📌Problem
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 -2Constraints:
- 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:
if self.__stack:
self.__stack.append((x,min(x,self.__stack[-1][1]))) # (x,minimum)
else:
self.__stack.append((x,x))
def pop(self) -> None:
if self.__stack:
self.__stack.pop()
def top(self) -> int:
if self.__stack:
return self.__stack[-1][0]
def getMin(self) -> int:
return self.__stack[-1][1]
# 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()
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