Problem Statement
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
- push(x) -- Push element x onto stack.
- pop() -- Removes the element on top of the stack.
- top() -- Get the top element.
- getMin() -- Retrieve the minimum element in the stack.
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:
- Methods
pop,topandgetMinoperations will always be called on non-empty stacks.
Solution:
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
def push(self, x: int) -> None:
self.stack.append(x)
def pop(self) -> None:
self.stack.pop()
def top(self) -> int:
return self.stack[-1]
def getMin(self) -> int:
return min(self.stack)
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()