155

155. Min Stack (Easy)

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.

 

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.

Companies:
Amazon, Microsoft, Bloomberg, Facebook, VMware, Apple, Yahoo, Goldman Sachs, Airtel

Related Topics:
Stack, Design

Similar Questions:

• Sliding Window Maximum (Hard)
• Max Stack (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/min-stack/
// Author: github.com/lzl124631x
// Time: O(1) for all
// Space: O(N)
class MinStack {
    stack<int> mn, val;
public:
    MinStack() {}
    void push(int x) {
        val.push(x);
        if (mn.empty() || mn.top() >= x) mn.push(x);
    }
    void pop() {
        if (val.top() == mn.top()) mn.pop();
        val.pop();
    }
    int top() {
        return val.top();
    }
    int getMin() {
        return mn.top();
    }
};