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We have $k$ sorted arrays, $A_1[1...n_1],...,A_k[1..n_k]$, where $n_1+n_2+...+n_k=n$.

How can we get the $m$ greatest elements in running time $O(k + m\lg k)$?

I have tried to use MIN-HEAP size of $k$ since we have $k$ arrays. I have to navigate through the ends of the $k$ array somehow. By extracting the min of heap and getting new element from respective array was my idea but it does not work. I could not figure out getting m greatest elements.

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  • $\begingroup$ Why does it not work? $\endgroup$ – Raphael Nov 19 '14 at 21:23
  • $\begingroup$ What do the arrows in the first line mean? $\endgroup$ – David Richerby Nov 19 '14 at 22:39
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Your idea works out. The idea is to maintain a heap of size $k$ which contains exactly one element from each array. For each array $A_i$, we maintain an index $p_i$ which keeps track of the position of the current element in the array. Initially, $p_i = n_i$, and we insert $A_i[n_i]$ into the heap. We then repeat the following $m$ times:

  1. Extract the maximum element from the heap.
  2. If the extracted element belonged to $A_i$, then we decrease $p_i$ by one and insert $A_i[p_i]$ into the heap.

The initialization phase takes time $O(k)$, and the extraction phase takes time $O(m\log k)$, for a total of $O(k+m\log k)$. You can prove that the extracted elements are indeed the $m$ largest ones.

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  • $\begingroup$ Since we can build a max-heap of $k$ elements in $O(k)$ time, the preprocessing is in $O(k)$ and the condition on $m$ is obsolete. $\endgroup$ – Raphael Nov 19 '14 at 21:25

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