Throw the endpoints of the intervals into an array, marking them as either start- or end-points. Sort them by breaking ties by placing end-points before start-points if the intervals are closed, or the other way around if they’re half-open. 1S, 2S, 3E, 4E, 5S, 10E, 12S, 13S, 14E, 15E Then iterate through the list, keeping … Read more
There’s no such thing as a “good hash function” for universal hashes (ed. yes, I know there’s such a thing as “universal hashing” but that’s not what I meant). Depending on the context different criteria determine the quality of a hash. Two people already mentioned SHA. This is a cryptographic hash and it isn’t at … Read more
I have done something similar, by decomposing images into signatures using wavelet transform. My approach was to pick the most significant n coefficients from each transformed channel, and recording their location. This was done by sorting the list of (power,location) tuples according to abs(power). Similar images will share similarities in that they will have significant … Read more
To understand the difference between polynomial time and pseudopolynomial time, we need to start off by formalizing what “polynomial time” means. The common intuition for polynomial time is “time O(nk) for some k.” For example, selection sort runs in time O(n2), which is polynomial time, while brute-force solving TSP takes time O(n · n!), which … Read more
For k=4, space complexity O(n), time complexity O(n2 * log(n)) Sort the array. Starting from 2 smallest and 2 largest elements, calculate all lesser sums of 2 elements (a[i] + a[j]) in the non-decreasing order and all greater sums of 2 elements (a[k] + a[l]) in the non-increasing order. Increase lesser sum if total sum … Read more
Everyone else comparing this to the Travelling Salesman Problem probably hasn’t read your question carefully. In TSP, the objective is to find the shortest cycle that visits all the vertices (a Hamiltonian cycle) — it corresponds to having every node labelled ‘mustpass’. In your case, given that you have only about a dozen labelled ‘mustpass’, … Read more
The simplest solution for this is maintaining a min heap of max size 5000. Every time a new number arrives – check if the heap is smaller then 5000, if it is – add it. If it is not – check if the minimum is smaller then the new element, and if it is, pop … Read more
You are looking for a derangement of your entries. First of all, your algorithm works in the sense that it outputs a random derangement, ie a permutation with no fixed point. However it has a enormous flaw (which you might not mind, but is worth keeping in mind): some derangements cannot be obtained with your … Read more
The key to these large-number modulus tasks is not to compute the full result before performing the modulus. You should reduce the modulus in the intermediate steps to keep the number small: 500! / 20! = 21 * 22 * 23 * … * 500 21 * 22 * 23 * 24 * 25 * … Read more
Regarding the minimum number of (not necessarily adjacent) swaps needed to convert a permutation into another, the metric you should use is the Cayley distance which is essentially the size of the permutation – the number of cycles. Counting the number of cycles in a permutation is a quite trivial issue. A simple example. Suppose … Read more