Given two strings s1 and s2, return true if s2 contains a permutation of s1 as a contiguous substring — same letters as s1, same counts, any order.
A fixed-size window of length s1.length slides across s2; compare letter-count tables instead of re-sorting each window.
Example
Input: s1 = "ab", s2 = "eidbaooo"
Output: true // "ba" is a permutation of "ab"