There are only 10,000 possible sequences of four digits (0000 to 9999), so at some point you will have found that every sequence has been duplicated, and there's no need to process further digits. If you assume that pi is a perfectly uniform random number generator, then each new digit...

c,algorithm,binary-search,upperbound,last-occurrence

n.m.'s 2-search approach will work, and it keeps the optimal time complexity, but it's likely to increase the constant factor by around 2, or by around 1.5 if you begin the second search from where the first search ended. If instead you take an "ordinary" binary search that finds the...

scala,generics,lower-bound,upperbound

This example may help scala> test.lowerBound(new Employee("Scala Jo")) res9: Test[Person] = [email protected] scala> test.lowerBound[Employee](new Employee("Scala Jo")) <console>:21: error: type arguments [Employee] do not conform to method lowerBound's type parameter bounds [U >: Person] test.lowerBound[Employee](new Employee("Scala Jo")) ^ In general, it's connected to the Liskov Substitution Principle - you can use...

algorithm,iteration,time-complexity,recurrence,upperbound

Look at step 4 T(n) = 2(2(2(2(2T(n-5) + 1) + 1) + 1) + 1) +1 ----- 4 T(n) = 2(2(2(2(2T(n-5))))) + 16 + 8 + 4 + 2 +1 = T(n) = 2^4* 2T(n-5) + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 = T(n) = 2^4* 2T(n-5)...

c++,binary-search,lower-bound,upperbound

The lower and upper bound of a binary search are the lowest and highest position where the value could be inserted without breaking the ordering. (In the C++ standard library, these bounds will be represented by iterators referencing the element before which the value could be inserted, but the concept...

c++,stl,lower-bound,upperbound

If you have multiple elements in the range [first, last) whose value equals the value val you are searching for, then the range [l, u) where l = std::lower_bound(first, last, val) u = std::upper_bound(first, last, val) is precisely the range of elements equal to val within the range [first, last)....