Algorithm analysis log linear solution

Author: ivanmmarkovic

analysis/anagrams-loglinear-solution.py

@@ -1,25 +1,22 @@
 
-# O(n2)
+# O(nlogn)
 def anagrams(string1: str, string2: str) -> bool:
 
-    is_anagram: bool = True
+    if len(string1) != len(string2):
+        return False
 
+    is_anagram: bool = True
+    list1: list = list(string1)
     list2: list = list(string2)
-    pos1: int = 0
-    while pos1 < len(string1) and is_anagram:
-        character = string1[pos1]
-        character_found: bool = False
-        pos2: int = 0
-        while pos2 < len(list2) and not character_found:
-            if list2[pos2] == character:
-                list2[pos2] = None
-                character_found = True
-            else:
-                pos2 += 1
-        if not character_found:
+    list1.sort() # sorting is O(nlogn)
+    list2.sort() # sorting is O(nlogn)
+    pos: int = 0
+    # loop is O(n)
+    while pos < len(list1) and is_anagram:
+        if list1[pos] != list2[pos]:
             is_anagram = False
         else:
-            pos1 += 1
+            pos += 1
 
     return is_anagram
 

analysis/anagrams-quadratic-solution.py

@@ -1,6 +1,8 @@
 
 # O(n2)
 def anagrams(string1: str, string2: str) -> bool:
+    if len(string1) != len(string2):
+        return False
 
     is_anagram: bool = True