4 ms·
changing a (theoretically) O(N) algorithm to an O(N log N) algorithm and increasing the number of operations by a factor of 10 nevertheless resulted in the code
by lutorm 18y ago
changing a (theoretically) O(N) algorithm to an O(N log N) algorithm and increasing the number of operations by a factor of 10 nevertheless resulted in the code running twice as fast
Your N is 10^10? That's a big problem.
- cperciva 18y agoThe implicit constants in the O() expressions are different.