My algorithm is O(n+n+..n) which is O(n), since there we also ignore addition fortunately :D
It would only be O(n) if the number of additions was constant. Here it varies with the size of the multiplier, giving us O(n*m).
alt Hacker News
It would only be O(n) if the number of additions was constant. Here it varies with the size of the multiplier, giving us O(n*m).