1.题目: A game for one player is played on a board consisting of N consecutive squares, numbered from 0 to N − 1. There is a number written on each square. A non-empty zero-indexed array A of N integers contains the numbers written on the squares. More…

How to solve this HARD issue 1. Problem: A non-empty zero-indexed array A consisting of N integers is given. A peak is an array element which is larger than its neighbours. More precisely, it is an index P such that 0 < P < N − 1 and A[P − 1] < A…

首先上题目: A DNA sequence can be represented as a string consisting of the letters A, C, G and T, which correspond to the types of successive nucleotides in the sequence. Each nucleotide has an impact factor, which is an integer. Nucleotides of types A,…

Task1: A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N. For example, number 9 has binary representation 1001 and contains a binary gap of…

没有宏观的架构设计,没有特定的框架语言.在Codility提出的一些小问题上,用最纯粹的方式测试你最基本的编码能力. Codility第一课:算法复杂度 各种算法书的开篇大多是算法分析,而复杂度(complexity)又是最基本的分析指标.所以Codility的第一课也不例外,直入复杂度主题.这里不再详述其概念,而只提及复杂度的几个注意点: 复杂度不是程序运行时间的准确度量,而是数量级. 重点是找到主要因子操作(dominant operation),忽略常量以及数量级更低的项,可理解为求极限.…

question:https://codility.com/programmers/lessons/9 To solve this question , I get each element's divsors which appearing in input Array A using Sieve of Eratosthenes method. Time complexity is O(nlogn); Then  we iterate array A to get the ith non-di…

Instant Complexity Time Limit: 1000MS Memory Limit: 10000K Description Analyzing the run-time complexity of algorithms is an important tool for designing efficient programs that solve a problem. An algorithm that runs in linear time is usually much f…

Runtime Complexity of .NET Generic Collection   I had to implement some data structures for my computational geometry class. Deciding whether to implement the data structures myself or using the build-in classes turned out to be a hard decision, as t…

O(1):constant - the operation doesn't depend on the size of its input, e.g. adding a node to the tail of a linked list where we always maintain a pointer to the tail node.int i=0;i++;++i;i+=6; O(n):linear - the run time complexity is proportionate to…

这一章比较短! 空间复杂度(space complexity)和辅助空间(auxiliary space)经常混用,下面是正确的辅助空间和空间复杂度的定义 辅助空间:算法需要用到的额外或者暂时的存储空间. 空间复杂度:是指算法所需要的所有存储空间,这是跟输入数据的大小决定的.空间复杂度包括辅助空间和保存输入的存储空间. 如果我们想比较几个标准的排序算法所需要的空间,那么用辅助空间来分析会比空间复杂度好.归并排序用了O(n)的辅助空间.而插入排序和堆排序用的O(1)的辅助空间.但是他们这些算法的空…

Instant Complexity Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other) Total Submission(s) : 8   Accepted Submission(s) : 7 Problem Description Analyzing the run-time complexity of algorithms is an important tool for desi…

THE ARCHITECTURE OF COMPLEXITY HERBERT A. SIMON* Professor of Administration, Carnegie Institute of Technology (Read April 26, 1962)…

A non-empty zero-indexed string S is given. String S consists of N characters from the set of upper-case English letters A, C, G, T. This string actually represents a DNA sequence, and the upper-case letters represent single nucleotides(核苷). You are…

A non-empty zero-indexed array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road. Array A contains only 0s and/or 1s: 0 represents a car traveling east, 1 represents a car traveling west. Th…

T(n)=aT(n/b)+f(n); where we can interpret n/b to mean either floor(b/n) or ceil(b/n), Then T (n) has the following asymptotic bounds: 1. If f (n)= O(nlogb a-c) for some constant c> 0, then T (n)=Θ(nlogb a)2.If f (n)= Θ(nlogb a), then T (n)=Θ(nlogb a…

The Brain vs Deep Learning Part I: Computational Complexity — Or Why the Singularity Is Nowhere Near July 27, 2015July 27, 2015 Tim Dettmers Deep Learning, NeuroscienceDeep Learning, dendritic spikes, high performance computing, neuroscience, singula…

