Min Stack

Question

Design a stack class that supports the push, pop, top, and getMin operations.

• MinStack() initializes the stack object.
• void push(int val) pushes the element val onto the stack.
• void pop() removes the element on the top of the stack.
• int top() gets the top element of the stack.
• int getMin() retrieves the minimum element in the stack.

Each function should run in O(1)O(1) time.

Example 1:

Input: ["MinStack", "push", 1, "push", 2, "push", 0, "getMin", "pop", "top", "getMin"]

Output: [null,null,null,null,0,null,2,1]

Explanation:
MinStack minStack = new MinStack();
minStack.push(1);
minStack.push(2);
minStack.push(0);
minStack.getMin(); // return 0
minStack.pop();
minStack.top();    // return 2
minStack.getMin(); // return 1

Constraints:

• -2^31 <= val <= 2^31 - 1.
• pop, top and getMin will always be called on non-empty stacks.

Solution

Two Stack

Keep both the minStack has the min value up to the current item in the stack. If we pop we pop both

class MinStack:

    def __init__(self):
        self.minStack = []
        self.stack = []

    def push(self, val: int) -> None:
        self.stack.append(val)
        self.minStack.append(min(val, self.minStack[-1] if self.minStack else val))

    def pop(self) -> None:
        self.stack.pop()
        self.minStack.pop()

    def top(self) -> int:
        return self.stack[-1]       

    def getMin(self) -> int:
        return self.minStack[-1]

One stack

Group the (val, minValue) together

class MinStack:

    def __init__(self):
        self.stack = []

    def push(self, val: int) -> None:
        if not self.stack:
            self.stack.append((val, val))
        else:
            _, prevMinValue = self.stack[-1]
            self.stack.append((val, min(prevMinValue, val)))

    def pop(self) -> None:
        self.stack.pop()

    def top(self) -> int:
        return self.stack[-1][0]

    def getMin(self) -> int:
        return self.stack[-1][1]