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The unusual $\Theta(n^2)$ implementation of Insertion Sort to sort an array uses linear search to identify the position where an element is to be inserted into the already sorted part of the array. If, instead, we use binary search to identify the position, the worst case running time will

1. remain $\Theta(n^2)$
2. become $\Theta(n (\log n)^2)$
3. become $\Theta(n \log n)$
4. become $\Theta(n)$

now for the best case, it will take O(N log N) time.
For best case...

searching takes logN and swap takes 1  , so total (logN + 1) becomes 1 , and for n elements  it becomes N (1) . so total for best case it becomes N.

correct me if i am wrong ?
best case input for insertion sort is a sorted array (already) hence we do only the find the place of the cureent element we are looking at in the sorted part of the array  we dont need to do swaps so it will be n*logn for n elements
$1)$ Linear search: $\left | \underbrace{sorted} \right |$ $\leftarrow\left | i \right |: comparisons=n,swaps=n$

For $n$ element : $n\times 2n=2n^2=O(n^2)$

$2)$ Binary search: $\left | \underbrace{sorted} \right |$ $\leftarrow\left | i \right |: comparisons=logn,swaps=n$

For $n$ element : $n\times (n+logn)=n^2+nlogn=O(n^2)$
Both vanilla insertion sort and binary insertion sort gives us O($n^{2}$)

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In insertion sort, with linear search, it takes

(worst case) $n$ comparisons for searching the right position, and $n$ swaps to make room to place the element.

Hence for n elements, a total of $n\times(n+n)$; $n$ for search and $n$ for swaps.

$= \Theta (2n^2) = \Theta (n^2)$

If we replace it with binary search, it takes

(worst case) $\log n$ comparisons for searching the right position, and $n$ swaps to make room to place the element.

Hence for n elements, a total of $n\times(\log n+n)$; $n$ for search and $n$ for swaps.

$= \Theta (n \times \log n + n^2) = \Theta (n^2)$

I think this explanation is correct and more clearer
replace with swap with word shifting
why you have take it for n elements. there is nothing specified for number of elements. please explain.
Then what do you think 'n' stands for in the given question?
edited

@ryan sequeira You mean for every element in the array we need to do max. n comparisons to search for the right position?

But if you see the code for each iteration, max no. of comparisons that can occur in an iteration will be from j to 1 where j=(i-1) so i.e. i-1-1+1=i-1 where i∈[2,n].

InsertionSort(arr,n)
{
for i=2 to n
{
key = arr[i]
j = i-1
while (j > 0 && arr[j] > key) // will run from j=(i-1) to max. j=1
{
arr[j+1] = arr[j]
j = j-1
}
arr[j+1] = key
}
}

So what I mean to say is no. of comparisons is not n for all elements and neither is the no. of swaps. It depends on the element's position in the array. Isn't it so? Correct me if it is wrong.

The answer is still the same though.

Explanation how we get θ(n2) time complexity in insertion sort-

T(n)=0+0+1+1+2+2......+(n-1)+(n-1) = 2 [0+1+2+.....(n-1] = θ(n2)

Why I have written every number twice? Since first one represent no of comparisons and second one no of swaps... is my understanding correct?

nice explanation
in this problem for time complexity comparison does not matter only swaps will derive the required result

Why question says θ(n2) and not O(n2)?

best case using nlogn ??

@Abhisek as per question TC in best case also O(n^2)

because we can find correct position for element in logn time.. but after inserting the element we need to adjust the array so finally it takes O(n)

My Question sounds lame but plz explain how we can use Binary search here when we know it can be applied on Sorted sequences only.? I mean what is question intending to say.?

@Suneel Padala best case when array is sorted no need of movement then O(n) ??

@SomeEarth binary search is used only to insert the current element in insertion sort. And all elements before the current element are sorted in Insertion sort.

Thank you very much sir I got it now. (i didn't think in this way)

@Arjun sir

i think average case O(n*n) not O(n*logn)

and best case O(n)

am i correct?

@Abhishek can you explain how average case is O(n$^{2}$)?
edited

For each element we need to find its position in the sorted part using binary search and then place the element in its correct position by swapping.

(Below I have taken sum, taking into consideration each iteration of insertion sort)

Worst case time complexity for binary search (taking all iterations into consideration)= log 1 + log 2 + log 3 + ... + log (n-1) + log (n) = log($1\times2\times3...\times(n-1)\times(n)$) = log n! = $\Theta(n log n)$ .(There is a nice proof for this here.)

Worst case time complexity for swaps (taking all iterations into consideration) = 1 + 2 + 3 + ... + (n-1) + (n) = $\frac{n\times(n+1)}{2}$ = $\Theta(n^{2})$

Total time complexity including swaps and binary search = $\Theta(n log n)$ + $\Theta(n^{2})$ = $\Theta(n^{2})$

edited

If we perform B-search instead of linear search( basically we are finding out the right positions in log n time) , no of swaps will be 1 each time. For n elements we will have n swaps in total .Isn't it ?

Then, what will be the number of comparisons in best case? as n X ((n-1) + 0) will again you O(n^2)…. Pl clarify.

A. Complexity remains same.θ(n2)

To place the element x in the correct position first we are finding its correct position in the sorted array using binary search but we have to make the space for it by shifting all elements to the right, which in worst case may be equal to the size of the sorted array.