Here is the way my own SuDoKu program does it: Start with a complete, valid board (filled with 81 numbers). Make a list of all 81 cell positions and shuffle it randomly. As long as the list is not empty, take the next position from the list and remove the number from the related cell. … Read more
You have heard about doing it in O(n) using dequeue. Well that is a well known algorithm for this question to do in O(n). The method i am telling is quite simple and has time complexity O(n). Your Sample Input: n=10 , W = 3 10 3 1 -2 5 6 0 9 8 -1 … Read more
Million points is a small number. The most straightforward approach works here (code based on KDTree is slower (for querying only one point)). Brute-force approach (time ~1 second) #!/usr/bin/env python import numpy NDIM = 3 # number of dimensions # read points into array a = numpy.fromfile(‘million_3D_points.txt’, sep=’ ‘) a.shape = a.size / NDIM, NDIM … Read more
Summary Type BST (*) Heap Insert average log(n) 1 Insert worst log(n) log(n) or n (***) Find any worst log(n) n Find max worst 1 (**) 1 Create worst n log(n) n Delete worst log(n) log(n) All average times on this table are the same as their worst times except for Insert. *: everywhere in … Read more
Your first mistake is thinking that ASCII encoding and Base64 encoding are interchangeable. They are not. They are used for different purposes. When you encode text in ASCII, you start with a text string and convert it to a sequence of bytes. When you encode data in Base64, you start with a sequence of bytes … Read more
It’s called Partitions. [Also see Wikipedia: Partition (number theory).] The number of partitions p(n) grows exponentially, so anything you do to generate all partitions will necessarily have to take exponential time. That said, you can do better than what your code does. See this, or its updated version in Python Algorithms and Data Structures by … Read more
Here’s the same code in Java without the 12.100000000000001 bug other answers have I also removed repeated code, changed power to a type integer to prevent floating issues when n – d is done, and made the long intermediate more clear The bug was caused by multiplying a large number with a small number. Instead … Read more
You can often entirely preserve the original structure of a recursive algorithm, but avoid the stack, by employing tail calls and changing to continuation-passing, as suggested by this blog entry. (I should really cook up a better standalone example.)
RGB distance in the euclidean space is not very similar to “average human perception”. You can use YUV color space, it takes into account this factor : | Y’ | | 0.299 0.587 0.114 | | R | | U | = | -0.14713 -0.28886 0.436 | | G | | V | | 0.615 … Read more
Most of these answers are horribly inefficient and/or will only give one-word solutions (no spaces). My solution will handle any number of words and is very efficient. What you want is a trie data structure. Here’s a complete Python implementation. You just need a word list saved in a file named words.txt You can try … Read more