LeetCode 155. Min Stack

Description

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:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.

Explanation

We can use two stacks to build a MinStack data structure. One of the stacks is used as a supporting stack to store minimum values in another stack. Whenever we push new element, we push minimum values to the supporting stack.

Java Solution

class MinStack {
    Stack<Integer> stack1;
    Stack<Integer> stack2;
    
    /** initialize your data structure here. */
    public MinStack() {
        stack1 = new Stack<>();
        stack2 = new Stack<>();
    }
    
    public void push(int x) {
        stack1.push(x);
        if (stack2.isEmpty()) {
            stack2.push(x);
        } else {
            stack2.push(Math.min(stack2.peek(), x));
        }
    }
    
    public void pop() {
        stack1.pop();
        stack2.pop();
    }
    
    public int top() {
        return stack1.peek();
    }
    
    public int getMin() {
        return stack2.peek();
    }
}

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack obj = new MinStack();
 * obj.push(x);
 * obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.getMin();
 */

Python Solution

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack1 = []
        self.stack2 = []

    def push(self, x: int) -> None:
        
        self.stack1.append(x)
        
        if not self.stack2:
            self.stack2.append(x)
        else:
            self.stack2.append(min(self.stack2[-1], x))
            
            
    def pop(self) -> None:
        self.stack1.pop()
        self.stack2.pop()

    def top(self) -> int:
        return self.stack1[-1]

    def getMin(self) -> int:
        return self.stack2[-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()

• Time Complexity: ~1
• Space Complexity: ~N