From 0 to 1 Data Structures & Algorithms in Java - Find The Minimum Element In A Stack In Constant Time

The video tutorial explains how to find the minimum element in a stack in constant time using a dual stack approach. It begins by discussing the limitations of traditional methods, which require iterating through the stack, and introduces a more efficient solution using two stacks: one for the actual elements and another for tracking the minimum elements. The tutorial includes a visual demonstration and a detailed code implementation, ensuring that the minimum element can be accessed in constant time regardless of stack operations.

What is the main challenge in finding the minimum element in a stack using traditional methods?

• It requires additional space.

• It takes constant time.

• It requires iterating through the entire stack.

• It can only be done with integers.

How does the dual stack approach help in maintaining the minimum element?

• By using two stacks, one for elements and one for minimums.

• By using a single stack to store all elements.

• By using a queue instead of a stack.

• By storing only the maximum elements.

What happens when a new element is pushed onto the stack in the dual stack method?

• The new element is always added to both stacks.

• The new element is added only if it is larger than the current minimum.

• The new element is compared with the current minimum and added accordingly.

• The new element replaces the current minimum.

In the visualization example, what happens when the number 3 is added to the stack?

• It replaces the top element of the min stack.

• It is added only to the original stack.

• It is ignored as it is not the minimum.

• It becomes the new minimum and is added to the min stack.

What is the role of the min stack during the pop operation?

• It stores the popped elements.

• It remains unchanged.

• It is used to find the maximum element.

• It is popped along with the original stack to update the minimum.

What is the key feature of the get minimum operation in the minimum stack implementation?

• It is an order of N operation.

• It requires additional space.

• It requires iterating through the stack.

• It is an order of 1 operation.

How does the minimum stack ensure that it always has the correct number of elements?

• By removing elements only when the stack is full.

• By adding elements only when the stack is empty.

• By ensuring both stacks have the same number of elements.

• By using a single stack for all operations.