add solution: encode-and-decode-strings

Author: dusunax

encode-and-decode-strings/dusunax.py

@@ -0,0 +1,65 @@
+'''
+# 271. Encode and Decode Strings
+
+## Time and Space Complexity
+
+Use ":" as a separator and also store the length of each string to decode the string correctly.
+
+
+### encode
+
+```
+TC: O(n * k)
+SC: O(m)
+```
+
+#### TC is O(n * k):
+
+- iterating through the list of strings and appending each string to the result. = O(n * k)
+  - f-string is O(k)
+
+#### SC is O(m):
+- storing the result in a string.
+
+### decode
+
+```
+TC: O(m)
+SC: O(m)
+```
+
+#### TC is O(m):
+- iterating over the string until the string length is 0. = O(m)
+- do list slicings for extract parts and removing the processed section = each operation takes O(k)
+
+#### SC is O(m):
+- storing the result in a list(total length of strings is m) = O(m)
+
+'''
+
+class Solution:
+    """
+    @param: strs: a list of strings
+    @return: encodes a list of strings to a single string.
+    """
+    def encode(self, strs):
+        result = ''
+        for str in strs: # TC: O(n)
+            result += f"{len(str)}:{str}" # TC: O(k)
+        return result
+
+    """
+    @param: str: A string
+    @return: decodes a single string to a list of strings
+    """
+    def decode(self, str):
+      result = []
+
+      while len(str) > 0: # TC: O(m)
+          length = int(str[:1]) # TC: O(k)
+          string = str[2:length+2] # TC: O(k)
+          str = str[length+2:] # TC: O(k)
+
+          result.append(string) # SC: O(m)
+        
+      return result