Is there another car around? If not the best way would be to put the car in neutral, get the car moving, than pop the clutch (or slam into drive if it's not a manual) the key should be on to do this. This has the same effect as a starter just doesn't require a battery.<span />
Answer:
where is the picture?????
Answer:
Pre-order: Lisa, Bart, Homer, Flanders, Marge, Maggie, Smithers, Milhouse
In-order: Bart, Flanders, Homer, Lisa, Maggie, Marge, Milhouse, Smithers
Post-order: Flanders, Homer, Bart, Maggie, Milhouse, Smithers, Marge, Lisa
Explanation:
The required orders are mentioned.
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
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As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
