With binary search you typically search in a sorted random access data structure like an array, by discarding half of the array with each comparison. Hence, in k steps you effectively cover 2^k entries. This yields a complexity of at most log2(n) of n elements. With landau symbols, the base of the logarithm disappears because … Read more
The answer is – you really don’t need to be worrying about this at this juncture. If your code is running too slow, then you can experiment with some optimization. However, “too slow” is subjective. Premature Optimization Is The Root Of All Evil. As angelatlarge points out, for the moment – go with what is … Read more
That would be a way to go on Linux: Compile your program with debug info (g++ -g merger_sort.cpp -o merger_sort) Load it in debuger: >>> gdb merge_sort Run it: (gdb) run. You will see: Program received signal SIGSEGV, Segmentation fault. 0x0000000000400b1e in merge_sort (A=0x7ffffffddda0, p=0, r=1) look at the position in the code: (gdb) layout … Read more
Here is a simple solution of your problem, you have a problem in your 3ed loop, you don’t need it! You just need to loop through your start(i) and end(j) indices as shown below: public static void main(String[] args) { String search_query = “one two three”; String[] queries = search_query.split(” “); List<String> liste = new … Read more
Here is a simple solution: String s = “Hello stack over flow”; List< String > words = Arrays.asList( s.split( ” ” ) ); Collections.reverse( words ); String reversed = words.get( 0 ); for ( int i = 1; i < words.size(); ++i ) { reversed += ” ” + words.get( i ); } I assume … Read more
To keep it simple, in case A, your program has to initialize i only once. Assign values to i only array.Length times. Increase i values only array.Length times. do the line execution and comparison 2*array.Length times. In case B, initialize i twice. Assign values to i 2*array.Length times. Increase i values 2*array.Length times. do line … Read more
You had some error about your indentation and element word, it was element def copy_sort(array): copy=array[:] sorted_copy=[] while len(copy)>0: minimum=0 for element in range(0,len(copy)): if copy[element] < copy[minimum]: minimum=element print(‘\nRemoving value’,copy[minimum], ‘from’,copy) sorted_copy.append(copy.pop(minimum)) return sorted_copy array=[5,3,1,2,6,4] print(‘Copy sort…\nArray:’,array) print(‘copy :’, copy_sort(array)) print(‘array’,array)` 1 solved Python sorting algorithm [closed]
I couldn’t find one…so I made one. Now I want to share it with anyone else who needs it. It has been tested and validated, and any additional comments are welcome. https://github.com/manas1213/hungarian_algorithm This is based on the comprehensive algorithm outlined at http://csclab.murraystate.edu/bob.pilgrim/445/munkres.html. solved PL/SQL Implementation of Hungarian/Kuhn-Munkres Algorithm [closed]
Disclaimer: This answer focuses on the practical aspects of dealing with the fact that there are problems for which no polynomial-time algorithm is known. To give an answer that is precise from a theoretical point of view, the terminology used in the question is not clear enough. There are two meanings of NP in computer … Read more
What you are asking for is a SAT algorithm and it has an exponential complexity, therefore unless you have some extra constraints you must do an exhaustive check (brute force as you said). You may be interested in the function itertools.combinations(iterable, combinations_length) to sum all possible combinations. also you can represent your elements like this: … Read more
Below example demonstrates the usage of erase-remove_if. limit is captured by reference and can thus be modified outside the lambda: #include <vector> #include <algorithm> #include <iostream> int main() { std::vector<int> vec{0,1,2,3,4,5,6,7,8,9}; int size = vec.size(); for (int limit = 0; limit <= size; limit++) { vec.erase(std::remove_if(std::begin(vec), std::end(vec), [&limit](int i) { return i < limit; }), … Read more
The toFixed() method converts a number into a string, keeping a specified number of decimals. Here is the JsFiddle link https://jsfiddle.net/zco2d5x1/ function fnum(x) { if (x < 1000000000) { alert((x / 1000000).toFixed(2) + “M”); } else if (x < 1000000000000) { alert((x / 1000000000).toFixed(2) + “B”); } else alert(“More”); } fnum(136866516683); 0 solved I have … Read more
Put the elements of set A into a structure that allows to test set membership in O(1) (for example a hash table or a bit vector if the range of elements is small). This will run in O(n). Now iterate through the set B and check if there is an element in A that is … Read more
As told in the comments, you can see how to check if a number is an integer or not, and perform an appropriate action: Checking if float is an integer. However, if you want more interesting way, without any conditions: float a; float result; cin >> a; result = std::floor(a) – !std::ceil(a – std::floor(a)); // … Read more
The line vector<int> list3(list1.size()+list2.size()); creates a vector of type int and inserts list1.size()+list2.size() default contructed elements. You want to create an empty vector of type int and reserve memory for list1.size()+list2.size() elements. Use vector<int> list3; list3.reserve(list1.size()+list2.size()); solved Some problem about the algorithm of merging two lists