time-complexity

问题 I implemented Merge Sort and Quick Sort in C++, and I wanna get the execution times using each of two with many of cases those are already Sorted or not & has different size. #include <iostream> #include <ctime> #include <vector> #include <algorithm> using namespace std; void Merge(vector<int>& s, int low, int mid, int high) { int i = low; int j = mid + 1; int k = low; vector<int> u(s); while (i <= mid && j <= high) { if (s.at(i) < s.at(j)) { u.at(k) = s.at(i); i++; } else { u.at(k) = s.at(j)

问题 How would the algorithm's (of Insertion Sort of O(n^2) ) complexity change if you changed the algorithm to use a binary search instead of searching previous values until you found where to insert your current value. Also, When would this be useful? 回答1: Your new complexity is still quadratic , since you need to move all of the sorted parts rightward. Therefore, using binary search is only marginally better. I would recommend a fast sorting algorithm (in O(n log n) time) for large arrays, the

问题 How would the algorithm's (of Insertion Sort of O(n^2) ) complexity change if you changed the algorithm to use a binary search instead of searching previous values until you found where to insert your current value. Also, When would this be useful? 回答1: Your new complexity is still quadratic , since you need to move all of the sorted parts rightward. Therefore, using binary search is only marginally better. I would recommend a fast sorting algorithm (in O(n log n) time) for large arrays, the

问题 I am taking a class on complexity analysis and we try to determin basic operations of algorithms. We defined it as the following: A basic operation is one that best characterises the efficiency of the particular algorithm of interest For time analysis it is the operation that we expect to have the most influence on the algorithm’s total running time: - Key comparisons in a searching algorithm - Numeric multiplications in a matrix multiplication algorithm - Visits to nodes (or arcs) in a graph

问题 I am using a JavaScript object as a dictionary, and wanted to make keys case-insensitive. I used Object.defineProperty() to implement this: Object.defineProperty(Object.prototype, "getKeyUpperCase", { value: function(prop) { for (var key in this) { if (key.toUpperCase() === prop.toUpperCase()) { return this[key]; }; }; }, enumerable: false }); What is the time complexity of searching an object via key in JavaScript? I'm expecting the dictionary to hold around 1 million keys. To me it looks

问题 Find an algorithm to recursively remove all adjacent duplicates in a given string this is the original question.I have thought of an algorithm using stacks. 1.Initialize a stack and a char_popped variable 2.Push the first char of str into the stack. 3.now iterate through the characters of string. if the top of stack matches with the character { pop the character from the stack, char_popped=character } else { if(character==char_popped) {DONT DO ANYTHING} else push the character on the stack

问题 I'm looking for the time complexity of these methods as a function of the number of rows in a dataframe, n. Another way of asking this question is: Are indexes for dataframes in pandas btrees (with log(n) time look ups) or hash tables (with constant time lookups)? Asking this question because I'd like a way to do constant time look ups for rows in a dataframe based on a custom index. 回答1: Alright so it would appear that: 1) You can build your own index on a dataframe with .set_index in O(n)

问题 Why does the Binet formula( O(LogN), but it is not exactly ) work worse in time than the iteration method( O(n) )? static double SQRT5 = Math.Sqrt(5); static double PHI = (SQRT5 + 1) / 2; public static int Bine(int n) { return (int)(Math.Pow(PHI, n) / SQRT5 + 0.5); } static long[] NumbersFibonacci = new long[35]; public static void Iteracii(int n) { NumbersFibonacci[0] = 0; NumbersFibonacci[1] = 1; for (int i = 1; i < n - 1; i++) { NumbersFibonacci[i + 1] = NumbersFibonacci[i] +

问题 sum =0; for (int i=1; i<n; i++) { for (int j=1; j< n/i; j++) { sum = sum +j; } } In the above outer loop , The variable i runs from 1 to n, thus making complexity of outer loop as O(n). This explains the n part of O(n logn) complexity. But for the outer part when we see then j runs from 1 to n/i, meaning whenever i is 1 , the complexity is n so i guess the inner time complexity should also be O(n) . Making the total time Complexity as O(n*n)=O(n^2). 回答1: This is what you can do using Sigma

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