Comparison Sorting Algorithm Complexity
-
19-09-2019 - |
Question
Why is the lower bound for the time complexity of comparison-based sort algorithms O(n log n)?
Solution
http://en.wikipedia.org/wiki/Comparison_sort#Number_of_comparisons_required_to_sort_a_list
answers that question quite well, I think.
OTHER TIPS
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)
Licensed under: CC-BY-SA with attribution
Not affiliated with StackOverflow