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Recursively partition into equal halves by rank, keeping the selected median in the lower half:All elements are distinct, so both halves have exactly $|A|/2$ elements. Since $|A|/k$ remains a power of two, every internal split occurs at a rank divisible by $k$.

For correctness, the desired ranks in the lower half are $k,2k,\dots ,|A|/2$. In the upper half they are the same local multiples of $k$, shifted by $|A|/2$. These are exactly all desired global ranks. At a leaf of size $k$, its maximum is its sole requested local rank. Induction therefore proves that the output contains every requested order statistic, in increasing order, including the original maximum. There is no separate internal-node output of p, which would duplicate a rank already supplied by the lower subtree.

Use a worst-case linear-time selection algorithm and linear partitioning. The divide-and-conquer algorithm hasThere are ${\log }_2(n/k)$ internal levels, each with total work $\Theta (n)$, and $n/k$ leaves with total work $\Theta (n)$. Since the source assumes $n/k\ge 2$, the running time isThis is multiselection by balanced rank partitioning. In-place selection and partitioning can use the original array; the output has $n/k$ values. The multiselection recursion has depth $O(\log (n/k))$; the selection routine contributes its own auxiliary storage.