If you use the binary search then the number of comparison = O(log n) ( here the number of comparisons got decreased) But the number of movements will be = O(n).

then O(n) + O(log n) = O(n)

This we have to do n-1 times then total time complexity will be = O (n^2).

What if we use the **double linked list **then

The number of comparisons will be = O(n)

And number movements = o(1) ( movement reduced)

Total = O(n) + O(1) = O(n)

We have to perform this n-1 times then Time Complexity will be = O(n^2).

There will not be any change in time complexity.