big-o

问题 I am stuck at this problem. I think it is equivalent to show 2m choose m is big theta of 4 to the power n, but still find difficult to prove it. Thanks of @LutzL's suggestion. I thought of stirling's approximation before. 回答1: The O -part should be easy. Choosing exactly n /2 elements out of n is a special case of choosing arbitrary combinations out of n elements, i.e. deciding for each of these n elements whether to choose it or not. The Ω -part is harder. In fact, plotting 4n / binomial(2 n

问题 I am stuck at this problem. I think it is equivalent to show 2m choose m is big theta of 4 to the power n, but still find difficult to prove it. Thanks of @LutzL's suggestion. I thought of stirling's approximation before. 回答1: The O -part should be easy. Choosing exactly n /2 elements out of n is a special case of choosing arbitrary combinations out of n elements, i.e. deciding for each of these n elements whether to choose it or not. The Ω -part is harder. In fact, plotting 4n / binomial(2 n

问题 I have a game where certain game objects spawn all at once and then despawn as they get destroyed/killed. The game objects are elements in an std::vector , and I'd like to minimize memory usage. I'm used to the swap trick, std::vector<gameObject>(gameObjectVector.begin(), gameObjectVector.end()).swap(gameObjectVector); but I noticed the inbuilt shrink_to_fit() from C++11. However, it has linear complexity while the swap trick is constant. Isn't the swap trick superior in every way? 回答1: The

问题 I just tried implementing code (in Java) for various means by which the nth term of the Fibonacci sequence can be computed and I'm hoping to verify what I've learnt. The iterative implementation is as follows: public int iterativeFibonacci(int n) { if ( n == 1 ) return 0; else if ( n == 2 ) return 1; int i = 0, j = 1, sum = 0; for ( ; (n-2) != 0; --n ) { sum = i + j; i = j; j = sum; } return sum; } The recursive implementation is as follows :- public int recursiveFibonacci(int n) { if ( n ==

问题 I just tried implementing code (in Java) for various means by which the nth term of the Fibonacci sequence can be computed and I'm hoping to verify what I've learnt. The iterative implementation is as follows: public int iterativeFibonacci(int n) { if ( n == 1 ) return 0; else if ( n == 2 ) return 1; int i = 0, j = 1, sum = 0; for ( ; (n-2) != 0; --n ) { sum = i + j; i = j; j = sum; } return sum; } The recursive implementation is as follows :- public int recursiveFibonacci(int n) { if ( n ==

问题 I just tried implementing code (in Java) for various means by which the nth term of the Fibonacci sequence can be computed and I'm hoping to verify what I've learnt. The iterative implementation is as follows: public int iterativeFibonacci(int n) { if ( n == 1 ) return 0; else if ( n == 2 ) return 1; int i = 0, j = 1, sum = 0; for ( ; (n-2) != 0; --n ) { sum = i + j; i = j; j = sum; } return sum; } The recursive implementation is as follows :- public int recursiveFibonacci(int n) { if ( n ==

问题 We know that this problem is np-complete, hence, no polynomial algorithm can be found for it. Also, we know that number of all partitions of a set is equal to the bell's number. I see there are few algorithms to generate all partitions of a set but couldn't find what is the time complexity to solve this problem. For example, this python code generates all partitions of a set recursively. What is the time complexity of this algorithm? Could this problem be solved with better time complexity?

问题 We know that this problem is np-complete, hence, no polynomial algorithm can be found for it. Also, we know that number of all partitions of a set is equal to the bell's number. I see there are few algorithms to generate all partitions of a set but couldn't find what is the time complexity to solve this problem. For example, this python code generates all partitions of a set recursively. What is the time complexity of this algorithm? Could this problem be solved with better time complexity?

问题 I was just working on a LeetCode problem, Roman to Integer, the conversion of Roman numerals to integers, and after finishing up and comparing solutions, I noticed a rather interesting nuance in how the solutions listed describe their computational complexity. I had described my solution as O(n) , linear with the number of input elements, as my solution iterated over the elements of a Roman numeral character by character. The official solutions, however, described how with numerals I , V , X

问题 I've seen explanations for why RAM is accessed in constant time ( O(1) ) and why it's accessed in logarithmic time ( O(n) ). Frankly neither makes much sense to me; what is n in the big-O notation and how does it make sense to measure the speed at which a physical device operates using big-O? I understand an argument for RAM being accessed in linear time is if you have an array a then the kth element would be at address a+k*size_of_type (point is address can be easily calculated). If you know