## What is it? Given a string, count the number of substrings that are anagrams of a given pattern. Two strings are anagrams if they have the same characters with the same frequencies. ## How it works **Fixed-size Sliding Window + frequency map:** - Build a frequency map for the pattern - Maintain a window of size `p.length()` over the string - As the window slides, add the incoming character and remove the outgoing character from the window map - Compare the window map with the pattern map — if equal, count it - Use a "match count" variable to avoid full map comparison each step ## When to use - Finding all anagram occurrences in a text - Problems involving permutation presence in a string ## Key Points - O(n) time — each character enters and exits the window once - O(1) space — frequency arrays of fixed size 26 - Key insight: two windows are anagrams iff their frequency arrays are identical