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Time complexity (in big-O notation) of the following code?


Time complexity (in big-O notation) of the following code?

By : Matt Sanders
Date : November 25 2020, 01:01 AM
Hope that helps I was just wondering what is the time complexity of the following code. , Looks like n^(2.75) to me:
code :


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Confused on whether to express time complexity with theta notation or Big Oh notation

Confused on whether to express time complexity with theta notation or Big Oh notation


By : MightyWave
Date : March 29 2020, 07:55 AM
may help you . Use Big Theta notation when you want to also specify a lower bound. f(n) = O(g(n)) says that f is bounded above by g, whereas f(n) = Theta(g(n)) says that f is bounded both above and below by g.
In other words, there are constants k1 and k2 such that k1 * |g(n)| <= |f(n)| <= k2 * |g(n)|
Time Complexity of this code with BigO Notation

Time Complexity of this code with BigO Notation


By : Jagannath S
Date : March 29 2020, 07:55 AM
it should still fix some issue Your fonction starts with a while loop (from 0 to N).
Inside this loop you have 2 for loops (from 0 to N).
What is the time complexity of the following code? in Big O notation

What is the time complexity of the following code? in Big O notation


By : Mike
Date : March 29 2020, 07:55 AM
Hope that helps The O(n) notation is the degree of the largest term in the polynomial describing the code, and describes how the time to run grows as the input size increases.
Therefore, small details such as coefficients and lower order terms are unimportant.
Time complexity (in big-O notation) of the following recursive code?

Time complexity (in big-O notation) of the following recursive code?


By : Deathzombee
Date : March 29 2020, 07:55 AM
it should still fix some issue Where I'm sitting...I think the runtime is O(n log n). Here's why.
Big O Notation and Time Complexity of C++ Code Snippet

Big O Notation and Time Complexity of C++ Code Snippet


By : 石榴花
Date : March 29 2020, 07:55 AM
will be helpful for those in need Yes it is in O(n log n) but the base does not matter in big O notation since f=n \cdot log_2(n) \in \mathcal{O}(log_2(n) * n ) \subseteq \mathcal{O}(\frac{ln(n)}{ln(2)} * n ) \subseteq \mathcal{O}(log(n) * n ) \ni f = n \cdot ln (n) i.e.
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