no, for eg.
array1 = { 1, 2, 5, 6, 7, 7, 7, 23};
array2 = { 1, 2, 2, 4, 8, 9, 12 };
then for
k = 2, answer = 1
k = 3, answer = 2
k = 4, answer = 2,
k = 6, answer = 4.
anyway to do it iteratively in logarithmic time
On Fri, Nov 11, 2011 at 2:27 AM, sravanreddy001 <sravanreddy...@gmail.com>wrote:
> Is it (k)th smallest element (distict integers)
> or the element at position k, when both are merged?
>
> 11111111111111113333333345555555555555556666666666666677799999999999999999999
> --> Is 3rd smallest element '1' or '4'
>
> If four, I am not able to think of a log complexity. Can u post your
> recursive solution only if u meant '4' in above case.
>
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