LeetCode 155. Min Stack

Question

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.

Source: https://leetcode.com/problems/min-stack/

Solution

We use a monotonic stack to maintain potential min elements in the future. Monotonic data structures are very effective in solving max/min problems.

This solution can be further optimized. For example, we can replace mins with an object stack that contains <value, times>, so that we can save space when there are many mins with the same value.

public class MinStack {
    private Deque<Integer> stack;
    // stack of candidate minimums
    // only keep candidates that can be the min after a series of operations
    private Deque<Integer> mins;

    public MinStack() {
        this.stack = new ArrayDeque<>();
        this.mins = new ArrayDeque<>();
    }

    public void push(int val) {
        stack.push(val);
        if (mins.isEmpty()) {
            mins.push(val);
        } else if (val <= mins.peek()) {
            mins.push(val);
        }
    }

    // @pre: stack is not empty
    public void pop() {
        if (stack.isEmpty()) {
            return;
        }
        int val = stack.pop();
        if (val == mins.peek()) {
            mins.pop();
        }
    }

    // @pre: stack is not empty
    public int top() {
        return stack.peek().intValue();
    }

    // @pre: stack is not empty
    public int getMin() {
        return mins.peek();
    }
}