Binary Insertion sort complexity for swaps and comparison in best case

Question

What could be the complexity for binary insertion sort? And how many swaps and comparisons are made?

It might be O(n(LG n)) comparisons, but I am not sure. For the worst case, it is indeed N^2 swaps. What about the best?

You are correct. O(NlogN) for comparisons. From 0 (sorted array) to O(N^2) (reversed sorted array) for swaps. — Stanislav Volodarskiy, Jan 18 '21 at 16:05

wsdookadr · Accepted Answer · 2021-01-19T14:04:08.860

You can write binary insertion sort easily by leveraging built-in functions such as bisect_left and list.pop(..) and list.insert(..) :

def bininssort(L):
    n = len(L)
    i,j=0,0
    for i in range(1,n):
        j=i-1
        x=L.pop(i)
        i1=bisect_left(L,x,0,j+1)
        L.insert(i1,x)
    return L

About the worst-case, since at the i-th iteration of the loop, we perform a binary search inside the sub-array A[0..i], with 0<=i<n, that should take log(i) operations, so we now know we have to insert an element at location i1 and we insert it, but the insertion means we have to push all the elements that follow it one position to the right, and that's at least n-i operations (it can be more than n-i operations depending on the insertion location). If we sum up just these two we get \sum_{i=1}^n log(i) + (n-i) = log(n!) + (n*(n+1))/2 ~ n*log(n) + (n*(n+1))/2

(in the above Stirling's approximation of log(n!) is being used)

Now the wiki page says

As a rule-of-thumb, one can assume that the highest-order term in any given function dominates its rate of growth and thus defines its run-time order

So I think the conclusion would be that in the worst-case the binary insertion sort has O(n^2) complexity.

Binary Insertion sort complexity for swaps and comparison in best case

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