Question

i’m just wondering what is efficiency (O(n)) of this algorithm:

1. Find the largest index k such that a[k] < a[k + 1]. If no such index exists, the permutation is the last permutation.
2. Find the largest index l such that a[k] < a[l]. Since k + 1 is such an index, l is well defined and satisfies k < l.
3. Swap a[k] with a[l].
4. Reverse the sequence from a[k + 1] up to and including the final element a[n].

As I understand the worst case O(n) = n (when k is the first element of previous permutation), best case O(n) = 1 (when k is last element of previous permutation).

Can I say that O(n) = n/2 ?

Answer 1

All steps in the algorithm takes O(n) asymptotically.

Your averaging is incorrect. Just because best case is O(1) and worst case is O(n), you can’t say the algorithm takes O(n)=n/2. Big O notation is simply for the upper bound of the algorithm.

So the algorithm is still O(n) irrespective of the best case scenario.