As you pointed out, the problem is equivalent to the problem of finding a maximum matching in a bipartite graph (one set of vertices is the set of people and the other on the set of slots, there is an edge between a person and a slot if the person is available for this slot). … Read more
The problem is that I can’t understand how the “official” JPG->PGM convertitors work in terms of what value to assign to the final pixel (i guess, 0->255) starting from the classic RGB format. There is likely a gamma adjustment in the conversion those “official” tools are using.That is, it is not just a linear transform. … Read more
In Ruby, the prime library gives you the factorization: require ‘prime’ 4800.prime_division #=> [[2, 6], [3, 1], [5, 2]] To get that list of yours, you take the cartesian product of the possible powers: require ‘prime’ def factors_of(number) primes, powers = number.prime_division.transpose exponents = powers.map{|i| (0..i).to_a} divisors = exponents.shift.product(*exponents).map do |powers| primes.zip(powers).map{|prime, power| prime ** … Read more
Big-O Notation describes the limiting behavior when n is large, also known as asymptotic behavior. This is an approximation. (See http://en.wikipedia.org/wiki/Big_O_notation) Insertion sort is faster for small n because Quick Sort has extra overhead from the recursive function calls. Insertion sort is also more stable than Quick sort and requires less memory. This question describes … Read more
It will have to go through all the elements in the array to do this. Array is a unique data structure where you have to specify a size when you initialize it. Order would be the source array’s size or in Big O terms its O(length). Infact this happens internally in an ArrayList. ArrayList wraps … Read more
It is certainly not possible with a plain singly-linked list. Sketch proof: to examine the last node of a singly-linked list, we must perform n-1 operations of following a “next” pointer [proof by induction on the fact that there is only one reference to the k+1th node, and it is in the kth node, and … Read more
This is an old question, but I had to solve it in order for a system I was working on to be backward compatible. The exact requirement was for a client-generated identifier that would be written to the database and stored in a 20-character unique column. It never got shown to the user and was … Read more
You need not to use std::sort. It can be done in a simpler way: if ( std::adjacent_find( myvector.begin(), myvector.end(), std::not_equal_to<>() ) == myvector.end() ) { std::cout << “All elements are equal each other” << std::endl; }
When I was trying to figure out how rotations would work for my tetris game, this was the first question that I found on stack overflow. Even though this question is old, I think my input will help others trying to figure this out algorithmically. First, I disagree that hard coding each piece and rotation … Read more
A linear search looks down a list, one item at a time, without jumping. In complexity terms this is an O(n) search – the time taken to search the list gets bigger at the same rate as the list does. A binary search is when you start with the middle of a sorted list, and … Read more