567. Permutation in String

 1class Solution:
 2    def checkInclusion(self, s1: str, s2: str) -> bool:
 3        s1m = [0] * 26
 4        s2m = [0] * 26
 5        n = len(s1)
 6
 7        def idx(c):
 8            return ord(c) - ord("a")
 9
10        for c in s1:
11            s1m[idx(c)] += 1
12
13        # let's slide
14        l = 0
15
16        for r, c in enumerate(s2):
17            if s1m == s2m:
18                return True
19
20            s2m[idx(c)] += 1
21
22            while (r - l + 1 > n) or (s2m[idx(c)] > s1m[idx(c)]):
23                s2m[idx(s2[l])] -= 1
24                l += 1
25
26        return s1m == s2m  # "adc" vs. "dcda"