[pyfan878]

This pretty standard interview question. You should keep track of the current minimum when you do a push: push 1: [(v=1,m=1)] push -1: [(v=1, m=1), (v=-1, m=-1)] push 3: [(v=1, m=1), (v=-1, m=-1), (v=3, m=-1)] get_min: [(v=1, m=1), (v=-1, m=-1), (v=3, m=-1)] returns m=-1 pop: [(v=1, m=1), (v=-1, m=-1)] returns v=3 get_min: returns -1

[masterof0]

Came here to say this, as far as LC questions go, this one is pretty straight forward. OP, the way you come up with a solution like this is practice and more practice, the same way you can solve an equation you never seen before, because you have a set of tools (patterns, "tricks", knowledge, ...) that allows you to come up with a solution. Keep practicing and you will see the progress.

[d--b]

OP meant O(1) in time and space (forgot the space but it’s in the linked article). That solution is O(N) in space.

[d--b]

problem is when you pop the min, you can't know which was the second to min unless you do the trick that's in the website quoted.

[ahmedtd]

You don't pop the min, you pop the top of the stack.