O( log* N ) is “iterated logarithm“: In computer science, the iterated logarithm of n, written log* n (usually read “log star”), is the number of times the logarithm function must be iteratively applied before the result is less than or equal to 1.
Just ask wolframalpha if you have doubts. In this case, it says Or you can also calculate the limit yourself: That means n^2 grows faster, so n log(n) is smaller (better), when n is high enough.
Hash tables are O(1) average and amortized case complexity, however it suffers from O(n) worst case time complexity. [And I think this is where your confusion is] Hash tables suffer from O(n) worst time complexity due to two reasons: If too many elements were hashed into the same key: looking inside this key may take O(n) time. Once a hash table has passed its load balance – it has … Read more
In simple words, a loop invariant is some predicate (condition) that holds for every iteration of the loop. For example, let’s look at a simple for loop that looks like this: In this example it is true (for every iteration) that i + j == 9. A weaker invariant that is also true is that i >= 0 && … Read more
I don’t know if you still need this problem solved, but http://www.ics.uci.edu/~eppstein/161/960130.html has an algorithm: Good luck!
In short: Hamiltonian cycle : O(N!) in the worst case WordBreak and StringSegment : O(2^N) NQueens : O(N!) Note: For WordBreak there is an O(N^2) dynamic programming solution. More details: In Hamiltonian cycle, in each recursive call one of the remaining vertices is selected in the worst case. In each recursive call the branch factor decreases by 1. Recursion … Read more
Simply use a new array for the shuffled cards and fill it using random cards removed from the prior array. Or just use Collections.shuffle():
This link might be helpful to you, as it details the use of the Haversine formula to calculate the distance. Excerpt: This script [in Javascript] calculates great-circle distances between the two points – that is, the shortest distance over the earth’s surface – using the ‘Haversine’ formula.
Each partitioning operation takes O(n) operations (one pass on the array). In average, each partitioning divides the array to two parts (which sums up to log n operations). In total we have O(n * log n) operations. I.e. in average log n partitioning operations and each partitioning takes O(n) operations.
a) Yes – in a perfect world you’d have to do log n merges of size n, n/2, n/4 … (or better said 1, 2, 3 … n/4, n/2, n – they can’t be parallelized), which gives O(n). It still is O(n log n). In not-so-perfect-world you don’t have infinite number of processors and context-switching … Read more