Optimize remove_duplicates from O(n²) to O(n) time complexity

Use a set for O(1) membership checks instead of checking membership in a list which is O(n). This reduces the overall time complexity from O(n²) to O(n).

Added documentation for time and space complexity.

Co-Authored-By: Keon <kwk236@gmail.com>

Author: devin-ai-integration[bot]

algorithms/arrays/remove_duplicates.py

@@ -6,13 +6,18 @@
 
 Input: [1, 1 ,1 ,2 ,2 ,3 ,4 ,4 ,"hey", "hey", "hello", True, True]
 Output: [1, 2, 3, 4, 'hey', 'hello']
+
+Time Complexity: O(n) where n is the length of the input array
+Space Complexity: O(n) for the seen set and result array
 """
 
 def remove_duplicates(array):
+    seen = set()
     new_array = []
 
     for item in array:
-        if item not in new_array:
+        if item not in seen:
+            seen.add(item)
             new_array.append(item)
 
-    return new_array
+    return new_array