I have found out that my algorithm will always do n!*4^n steps . I’d like to know wheter its complexity will be O(n!*4^n) or will it be something else? Thanks.
I have found out that my algorithm will always do n!*4^n steps . I’d
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I have found out that my algorithm will always do n!*4^n steps . I’d like to know wheter its complexity will be O(n!*4^n) or will it be something else? Thanks.
If it does exactly
n!*4^nsteps, there’s no real need for big oh notation.And yes, that means it has
O(n!*4^n)complexity.