155. Min Stack

Medium

Problem Description

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

Implement the MinStack class:

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

Examples

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
Jump to Solution: Python Java C++ JavaScript C#

Python Solution


class MinStack:
    def __init__(self):
        self.stack = []
        self.min_stack = []

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

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

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

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

Java Solution


class MinStack {
    private Stack stack;
    private Stack minStack;
    
    public MinStack() {
        stack = new Stack<>();
        minStack = new Stack<>();
    }
    
    public void push(int val) {
        stack.push(val);
        if (minStack.isEmpty() || val <= minStack.peek()) {
            minStack.push(val);
        }
    }
    
    public void pop() {
        if (stack.peek().equals(minStack.peek())) {
            minStack.pop();
        }
        stack.pop();
    }
    
    public int top() {
        return stack.peek();
    }
    
    public int getMin() {
        return minStack.peek();
    }
}

C++ Solution


class MinStack {
private:
    stack stack;
    stack minStack;
    
public:
    MinStack() {
        
    }
    
    void push(int val) {
        stack.push(val);
        if (minStack.empty() || val <= minStack.top()) {
            minStack.push(val);
        }
    }
    
    void pop() {
        if (stack.top() == minStack.top()) {
            minStack.pop();
        }
        stack.pop();
    }
    
    int top() {
        return stack.top();
    }
    
    int getMin() {
        return minStack.top();
    }
};

JavaScript Solution


/**
 * initialize your data structure here.
 */
var MinStack = function() {
    this.stack = [];
    this.minStack = [];
};

/** 
 * @param {number} val
 * @return {void}
 */
MinStack.prototype.push = function(val) {
    this.stack.push(val);
    if (this.minStack.length === 0 || val <= this.minStack[this.minStack.length - 1]) {
        this.minStack.push(val);
    }
};

/**
 * @return {void}
 */
MinStack.prototype.pop = function() {
    if (this.stack[this.stack.length - 1] === this.minStack[this.minStack.length - 1]) {
        this.minStack.pop();
    }
    this.stack.pop();
};

/**
 * @return {number}
 */
MinStack.prototype.top = function() {
    return this.stack[this.stack.length - 1];
};

/**
 * @return {number}
 */
MinStack.prototype.getMin = function() {
    return this.minStack[this.minStack.length - 1];
};

C# Solution


public class MinStack {
    private Stack stack;
    private Stack minStack;
    
    public MinStack() {
        stack = new Stack();
        minStack = new Stack();
    }
    
    public void Push(int val) {
        stack.Push(val);
        if (minStack.Count == 0 || val <= minStack.Peek()) {
            minStack.Push(val);
        }
    }
    
    public void Pop() {
        if (stack.Peek() == minStack.Peek()) {
            minStack.Pop();
        }
        stack.Pop();
    }
    
    public int Top() {
        return stack.Peek();
    }
    
    public int GetMin() {
        return minStack.Peek();
    }
}

Complexity Analysis

Solution Explanation

This solution uses two stacks to maintain minimum element:

Key points: