The 'aba' case was obvious but I fumbled explaining why 'aaab' is impossible.
Start by explaining that this is a greedy problem where you always place the most frequent remaining character that isn't the same as the previously placed one. Use a max-heap to efficiently track character frequencies, and after placing a character, reinsert it if it still has remaining count. If at any point the most frequent character exceeds (n+1)/2, no valid arrangement exists.
Pro tip: Mention that you can optimize by using a max-heap of counts and a 'cooldown' variable to avoid reinserting the same character immediately, but the standard greedy with a heap is sufficient. Also, clarify that the problem is equivalent to reorganizing a string such that no two adjacent characters are the same, which is a common Amazon interview question.
Restate the problem: rearrange a string so no two adjacent characters are identical. Identify edge cases: empty string, single character, and strings where the maximum frequency exceeds (n+1)/2.
Count the frequency of each character. If any character appears more than (n+1)/2 times, return an empty string immediately.
Use a max-heap (priority queue) to store characters by frequency. This allows efficient retrieval of the most frequent character that can be placed next.
While the heap is not empty, pop the most frequent character, append it to the result, decrement its count, and if count > 0, hold it aside. Then pop the next most frequent character, append it, decrement, and reinsert the held character if its count > 0. Repeat until heap is empty.
Time complexity: O(n log k) where k is the number of distinct characters (at most 26). Space: O(k). Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.