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Consider this algorithm iterating over $2$ arrays $(A$ and $B)$

size of $ A = n$

size of $ B = m$

Please note that $m \leq n$

The algorithm is as follows

for every value in A:
    // code

for every value in B:
    // code

The time complexity of this algorithm is $O(n+m)$ But given that $m$ is strictly lesser than or equal to $n$, can this be considered as $O(n)$?

S. M.
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Shylajhaa
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3 Answers3

40

Yes:

$n+m \le n+n=2n$ which is $O(n)$, and thus $O(n+m)=O(n)$

For clarity, this is true only under the assumption that $m\le n$. Without this assumption, $O(n)$ and $O(n+m)$ are two different things - so it would be important to write $O(n+m)$ instead of $O(n)$.

nir shahar
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Yes, since $n + m \leq 2n$ the algorithm is $O(n)$. However, you may wish to write $O(m + n)$ because it clearly shows which variables the algorithm depends on, and what each variable does to the complexity.

BBrooklyn
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0

(expanding on my comment)

Technically No

You need to be very careful here, as there is a difference between algorithmic time complexity and runtime. In the case you have described, the runtime may be O(n) but the algorithm itself is O(n+m).

This is because, without your external constraints, the algorithm's time complexity cannot be determined without knowing both m and n.

If you wanted to make the algorithm itself O(n), you would need to explicitly encode your external constraint that m<n within the algorithm itself. Adding a check that aborts if m>n would work.

Unless...

The above does not hold if m < n by definition (ie it arises from a fundamental property of your data structures).

Say, for example, A is an array and B is an array composed of only the elements at the even indices of A. Then m is indeed < n by definition, and no check is required.

In cases like these the answer is yes, the algorithm is O(n). In fact, it would be 'incorrect' (assuming you are going for a tight upper bound) to say it is O(n+m) since the asymptotic performance depends entirely on n.

Cole
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