Sort Complexities | LeetCode Discuss



Sort Complexities | LeetCode Discuss

Yeah, that's correct!

You can see this by breaking the multiplication down to individual steps:

e^n = e x  e  x  e  x ... x e x e  n!  = n x n-1 x n-2 x ... x 2 x 1

As you can see above, when n grows, n! easily overtakes e^n.


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