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vector1.insert( vector1.end(), vector2.begin(), vector2.end() );
What will be the Time complexity of concatenating/copying two vectors?
From the standard [vector.modifiers]/2:
Complexity: If reallocation happens, linear in the number of elements
of the resulting vector; otherwise, linear in the number of elements
inserted plus the distance to the end of the vector.
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I am stuck on finding a solution for finding all the contiguous subarrays of a given array in minimum time complexity O(n). For example: [1,2,3,4] Subarrays are: [1][2][3][4][1,2][2,3][3,4][1,2,3][2,3,4][1,2,3,4] I have done it with time complexity O(n^3) but question wants in O(N).
There are subarray as you described. Therefore, There is a lower bound of to your problem.
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Does Eigen contains an algorithm for solving nearest correlation matrix problems?
You can use the SelfAdjointEigenSolver to decompose your non-positive definite matrix, edit the negative eigenvalues to something slightly positive and (manually) re-compose the matrix.
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Which of the following containers have average search by value complexity equal to O(log(n))?
Which ones have O(n) and O(n^2)?
std::vector
std::list
std::deque
std::set
std::multiset
std::unordered_set
std::unordered_multiset sorted
std::vector sorted
std::list sorted
std::deque sorted
The standard specifies the complexity of the relevant functions. Get a copy, or read a good reference. (See e.g. http://en.cppreference.com/w/cpp/container/unordered_set/find.)
They're all pretty much what you'd expect.
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I have a SORTED singly linked list of strings. I need to find occurrences of a particular string in the list. How can I do it in logarithmic time?
Not. Lists support only linear access, and since your string can be anywhere you must check O(N) elements.
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I found implementation using 2D array http://www.geeksforgeeks.org/dynamic-programming-set-10-0-1-knapsack-problem/. But how to implement it using 1D array and if it is not possible then why?
Note that each of values in the current row uses only elements of the current and previous rows. Hence you can implement the algorithm with an array K[2][W], which is the same as using an array K[2*W] with some additional trivial index calculations