I overthought this at first and started sketching out some greedy approach with sorting and adjacent swaps.
First, clarify that the problem asks for the maximum number of adjacent pairs (i, i+1) such that arr[i] < arr[i+1] after reordering. Then, recognize that the optimal strategy is to sort the array and interleave the smaller half with the larger half to maximize the number of such pairs. Finally, derive the formula: for n elements, the maximum is floor(n/2) if there are at least two distinct values, otherwise 0.
Pro tip: Mention that this is equivalent to maximizing the number of 'ascents' in a permutation, and that sorting plus interleaving achieves the theoretical upper bound of floor(n/2). Also, note that if all elements are equal, no pair satisfies the strict inequality.
Confirm that we are counting adjacent index pairs (i, i+1) where the left value is strictly less than the right value, and that we can reorder the array arbitrarily.
We want to maximize the number of such pairs, which is equivalent to maximizing the number of ascents in the rearranged sequence.
In any sequence of length n, there are n-1 adjacent pairs. However, not all can be ascents because the sequence cannot be strictly increasing throughout if there are duplicate values or if the maximum is not at the end. The maximum possible ascents is floor(n/2).
Sort the array. Split it into two halves: the smaller half (first floor(n/2) elements) and the larger half (remaining elements). Interleave them: place a smaller element, then a larger element, alternating. This ensures each smaller element is followed by a larger one, yielding floor(n/2) ascents.
If all elements are equal, no pair satisfies the strict inequality, so the answer is 0. If n < 2, answer is 0. Otherwise, the answer is floor(n/2) as long as there are at least two distinct values.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.