Given two lowercase strings s1 and s2, return true if s2 contains a contiguous substring that is a permutation (an anagram, rearranged) of s1. A permutation doesn't care about order — only about having exactly the same letters, with exactly the same counts. That turns this from a string-matching problem into a counting problem: a window of s2 is a match precisely when its 26-letter frequency count equals s1's. Rebuilding that count for every window works, but the sliding windowSliding WindowMaintaining a running result over a contiguous range that grows or shrinks one element at a time, instead of recomputing the result for every range from scratch. technique keeps the count updated in O(1) per slide — bump the letter entering, drop the letter leaving — instead of recounting all of s1.length letters every time.