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Keep two pointers to the arrays, go as in merge-sort's merge step, stop at the Nth entry. It will be in either of the two arrays. Return the value.

- Ananamas October 22, 2009 | Flag Reply
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Almost there. Since there are 2N elements, the median will be the average of the Nth and (N+1)st entries.

- Bullocks December 31, 2009 | Flag
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Well...

Depends on your definition of median.

The way I know it is that the median(s) of a set belong to it. So in case of even number of elements, there are two medians, the Nth and N+1st.

- Anonymous December 31, 2009 | Flag
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In fact the median is well defined for any ordered set. There is no need for the concept of 'average' to hold.

- Anonymous December 31, 2009 | Flag
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cormen et. al exercise 9.3.7.

Instructor's manual has the solution for it.

- Anonymous February 04, 2010 | Flag Reply
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cormen et. al exercise 9.3.8.

Instructor's manual has the solution for it.

- Anonymous February 04, 2010 | Flag Reply


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