Min Stack
Design a stack that supports push, pop, top, and retrieving the minimum element in O(1) time.
Input: operations separated by semicolons. Output results of top, getMin operations.
Example:
push,-2;push,0;push,-3;getMin;pop;top;getMin
-3 0 -2
- We start with an empty stack and apply the operations in sequence: push -2, push 0, push -3.
- The
getMinoperation returns the current minimum element, which is -3, so the first output is −3. - We then
popthe top element (-3), and thetopoperation returns the new top element, which is 0. - After the
topoperation, anothergetMinoperation is performed, returning the new minimum element, which is -2.
Constraints:
- -2^31 <= val <= 2^31 - 1
- pop, top, getMin always called on non-empty stack
Background Knowledge
The Min Stack problem involves designing a data structure that supports standard stack operations like push, pop, and top, along with an additional operation to retrieve the minimum element in constant time, O(1). To approach this problem, it's essential to understand the basics of stacks and how they can be implemented using arrays or linked lists. A stack is a Last-In-First-Out (LIFO) data structure, meaning the last element added to the stack will be the first one to be removed.
Key concepts to grasp include the time complexity of various operations on stacks, such as push, pop, and accessing the top element, all of which are typically O(1) for a basic stack implementation. However, finding the minimum element in a standard stack would require iterating through all elements, resulting in a time complexity of O(n). To achieve O(1) time complexity for retrieving the minimum element, we need to consider additional data structures or modifications to the standard stack implementation.
Understanding auxiliary data structures like additional stacks or arrays can also be beneficial. These can help keep track of the minimum element at each step, ensuring that the minimum can be retrieved in constant time. Familiarity with trade-offs between time and space complexity is also important, as optimizing for one often affects the other.
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