The problem is in the step where you say to find the true median of the medians. In your example, you had these medians: 50 45 40 35 30 25 20 15 10 The true median of this data set is 30, not 15. You don’t find this median by splitting the groups into blocks … Read more
One way of thinking about this problem is to think about doing a binary search over a transformed version of the array, where the array has been modified by applying the function f(x) = 1 if x > target 0 else Now, the goal is to find the very first place that this function takes … Read more
There you go. This is probably the most trivial example of a function that runs in O(n!) time (where n is the argument to the function): void nFacRuntimeFunc(int n) { for(int i=0; i<n; i++) { nFacRuntimeFunc(n-1); } }
I don’t know if Boost has more specific functions, but you can do it with the standard library. Given std::vector<double> v, this is the naive way: #include <numeric> double sum = std::accumulate(v.begin(), v.end(), 0.0); double mean = sum / v.size(); double sq_sum = std::inner_product(v.begin(), v.end(), v.begin(), 0.0); double stdev = std::sqrt(sq_sum / v.size() – mean … Read more
(n & (n – 1)) == 0 is best. However, note that it will incorrectly return true for n=0, so if that is possible, you will want to check for it explicitly. http://www.graphics.stanford.edu/~seander/bithacks.html has a large collection of clever bit-twiddling algorithms, including this one.
This is a pretty clever solution — it exploits the fact that there is a pattern of results in the running XORs. The f() function calculates the XOR total run from [0, a]. Take a look at this table for 4-bit numbers: 0000 <- 0 [a] 0001 <- 1 [1] 0010 <- 3 [a+1] 0011 … Read more
What you are talking about is lifetime dependency chaining, that one thing is dependent on another which can be modified outside of it’s control. If you have an idempotent function from a, b to c where, if a and b are the same then c is the same but the cost of checking b is … Read more
This problem is known as the subset sum problem, which is a special case of the Knapsack problem. Wikipedia is a good starting point for some algorithms.
Here is a solution to unbounded prime sieving in C#, which can be implemented using the Sieve of Eratosthenes (SoE) or the Sieve of Atkin (SoA) algorithms; however, I maintain that it is hardly worth the extreme complexity of an optimized SoA solution given than the true SoE gives about the same performance without as … Read more
The simplest approach is as follows: Sort the list: O(n lg n) The sorted list is the first permutation Repeatedly generate the “next” permutation from the previous one: O(n! * <complexity of finding next permutaion>) Step 3 can be accomplished by defining the next permutation as the one that would appear directly after the current … Read more