The body of the inner loop is executed n times. The body of the outer loop is executed for all powers of 2 from 1 to n and there are about log2(n) of them, hence the global complexity O(n.log(n)) 1 solved Showing the order of growth of the function
Provided your intentions are not nefarious– As mentioned in the comment, one way to achieve this is executing the crawler in parallel (multithreading)—as opposed to doing one domain at a time. Something like: exec(‘php crawler.php > /dev/null 2>&1 &’); exec(‘php crawler.php > /dev/null 2>&1 &’); exec(‘php crawler.php > /dev/null 2>&1 &’); exec(‘php crawler.php > /dev/null … Read more
The overall semantics of all three iteration statements is the same once they have been compiled into binary code. They just provide a different taste over the same thing. Performance depends on: the complexity of the body of the loop the complexity of calculating the end condition / next iteration step Since they all provide … Read more
Here is the example, which reads sample XML and shows comparison between Linq/XMlReader and XmlDocument Linq is fastest. Sample Code using System; using System.Diagnostics; using System.Linq; using System.Xml; using System.Xml.Linq; namespace ReadXMLInCsharp { class Program { static void Main(string[] args) { //returns url of main directory which contains “/bin/Debug” var url=System.IO.Path.GetDirectoryName( System.Reflection.Assembly.GetExecutingAssembly().GetName().CodeBase); //correction in path … Read more
Based on the referenced answer I tried something that doesn’t create Lookup‘s but Dictionary‘s instead. IMHO that’s the only thing you can cut that will actually save some time. Skipping creation of the HashSet (as proposed in my rushed comment) is not an option as this will lead to the duplication of all joined pairs. … Read more
Well designed test shows that first function is slowest on Python 2.x (mostly because two lists have to be created, first one as a increasing range, second one as a reverted first one). I also included a demo using reversed. from __future__ import print_function import sys import timeit def iterate_through_list_1(arr): lala = None for i … Read more
use list comprehension in python l = [3, 11, 4, 9, 1, 23, 5, 11] indexes = [index for index in range(len(l)) if l[index] == 11] output: [1, 7] use numpy for finding the matching indices. import numpy as np l = [3, 11, 4, 9, 1, 23, 5, 11] np_array = np.array(l) item_index = … Read more
The answer is “it depends”, or perhaps better put as “nobody knows”, as this type of question is always tightly coupled with compiler optimisations, processor architecture and many other factors. I would also like to point out that if you find one better on one system, it may not reflect that it’s better in a … Read more
If loading a module while navigating takes too much time, you could consider eagerly preloading modules instead. What this does is load your main application module first and display your view, and in the background load all the other modules even before you navigate to them. You can use the following RouterModule configuration to make … Read more
You have jank, because you have one massive layer being repainted each frame. It is almost 12000 pixels long. You need to find a way to minimize this. To figure this out, turn on “Paint” profiling and turn off the rest. Then just run a quick 2-3 second profile. It will take a bit to … Read more
There are two ways to look at this: Integer.toBinaryString(n) produces a string of length ceiling(log2(n)). That computation will take time that is roughly proportional to log(n). The maximum value for n is 2^31 – 1, and that computation will always take less than some fixed constant time. In fact, the question of the big O … Read more
The best way to exit an app is using finishActivity(); You don’t want to kill the whole process, it is not good practice. solved Android close an application
template<size_t stride, class T, size_t N, size_t count = N/stride> std::array<T*, count> make_2d( T(&raw)[N] ) { std::array<T*, count> retval; for (size_t i = 0; i < count; ++i) retval[i] = raw + i*stride; return retval; } this will return a array of pointers to the lower dimensions To call func2, simply do: func2( 10, make_2d<10>(arr).data() … Read more
1.Why “rails” command is very slower than “ruby”? ruby invokes the Ruby interpreter, which is a compiled executable. rails invokes the Ruby interpreter plus loads many Ruby libraries that need to be located then parsed by the interpreter, before executing your Rails command. So rails –version will always take longer than ruby –version because it … Read more
Basically it stores deltas rather than the final list; that means each operation only takes 2 reads and writes rather than (b – a + 1). Then the final max scan adds the deltas as it goes along, which is still an O(n) operation which you would have had to do anyway. n, inputs = … Read more