Most red black tree packages will have an “insert element” method. If you’re not using one already, it might be good to start. If you’re married to a red-black tree implementation that doesn’t have an insert element operation, it’d be a good idea to add such a method, possibly from some good red-black tree doc: … Read more
W.l.o.g. we say n>=m and choose m to be of the form 2^x. (If m is arbitrary <= n as well it only means that we apply the same approach to a second rectangle that is n times m-2^x in size, where x=int_floor(ld(m)) and ld is log to the base of 2 of course.) The … Read more
Edit: Original supposition on the problem completely wrong. The cause of you getting all zeros is that the state is not seeded. You need to fill state with something ‘random’. This code work. Note that the function seed() is in absolutely no way scientifically proven to be good – in fact, I just made it … Read more
It can be done with a bit of recursion, provided you don’t have too many items in source and also don’t get a combinatorial explosion from how long each sublist is. import java.util.ArrayList; import java.util.List; public class AllCombinations { static void AddCombination(List<List<String>> source, int depth, String prefix, List<String> output) { for (String layer : source.get(depth)) … Read more
A pair of lines associated with two number can only cross if those numbers appear in opposite orders in a and b. Since you state that the first list – a – is sorted (in increasing order) we will only have lines crossing if we pick numbers that are in de-creasing order in b. You … Read more
Let’s construct our number that doesn’t have three consecutive ones from left to right. It necessarily starts with at most two 1s, so it starts with either no 1s, a single 1, or two 1s. In other words, our number starts with either 0, 10 or 110. In all of these cases, the only restriction … Read more
For an input {x1, …, xn}, return 2n-1 (x1 + … + xn), since each term appears in that many sums. 1 solved Finding the sum of all subsets of a given set
The sorting algorithm for Arrays.sort(int[] a) is a tuned quicksort. Ref: https://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html#sort(int[]) Read this article: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.14.8162&rep=rep1&type=pdf for understanding the Quick-sort used in Array.sort() by JON L. BENTLEY & M. DOUGLAS McILROY In Java 7 pivot dual quick-sort algorithm is used for Arrays.sort(int[] a) i.e. refer this: What’s the difference of dual pivot quick sort and … Read more
It is a variant of the “Connected components of a graph” problem. Google for it to see answers like this and this. solved Request a popular name for this or its solution algorithm
In-place? Probably something like this: byte[] Reverse( byte[] b ) { for ( int i = 0 , int j = b.Length-1 ; i < j ; ++i, –j ) { arr[i] ^= arr[j] ; arr[j] ^= arr[i] ; arr[i] ^= arr[j] ; } return b; } Idempotent, where you allocate a new array to … Read more
This looks like a very complex problem for me. Its for sure a optimization problem with many constraints, so i hope you are good at math ;-). I would suggest to have a look at the Simplex Algorithm which will work very well for your problem if you have the mathematical know how to use … Read more
Quantum algorithms have the possibility of searching a complete number space for an answer in one go, that much is true. However, the algorithm you choose will still determine how many of these steps are needed, and whether you can pack the whole space with useful inputs. solved Quantum computers and algorithm performance [closed]
From your description, I think you are probably looking for topological sorting. It is based on the assumption that ‘impossible situation’ occurs when one connections suggests that A comes before B but there is some another connection which suggests that B comes before A. Link for topological sort: Topological Sorting solved Looking for a sorting … Read more
Since you’re looking for continuous sequences: $string = ‘aaabababbbbbaaaaabbbbbbbbaa’; $count = strlen($string); if ($count > 0) { $mostFrequentChar = $curChar = $string[0]; $maxFreq = $curFreq = 1; for ($i = 1; $i < $count; $i++) { if ($string[$i] == $curChar) { $curFreq++; if ($curFreq > $maxFreq) { $mostFrequentChar = $curChar; $maxFreq = $curFreq; } } … Read more
You should almost always use an algorithm like std::count_if if one is available. The reason is that the compiler vendor can put optimizations in that are not portable if you were to put them in manually in your own loop. For example there are intrinsic functions that could be CPU specific that speed up even … Read more