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I'm given a binary min-heap (implemented with an array) and need to come up with a (simple) efficient (no more than $k$ comparisons) to find the $k$-th minimal element.

My attempt was as follows:

  1. check who is the smallest among the root children
  2. scan the corresponding sub-heap maintaining a counter counting how many nodes are smaller than the larger child of the root (but larger than the smaller child). If the counter reaches $k-1$ return the value of the current node. other-wise after the scan is finished, call this method recursively on the larger root child to find the ($k$ $-$ couter_value + 1)-th minimal element of the larger child.

I just can't put this together formally and not sure this can be implemented with no more than $k$ comparisons.

Thanks for any help.

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1 Answer 1

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The minimum element will the root. So in constant time you can find the minimum element. Delete the minimum element from the root. Now min-heapify (make it minheap) the resultant tree.

The second minimum element will be the root again. Repeat the procedure mentioned above.

By this procedure, you will find $k$-th minimum element in time $O(k \log n)$ where $n$ is the number of elements in the given min-heap.

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  • $\begingroup$ This does not use at most $k$ comparisons in the general case (as stated in the question). $\endgroup$
    – Nathaniel
    Apr 19, 2021 at 12:24
  • $\begingroup$ One can't use at most $k$ comparisons as otherwise one can sort $n$ element in O(n) time. $\endgroup$
    – old
    Apr 19, 2021 at 12:28
  • $\begingroup$ This is not clear, as you would need $O(\sum\limits_{k=1}^n k) = O(n^2)$ comparisons to sort $n$ elements. If you think that's the case, please explain why in your answer. $\endgroup$
    – Nathaniel
    Apr 19, 2021 at 12:33

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