time-complexity

问题 What would be the time complexity of Array.from() . For example: const set = new Set(); set.add('car'); set.add('cat'); set.add('dog'); console.log(Array.from(set)); // time complexity of making this convertion from Set to Array 回答1: It's O(n) . When used on an iterable (like a Set ), Array.from iterates over the iterable and puts every item returned into the new array, so there's an operation for every item returned by the iterable. 回答2: It is always going to be O(n) as the number of

问题 What would be the time complexity of Array.from() . For example: const set = new Set(); set.add('car'); set.add('cat'); set.add('dog'); console.log(Array.from(set)); // time complexity of making this convertion from Set to Array 回答1: It's O(n) . When used on an iterable (like a Set ), Array.from iterates over the iterable and puts every item returned into the new array, so there's an operation for every item returned by the iterable. 回答2: It is always going to be O(n) as the number of

问题 What would be the time complexity of Array.from() . For example: const set = new Set(); set.add('car'); set.add('cat'); set.add('dog'); console.log(Array.from(set)); // time complexity of making this convertion from Set to Array 回答1: It's O(n) . When used on an iterable (like a Set ), Array.from iterates over the iterable and puts every item returned into the new array, so there's an operation for every item returned by the iterable. 回答2: It is always going to be O(n) as the number of

问题 What is the time complexity of the MD5 algorithm? I couldn't find a definitive answer online. I think the complexity is O(n) but I'm not really sure. 回答1: O(n) as mentioned by DuBuisson and osgx in the comments. 来源： https://stackoverflow.com/questions/43625569/time-complexity-of-md5

问题 Is it O(n) or O(1) (By saving length in a private variable during string allocation to the object). if it is O(n), does it mean that the complexity of following code is O(n^2)? for(int i=0; i<s.length()-1;i++){ //some code here! } 回答1: It is O(1) as the length is already known to String instance. From JDK 1.6 it is visible. public int length() { return count; } Update It is important to understand why they can cache the value of count and keep using same value for count . The reason lies in a

问题 Is it O(n) or O(1) (By saving length in a private variable during string allocation to the object). if it is O(n), does it mean that the complexity of following code is O(n^2)? for(int i=0; i<s.length()-1;i++){ //some code here! } 回答1: It is O(1) as the length is already known to String instance. From JDK 1.6 it is visible. public int length() { return count; } Update It is important to understand why they can cache the value of count and keep using same value for count . The reason lies in a

问题 My function should return the missing element in a given array range. So i first sorted the array and checked if the difference between i and i+1 is not equal to 1, i'm returning the missing element. // Given an array A such that: // A[0] = 2 // A[1] = 3 // A[2] = 1 // A[3] = 5 // the function should return 4, as it is the missing element. function solution(A) { A.sort((a,b) => { return b<a; }) var len = A.length; var missing; for( var i = 0; i< len; i++){ if( A[i+1] - A[i] >1){ missing = A[i

问题 My function should return the missing element in a given array range. So i first sorted the array and checked if the difference between i and i+1 is not equal to 1, i'm returning the missing element. // Given an array A such that: // A[0] = 2 // A[1] = 3 // A[2] = 1 // A[3] = 5 // the function should return 4, as it is the missing element. function solution(A) { A.sort((a,b) => { return b<a; }) var len = A.length; var missing; for( var i = 0; i< len; i++){ if( A[i+1] - A[i] >1){ missing = A[i

问题 My function should return the missing element in a given array range. So i first sorted the array and checked if the difference between i and i+1 is not equal to 1, i'm returning the missing element. // Given an array A such that: // A[0] = 2 // A[1] = 3 // A[2] = 1 // A[3] = 5 // the function should return 4, as it is the missing element. function solution(A) { A.sort((a,b) => { return b<a; }) var len = A.length; var missing; for( var i = 0; i< len; i++){ if( A[i+1] - A[i] >1){ missing = A[i

问题 My professor showed the following problem in class and mentioned that the answer is O(1) while mine was quit different, I hope to get some help knowing of what mistakes did I made. Question: Calculate the Amortized Time Complexity for F method in AVL tree, when we start from the minimal node and each time we call F over the last found member. Description of F: when we are at specific node F continues just like inorder traversal starting from the current one until the next one in inorder