Given two strings `text` and `pattern`, return the starting index of the first substring in `text` that is an anagram of `pattern`. If no such substring exists, return `-1`. A substring is an anagram of `pattern` if it has exactly the same character frequencies as `pattern`, possibly in a different order. Assume the input strings contain lowercase English letters unless otherwise specified. Example: ``` text = "cbaebabacd" pattern = "abc" ``` The substrings of length `3` are checked from left to right. The first substring that is an anagram of `"abc"` is `"cba"`, so the answer is `0`. Use a fixed-size sliding window of length `len(pattern)`: maintain a character-frequency count for the current window and compare it against the frequency count of `pattern`. Slide one character at a time, adding the incoming character and removing the outgoing one in O(1). Follow-up: If `text` were extremely large, streamed in chunks, or distributed across machines, the sliding window only needs O(26) state per position, so you can process the stream incrementally; for a distributed split, each shard scans its slice plus an overlap region of `len(pattern) - 1` characters from the next shard so anagrams straddling a boundary are not missed.