Solution Review: Min Stack
Let's discuss the solution of the challenge posed in the previous lesson.
We'll cover the following...
We'll cover the following...
Solution: A two-stack class
We’ll make a new class consisting of two stacks: mainStack and minimumStack. The class will have the following functions:
Push()Pop()Min()
Let’s see what each function will do:
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Push: Push an element to the top ofmainStack. It compares the new value with the value at the top ofminimumStack. If the new value is smaller, then we push the new value intominimumStack. Otherwise, we ...