Refactor matches calculation to one-line.

We don't need l and r pointers, r is always l + len(s1).

Author: Chris Cooper

python/0567-permutation-in-string.py

@@ -1,5 +1,5 @@
 class Solution:
-    def checkInclusion(self, s1: str, s2: str) -> bool:
+    def checkInclusion(self, s1, s2):
         if len(s1) > len(s2):
             return False
 
@@ -8,27 +8,24 @@ def checkInclusion(self, s1: str, s2: str) -> bool:
             s1Count[ord(s1[i]) - ord("a")] += 1
             s2Count[ord(s2[i]) - ord("a")] += 1
 
-        matches = 0
-        for i in range(26):
-            matches += 1 if s1Count[i] == s2Count[i] else 0
+        matches = sum(map(lambda i: s1Count[i] == s2Count[i], range(26)))
 
-        l = 0
-        for r in range(len(s1), len(s2)):
+        for l in range(len(s2) - len(s1)):
             if matches == 26:
                 return True
 
-            index = ord(s2[r]) - ord("a")
-            s2Count[index] += 1
-            if s1Count[index] == s2Count[index]:
+            i = ord(s2[l]) - ord("a")
+            s2Count[i] -= 1
+            if s2Count[i] == s1Count[i]:
                 matches += 1
-            elif s1Count[index] + 1 == s2Count[index]:
+            elif s2Count[i] == s1Count[i] - 1:
                 matches -= 1
 
-            index = ord(s2[l]) - ord("a")
-            s2Count[index] -= 1
-            if s1Count[index] == s2Count[index]:
+            i = ord(s2[l + len(s1)]) - ord("a")
+            s2Count[i] += 1
+            if s2Count[i] == s1Count[i]:
                 matches += 1
-            elif s1Count[index] - 1 == s2Count[index]:
+            elif s2Count[i] == s1Count[i] + 1:
                 matches -= 1
-            l += 1
-        return matches == 26
+
+        return matches == 26