values()
 
 
 # Leetcode
@@ -20,3 +33,26 @@ def groupAnagrams(self, strs):
         :type strs: List[str]
         :rtype: List[List[str]]
         """
+        # # Step 1
+        # anagrams = defaultdict(list)  # Initialize a dictionary to hold lists of anagrams
+        
+        # # Iterate over each word in the list
+        # for word in strs:
+        #     # Sort the characters of the word to use as the key
+        #     sorted_word = ''.join(sorted(word))
+        #     # Group anagrams together using the sorted word as the key
+        #     anagrams[sorted_word].append(word)
+        
+        # # Convert the dictionary values (which are lists of anagrams) to a list of lists
+        # return list(anagrams.values())
+
+        # # Step 2:
+        ans = collections.defaultdict(list)
+
+        for s in strs:
+            count = [0] * 26
+            for c in s:
+                count[ord(c) - ord("a")] += 1
+            ans[tuple(count)].append(s)
+        return ans.values()
+