My first instinct was to just normalize everything to start from 'a', which works, but I fumbled explaining why wrap-around matters until they pushed back.
Clarify that grouping is based on the shift difference between characters, which is invariant under uniform shifts. For each string, compute a canonical representation by normalizing the first character to 'a' and applying the same shift to all characters. Use a hash map to group strings by this canonical key.
Pro tip: Mention that the canonical key can be computed in O(n) per string by shifting each character relative to the first, and that using a tuple of differences avoids modulo issues. Also note that empty strings and single-character strings form their own groups.
Confirm that two strings are in the same group if there exists a fixed shift k such that shifting each character of one string by k (mod 26) yields the other. This means the relative differences between consecutive characters are invariant.
For each string, compute a key that is identical for all strings in the same group. One approach: shift the first character to 'a' and apply the same shift to all characters. Another: compute the sequence of differences between consecutive characters modulo 26.
Iterate through the array, compute the canonical key for each string, and use a hash map to map the key to a list of strings. Finally, return the lists as the groups.
The time complexity is O(N * L) where N is the number of strings and L is the average length. Space is O(N * L) for the hash map. Handle edge cases: empty strings, single-character strings, and strings of different lengths (which cannot be in the same group).
Walk through examples to verify the grouping. For instance, ['abc', 'bcd', 'xyz'] should group 'abc' and 'bcd' together, while 'xyz' forms its own group. Also test with strings that wrap around, like 'zab' and 'abc'.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.