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As per Big O notations, if time complexity of one algorithm is O(2^n) and the other is O(n^1000), then which would be faster one?
How to recognize overall behavior for some non-obvious cases: get logarithm of both functions.
(Sometimes we can also get ratio of the functions and evaluate ratio limit for large n's, here this approach is not good)
log(2^n) = n*log(2)
log(n^1000) = 1000*log(n)
The first result is slanted line with positive coefficient. The second one's plot is convex curve with negative second derivative, so the first function becomes larger at some big n value.
How plot looks
O(n^1000) is in the same class as (n^2) and O(n^777777777) which is Polynomial time, whereas O(2^n) is Exponential time which is way slower than Polynomial
https://www.bigocheatsheet.com/
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Which time complexity is better N*(2^N) or N^2 and why?
N*(2^N) or N^2
N*(2^N) is exponential.
If you take n=10, for example, you get 10240
N^2 is merely polynomial.
If you take n=10, for example, you get 100
Exponential is worse than polynomial for large N, and even for reasonable Ns, in your case. To see it intuitively, imagine growing N by 1. In the polynomial case, the result grows by a fraction ((N+1) / N) ^ 2. It grows, but not much. In the exponential case, growing N by 1 doubles the result.
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Is there any advantage of using Dinic's O((V^2)E) algorithm over Edmond-Karp algorithm O(V(E^2))?
In other words, I want to know how is O((V^2)E) better than O(V(E^2)) if it is from a Competitive Programming point of view.
Let's say the total number of vertices in n. "Usually", number of edges in a connected graph tend to be between n and n^2.
Mostly the input graphs are not very sparse, so the number of edges in maximum percentage of the cases would be greater than n (might be O(n log n), or in the worst case, O(n^2)).
So, if you consider the worst case scenario, O(V^2 * E) is O(n^4), whereas O(V*E^2) is O(n^5). Hence you see the advantage of using an O(V^2*E) time algorithm over O(V*E^2).
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I had made the comparison in between 2 algorithms, A and B. I found that, the time complexity of these 2 algorithms is different, which are O(n^3) and O(n^4) respectively, but the space complexity of both A and B is same, which is O(n). Is that have probability?
That is possible, yes. The time complexity and the space complexity are not necessarily related to each other.
As an example, think of various sorting algorithms, such as the ones in this list - as you see, both insertion sort and heapsort have a space-complexity (column memory) of O(1), while their time complexities are different.
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I just wrote the quick and merge sort algorithms and I want to make a log-log plot of their run time vs size of array to sort.
As I have never done this my question is does it matter if I choose arbitrary numbers for the array length (size of input) or should I follow a pattern (something like 10^3, 10^4, 10^5, etc)?
In general, you need to choose array lengths, for each method, that are large enough to display the expected o(n log n) or O(n^2) type behavior.
If your n is too small the run time may be dominated by other growth rates, for example an algorithm with run time = 1000000*n + n^2 will look to be ~O(n) for n < 1000. For most algorithms the small n behavior means that your log-log plot will initially be curved.
On the other hand, if your n is too large your algorithm may take too long to complete.
The best compromise may be to start with small n, and time for n, 2n, 4n,..., or n, 3n, 9n,... and keep increasing until you can clearly see the log log plots asymptoting to a straight lines.
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I'm not sure if this is a problem with my understanding but this aspect of Big Oh notation seems strange to me. Say you have two algorithms - the first preforms n^2 operations and the second performs n^2-n operations. Because of the dominance of the quadratic term, both algorithms would have complexity O(n^2), yet the second algorithm will always be better than the first. That seems weird to me, Big Oh notation makes it seem like they are same. I dunno...
Big O is not about the time it takes to execute your algorithm, it is about how well it will scale when presented with large data sets (large values of n).
When presented with a large data set, the n^2 term will quickly overshadow any linear term. So the linear term becomes insignificant.
When n grows towards infinity n^2 will be much greater then n so the -n won't have any significant difference on the outcome.