todos removal

Author: Coderbeep

leetcode/models/graphql_submission_list.py

@@ -31,8 +31,6 @@ def from_dict(cls, data):
 
 
 
-# TODO: Emojis/colors for status display (Accepted, Wrong Answer, Runtime Error)
-# TODO: Handle empty submissions table
 class SubmissionList(QueryTemplate):
     def __init__(self):
         super().__init__()
@@ -66,11 +64,11 @@ def execute(self, args):
                     raise ValueError("Apparently you don't have any submissions for this problem.")
             self.show()
 
-            # if self.show_terminal:
-            #     self.show_code()
+            if self.show_terminal:
+                self.show_code()
 
-            # if self.submission_download:
-            #     self.download_submission()
+            if self.submission_download:
+                self.download_submission()
         except Exception as e:
             console.print(f"{e.__class__.__name__}: {e}", style=ALERT)
 
@@ -134,7 +132,7 @@ def download_submission(self):
                 with open(file_name, 'w') as file:
                     file.write(code)
 
-            console.print(f"File saved as {file_name}")
+            console.print(f"✅ File saved as '{file_name}'")
         except Exception as e:
             console.print(f"{e.__class__.__name__}: {e}", style=ALERT)
 

leetcode/models/graphql_user_problems_solved.py

@@ -50,8 +50,6 @@ def parse_args(self, args):
                 console.print("Username neither provided nor configured. Head to --help.", style=ALERT)
                 sys.exit(1)
 
-    # TODO: add a bar chart for the stats
-    # TODOL add last submissions
     def execute(self, args):
         try:
             with Loader('Fetching user stats...', ''):            

smallest-subsequence-of-distinct-characters.1059514138.py

@@ -0,0 +1,24 @@
+class Solution:
+    def smallestSubsequence(self, s: str) -> str:
+                
+        # This will store the last occurences of letters
+        occurences = {}
+        for i in range(len(s)):
+            occurences[s[i]] = i
+
+        stack = []
+        used = set()
+
+        for i in range(len(s)):
+            # If letter not used, 
+            if s[i] not in used:
+                while stack and s[i] < stack[-1] and occurences[stack[-1]] > i:
+                    used.remove(stack.pop())
+                
+                stack.append(s[i])
+                used.add(s[i])
+            # If letter already used, go to the next iteration    
+            else:
+                continue
+        
+        return ''.join(stack)