The Password complexity is a Local Policy setting named "Passwords must meet complexity requirements" under Computer Configuration/Windows Settings/Security Settings/Account Policies/Password Policy. In a Server Core installation there is no gra…

https://codility.com/demo/take-sample-test/peaks http://blog.csdn.net/caopengcs/article/details/17491791 其实可以做到O(n) #include <iostream> #include <sstream> using namespace std; int solution(vector<int> &A) { int n = A.size(); int prev…

https://codility.com/programmers/challenges/fluorum2014 http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1273 http://blog.csdn.net/caopengcs/article/details/36872627 http://www.quora.com/How-do-I-determine-the-order-of-visiting-all-leave…

https://codility.com/programmers/challenges/magnesium2014 图形上的DP,先按照路径长度排序,然后依次遍历,状态是使用到当前路径为止的情况:每个节点记录以该节点结束的最长路径,这样加入新的路径时去更新.注意路径是双向的~ #include <vector> #include <algorithm> using namespace std; struct Road { int start; int end; int val; }…

https://codility.com/demo/take-sample-test/max_double_slice_sum 两个最大子段和相拼接,从前和从后都扫一遍.注意其中一段可以为0.还有最后和最前面一个不可能取到~ #include <vector> using namespace std; int solution(vector<int> &A) { vector<int> maxEnd; maxEnd.resize(A.size()); maxEn…

https://codility.com/demo/take-sample-test/fish 一开始习惯性使用单调栈,后来发现一个普通栈就可以了. #include <stack> using namespace std; int solution(vector<int> &A, vector<int> &B) { int size = A.size(); int deadNum = 0; stack<int> stk; // downst…

https://codility.com/programmers/challenges/upsilon2012 求笛卡尔树的高度,可以用单调栈来做. 维持一个单调递减的栈,每次进栈的时候记录下它之后有多少元素,就是以它为根的子树的高度.出栈的时候再更新一次供新进栈者使用. int solution(vector<int> &A) { A.push_back(1000000001); // an element larger than any one in A int size = A.…

https://codility.com/demo/take-sample-test/count_div 此题比较简单,是在O(1)时间里求区间[A,B]里面能被K整除的数字,那么就计算一下就能得到. int solution(int A, int B, int K) { // write your code in C++11 int diff = (B-A+1); int result = diff / K; for (int i = A + result * K; i <= B; i++)…

https://codility.com/demo/take-sample-test/min_avg_two_slice 此题要求一个数组子段的最小的平均数(返回第一个数字的index).刚开始想记录sum,也没用,因为所有子段的组合是O(n^2)的.后来看解释发现,最小值必然存在于长度为2或3的子段中(长度为3的子段也无法分解).可以用反证法,如果有一个长度大于3的子段,均值小于之前找出的长度2和3的子段均值,那么必然可以在这个子段中找到均值更小的长度2和3的子段,矛盾. int soluti…

今天开始刷刷codility上respectable的题目,难度适中. https://codility.com/demo/take-sample-test/missing_integer 本题是找出数组里第一个非负的整数,要求复杂度O(n).那么首先想到的做法是排序,但这样负责度就上去了.要从nlogn降到n,常见的一个做法是用hashtable,这里就可以用set记录.要注意的是全负的情况,所以这里用了maxVal=0作为初值. #include <unordered_set> using…

今天在图书馆闲逛的时候偶然看见<Think Complexity>(复杂性思考)这本书,下午看了一会儿觉得很有意思.本书第二章讲的是用Python实现的图,特别写篇博客记录.   首先,图有两大元素:顶点和边.分别用Vertex和Edge类来描述顶点集和边集.   顶点集:Vertex class Vertex(object): """A Vertex is a node in a graph.""" def __init__(sel…

在写上一随笔之前,在Codility网站上还做了一个道题(非Demo题):CountBoundedSlices,得了60分(害臊呀).今天又重新做了一下这个算法,性能提高了不少,但由于此题不是Demo题,不能重新在Codility网站测试了. 可以从http://codility.com/cert/view/certAWY5VP-D46CA7989XU4XEZT/details看到题目的详细信息. 这里把题目copy出来: An integer K and a non-empty zero-in…

https://codility.com/demo/take-sample-test/delta2011/ 0-1背包问题的应用.我自己一开始没想出来.“首先对数组做处理,负数转换成对应正数,零去掉,计数排序统计有多少个不同元素及其对应个数,并累加所有数的和sum,不妨记b=sum/2,不同元素个数为m,则目标转换为在m个不同元素中挑出若干个元素(每个元素可以重复多次,但少于它们的出现次数),使得它们的和不大于b并尽量接近.到了这里,应该有点熟悉的感觉了吧.对了,其实这就是0-1背包问题!” 参…

Original Link: http://www.ehow.com/how_4812793_password-complexity-requirements-windows-xp.html#ixzz32PEZAbOn When you create a new account in Windows XP, you choose a username and a password, which must be a certain length. If you want to get rid of…