2010年2月22日 · The binary search is a classical example. At each step we throw away 1/2 of the problem space. But binary search is not the only such example. Suppose, you proved …

2012年4月29日 · O(n) means that the algorithm's maximum running time is proportional to the input size. basically, O(something) is an upper bound on the algorithm's number of instructions …

2019年4月27日 · You still need to study a lot. O(..) describes the complexity of your algorithm. To be easy, you can imagine as the time to take to finish you algorithm for an n input, if O(n) it will …

2009年10月20日 · O(logn) - finding something in your telephone book. Think binary search. O(n) - reading a book, where n is the number of pages. It is the minimum amount of time it takes to …

2015年8月24日 · You want to know if there is an easy way to identify if an algorithm is O(log N). Well: just run and time it. Run it for inputs 1.000, 10.000, 100.000 and a million. If you see like …

2013年12月11日 · Since logs at different bases differ only a by a constant ratio, and Big O is indifferent to minor things like constant multiplying factors, you will often find O(logN) actually …

2018年4月2日 · So O(log(n^2)) = O(2*log(n)). With complexity calculations, the focus is on convergence behaviour in the limit, so constant multipliers are cancelled out. This means …

2013年6月15日 · is in O(1), but also in O(n), O(n²), and, if you wanted to be clever, O(log(n)).Why? Because all constant time algorithms are bounded by some linear, quadratic, …

2017年5月23日 · Compute SSSP for each part: again because we have O(|G'|) part and we can compute SSSP for all parts in time |n/logn| * |log n/log logn * log (logn /log log n). update …

A lookup for a node with value 1 has O(n) time complexity. To make a lookup more efficient, the tree must be balanced so that its maximum height is proportional to log(n). In such case, the …