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If we have an algorithm which is order N^2*logN, and if it takes 1 ms with input size 64; does it take 2^10*(11/6) ms to run this algorithm with input size 2048? I am using direct proportion here, that’s why it seemed defective to me.
Easiest way to solve is probably to divide 2048 by 64, plug the resulting number into the complexity equation, and the result is the number of milliseconds for Input size 2048.