155 Min Stack

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Implement the MinStack class:

• 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.

You must implement a solution with O(1) time complexity for each function.

Example 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2

Constraints:

• -231 <= val <= 231 - 1
• Methods pop, top and getMin operations will always be called on non-empty stacks.
• At most 3 * 104 calls will be made to push, pop, top, and getMin.

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class MinStack:

    def __init__(self):
        self.data = []
        self.min_stack = []

    def push(self, val: int) -> None:
        self.data.append(val)
        if not self.min_stack or val < self.min_stack[-1][0]:
            self.min_stack.append([val, 1])
        elif val == self.min_stack[-1][0]:
            self.min_stack[-1][1] += 1
            

    def pop(self) -> None:
        if self.min_stack[-1][0] == self.data[-1]:
            self.min_stack[-1][1] -= 1

        if self.min_stack[-1][1] == 0:
            self.min_stack.pop()
            
        self.data.pop()

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

    def getMin(self) -> int:
        return self.min_stack[-1][0]


# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(val)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()