time-complexity

问题 so when there's an index on a column, and you do a simple SELECT * FROM table WHERE indexed_column = value, is that a O(1) search? does it matter whether the contents indexed are integers or string? 回答1: None of the lookups in MySQL's MyISAM or InnoDB storage engines are O(1) searches. Those storage engines use B+Trees to implement indexes. The best they can do is O(log 2 n) searches. The MEMORY storage engine uses a HASH index type by default, as well as the B+Tree index type. Only the HASH

问题 can anyone tell me what exactly is the time complexity of this Heap's algorithm shown in wikipedia, https://en.wikipedia.org/wiki/Heap%27s_algorithm ? I searched several websites and the answers are all vague, some of them say the time complexity is O(N!) some of them say it's O(NlogN). Which one is the correct answer? And why? Thank you. 回答1: There are N ! permutations in all and generating all of them requires Θ( N !) time and Θ(N) space. In other words, each permutation requires amortised

问题 can anyone tell me what exactly is the time complexity of this Heap's algorithm shown in wikipedia, https://en.wikipedia.org/wiki/Heap%27s_algorithm ? I searched several websites and the answers are all vague, some of them say the time complexity is O(N!) some of them say it's O(NlogN). Which one is the correct answer? And why? Thank you. 回答1: There are N ! permutations in all and generating all of them requires Θ( N !) time and Θ(N) space. In other words, each permutation requires amortised

问题 can anyone tell me what exactly is the time complexity of this Heap's algorithm shown in wikipedia, https://en.wikipedia.org/wiki/Heap%27s_algorithm ? I searched several websites and the answers are all vague, some of them say the time complexity is O(N!) some of them say it's O(NlogN). Which one is the correct answer? And why? Thank you. 回答1: There are N ! permutations in all and generating all of them requires Θ( N !) time and Θ(N) space. In other words, each permutation requires amortised

问题 I have a list of 100 random integers. Each random integer has a value from 0 to 99. Duplicates are allowed, so the list could be something like 56, 1, 1, 1, 1, 0, 2, 6, 99... I need to find the smallest integer (>= 0) is that is not contained in the list. My initial solution is this: vector<int> integerList(100); //list of random integers ... vector<bool> listedIntegers(101, false); for (int theInt : integerList) { listedIntegers[theInt] = true; } int smallestInt; for (int j = 0; j < 101; j++

问题 I have a list of 100 random integers. Each random integer has a value from 0 to 99. Duplicates are allowed, so the list could be something like 56, 1, 1, 1, 1, 0, 2, 6, 99... I need to find the smallest integer (>= 0) is that is not contained in the list. My initial solution is this: vector<int> integerList(100); //list of random integers ... vector<bool> listedIntegers(101, false); for (int theInt : integerList) { listedIntegers[theInt] = true; } int smallestInt; for (int j = 0; j < 101; j++

问题 I have a list of 100 random integers. Each random integer has a value from 0 to 99. Duplicates are allowed, so the list could be something like 56, 1, 1, 1, 1, 0, 2, 6, 99... I need to find the smallest integer (>= 0) is that is not contained in the list. My initial solution is this: vector<int> integerList(100); //list of random integers ... vector<bool> listedIntegers(101, false); for (int theInt : integerList) { listedIntegers[theInt] = true; } int smallestInt; for (int j = 0; j < 101; j++

问题 I have a list of 100 random integers. Each random integer has a value from 0 to 99. Duplicates are allowed, so the list could be something like 56, 1, 1, 1, 1, 0, 2, 6, 99... I need to find the smallest integer (>= 0) is that is not contained in the list. My initial solution is this: vector<int> integerList(100); //list of random integers ... vector<bool> listedIntegers(101, false); for (int theInt : integerList) { listedIntegers[theInt] = true; } int smallestInt; for (int j = 0; j < 101; j++

问题 Question: Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such that: Only one letter can be changed at a time. Each transformed word must exist in the word list. Note that beginWord is not a transformed word. Example 1: Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] Output: [ ["hit","hot","dot","dog","cog"], ["hit","hot","lot","log","cog"] ] My solution

问题 I was solving the following job interview question and solved most of it but failed at the last requirement. Q: Build a data structure which supports the following functions: Init - Initialise Empty DS. O(1) Time complexity. SetPositiveInDay(d,x) - Add to the DS that in day d exactly x new people were infected with covid-19. O(log n)Time complexity. WorseBefore(d) - From the days inserted into the DS and smaller than d return the last one which has more newly infected people than d. O(log n