Question

I am having a hard time understanding why best case of insertion sort in o(n) ?

     for (int i = 0; i < size; i++) {

                for (int j = i; j > 0; j--) {
                    int k = j-1;
                        if( a[j] < a[k]){
                            int temp = a[j];
                            a[j] = a[k];
                            a[k] = temp;
                        }

                }
     }

Lets consider an example initial array [1,2,3,4,5] size = 5
first loop will go from i = 0 to size – 1
and second loop will go from i to 1 but lets assume, inner for loop also goes from 0 to size – 1 in other words inner for loop also executes (n-1) times similar to outer for loop
I agree there will be no swaps but there will be Comparison’s, & it will be exactly equal as unsorted array ?
then n-1 (outer loop) * n – 1(inner loop) = n^2 – n + 1 = O(n^2)
can any one explain me where i m wrong ?

Answer 1

Your code always runs in O(n^2). You have to break the inner for loop at the time you have found the place where the element should be.