Binary Insertion Sort

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We know of Insertion Sort, and we also know of Binary Search. What if we put the two together? Could we create something more efficient?

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I'm taking the Algorithm and Complexity course at school and this is so helpful! Thank you so much!

vyhuynh

Good explanation of why it's not used in practical application! Thanks for the video!

tedlava

Of course the algorithm is still O(n²), but that doesn't matter. What binary searching does, is to minimize the number of comparisons needed, without changing any of the read/writes. So if the comparison routine is way slower than the read/write, it will make the algorithm faster by a constant factor. (This is of course assuming the array is small enough to fit within cache, so it doesn't have to reload multiple times, which could be a big slow down, but if you really are insertion sorting like 64k items without any a priori knowledge, you should really reconsider everything)

And when comparing sorting algorithms of the same asymptotic runtime, we really do care about the constant factor. (i.e. the reason you would use insertion sort over bubble sort).

P.S. Of course normal Insertion sort is faster on already sorted arrays, because it will only need exactly 1 comparison each time.

Eknoma

Thank you a lot for this very clear video !

ARMofficial

Another variant of Insertion Sort is Shell Sort. I wonder if Binary Search can be applied to Shell Sort. I hope you will upload a video about that if it can!

thanhonguyen

does it mean that if i use a singly linked list to store the items, i can avoid the linear time swapping (i do O(logn)binary search and basically just move some pointers which takes O(1) ) and get a O(nlogn) time complexity in this case?

nanayang

a very detailed explanation.. good work bro

susmitsingh

Wouldn't this depend on how the data structure you are working with is actually stored in memory? For example, what if you are working in a language that stores arrays as linked lists and performing insertions is as simple as reassigning a couple of pointers? Obviously if you're working with simple arrays in a low-level language such as C this wouldn't be the case, but isn't it good practice to optimize your code to be as efficient as possible across as large a variety of data structures as possible?

AndrewPreston-yhjm

This help me a lot...Thank you so much

s

So, for the normal insertion sort, it takes O(n^2) because you have n elements, and for each element you do n comparisons and n shifting. Therefore: n(O(n) + O(n)) = O(n^2) + O(n^2) = O(n^2)

And for binary insertion sort, it takes O(n^2) because you have n elements, and for each element you spend log n time finding the proper position, and then n shifting. Therefore: n(O(logn) + O(n)) = O(n logn) + O(n^2) = O(n^2)

Did I get it right?

inzanity

Hey. Great video.
If i understand it correctly Insertion sort has O(n) comparisons and O(n) swaps which makes it a O(n^2) Algorithm.
Since Binary Insertion Sort has O(log(n)) comparisons and O(n) swaps why is it still O(n^2)?
The amount of swaps i have to do in the normal Insertion Sort are the same like with the Binary Insertion Sort. But the amount of comparisons changes. So why doesnt it affect the Time Complexity?

ikaros

not shifting any element but overwriting the array itself with its portions plus the element to insert, does the complexity remain O(nlogn)?

matteomanuelli

But what if we don't use in-place sorting method? we can just initialize list of the same size of the original source, and we put elements to new list in order to the result of the binary search??

popcorn