The problem sounds straightforward until you actually try to code it.
Use a max-heap to greedily place the most frequent remaining element at each position, ensuring it differs from the previously placed element. This approach guarantees a valid arrangement when one exists and runs in O(n log n) time.
Pro tip: Mention that you can optimize space by using a frequency map and a heap, and that the greedy choice is safe because the problem guarantees a solution. Also, briefly discuss an alternative O(n) approach using the majority element if it exists.
Clarify that the input is an integer array, a valid arrangement always exists, and we need to return any valid rearrangement. Note that the array can have duplicates.
Use a frequency map (hash map) to count occurrences, and a max-heap (priority queue) to efficiently retrieve the most frequent element. This allows O(log n) operations per element.
At each step, pop the most frequent element from the heap. If it's the same as the previously placed element, pop the next most frequent instead. Place the chosen element, decrement its count, and push it back if count > 0.
Ensure the loop runs until the heap is empty. If at any point the heap is empty but we still need to place an element, the arrangement is impossible (but guaranteed not to happen).
State that time complexity is O(n log n) due to heap operations, and space is O(n) for the frequency map and heap. Mention that if the maximum frequency is ≤ (n+1)/2, a solution exists, and an O(n) approach using the majority element is possible.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.