I was thinking the time complexity will be O(n^2*logn). Is it correct. If not, What will be time complexity?
for(i = 0; i < n; i++)
for(j = 0; j < n; j++)
for(k = 0; k < 100000; k++)
a++;
Asked
Active
Viewed 79 times
-2
-
Where did you get the "log n" part from? – kaya3 Apr 28 '21 at 12:27
-
**Hint:** will `k` grow as `n` increases? – Robert Harvey Apr 28 '21 at 12:29
-
Same question as above, this is O(n^2), why log n? – Rocco Apr 28 '21 at 12:29
-
I think the logn part is just a guess. – Robert Harvey Apr 28 '21 at 12:30
-
kaya 3, I asked for an explanation of the code. "log n" is just a guess. If I know the answer why would I ask you in the first place? – Lucas Apr 28 '21 at 12:35
-
Here the explanation: n * n * 100000 iterations, executing an O(1) task, 1000000 is a constant and not relevant for complexity, so in total O(n^2) – Rocco Apr 28 '21 at 12:47
