Add documentation to sliding window permutation solution

.gitignore

@@ -1,2 +1,7 @@
 # macOS system files
 .DS_Store
+
+# Python cache files
+__pycache__/
+*.py[cod]
+*$py.class

0567-permutation-in-string/0567-permutation-in-string.py

@@ -1,34 +1,61 @@
 class Solution:
     def checkInclusion(self, s1: str, s2: str) -> bool:
+        """
+        Check if s2 contains a permutation of s1 using sliding window approach.
+
+        Time Complexity: O(len(s1) + len(s2)) for initial setup and sliding window
+        Space Complexity: O(1) - arrays of fixed size 26
+
+        Algorithm:
+        1. Use frequency arrays to count character occurrences
+        2. Use a sliding window of size len(s1) over s2
+        3. Track number of matching character frequencies between windows
+        4. If all 26 letters match, we found a permutation
+        """
+        # Edge case: if s1 is longer than s2, no permutation can exist in s2
         if len(s1) > len(s2):
             return False
+        # Initialize frequency arrays for 26 lowercase letters
         s1Count, s2Count = [0]*26, [0]*26
 
+        # Build frequency count for s1 and the first window of s2
         for i in range(len(s1)):
             s1Count[ord(s1[i]) - ord('a')] += 1
             s2Count[ord(s2[i]) - ord('a')] += 1
 
+        # Count how many of the 26 letters have matching frequencies
+        # This optimization avoids comparing entire arrays repeatedly
         matches = 0
         for i in range(26):
             matches += (1 if s1Count[i] == s2Count[i] else 0)
 
+        # Slide the window across s2
         l = 0
         for r in range(len(s1), len(s2)):
+            # If all 26 characters match, we found a permutation
             if matches == 26: return True
 
+            # Add the new character entering the window (right side)
             index = ord(s2[r]) - ord('a')
             s2Count[index] += 1
             if s1Count[index] == s2Count[index]:
+                # This character now matches, increment matches
                 matches += 1
             elif s1Count[index] + 1 == s2Count[index]:
+                # This character was matching before, but now exceeds count
                 matches -= 1
 
+            # Remove the character leaving the window (left side)
             index = ord(s2[l]) - ord('a')
             s2Count[index] -= 1
             if s1Count[index] == s2Count[index]:
+                # This character now matches after removal
                 matches += 1
             elif s1Count[index] - 1 == s2Count[index]:
+                # This character was matching before removal
                 matches -= 1
             l += 1
+
+        # Check the final window
         return matches == 26