Comparison Sorting Algorithm Complexity

Question

Why is the lower bound for the time complexity of comparison-based sort algorithms O(n log n)?

meriton · Accepted Answer · 2009-12-11 23:24:04Z

7

http://en.wikipedia.org/wiki/Comparison_sort#Number_of_comparisons_required_to_sort_a_list

answers that question quite well, I think.

answered Dec 11, 2009 at 23:24

meriton

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Brad · Accepted Answer · 2009-12-11 23:25:49Z

1

In short, because you must look at every element which is O(n). For each of those elements you look at, you must find out if its in the right order, which is at best O(log n) (binary search for example). So the net sum becomes O(n log n)

answered Dec 11, 2009 at 23:25

Brad

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