Classic sliding window with frequency counts.
Use a sliding window of length equal to the pattern, maintaining a frequency count of characters in the window and comparing it to the pattern's frequency. This yields O(n) time by avoiding recomputation for each window. Alternatively, sort each substring and compare, but that is less efficient.
Pro tip: Mention that you can optimize the frequency comparison by tracking the number of matches between the window and pattern frequencies, updating it incrementally as the window slides. This avoids comparing the entire frequency array at each step, reducing constant factors.
Restate the problem to ensure understanding: find all start indices where a substring of length equal to pattern is an anagram of pattern. Confirm assumptions like case sensitivity and character set.
Decide between sorting each substring (O(n * m log m)) or sliding window with frequency counts (O(n)). Explain why sliding window is optimal.
Initialize frequency arrays for pattern and first window. Slide the window one character at a time, updating frequencies and checking for anagram condition.
Instead of comparing full frequency arrays each time, maintain a count of matching characters or use a variable to track differences. Update it incrementally.
State time O(n) and space O(1) (since alphabet size is constant). Discuss edge cases: pattern longer than text, empty strings, repeated characters.
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