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155. Min Stack

Solve in Leetcode

Description

Static Badge

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.

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

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.

Solution: Singly Linked List

Each node stores its value and the current minimum at the time it was pushed. The head of the list is always the top of the stack.

  • push(val) — Create a new node with value = min(val, current min) and prepend it to the list.
  • pop() — Move the head pointer to head.next.
  • top() — Return head.key.
  • getMin() — Return head.value (the tracked minimum).

Since the minimum is baked into every node at push time, removing the top always reveals the correct minimum for the remaining stack — no auxiliary structure needed.

  • Time Complexity: O(1) for all operations
  • Space Complexity: O(n) where n is the number of elements in the stack — each push allocates one node
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interface SListNode {
    key: number;
    value: number; // store min value
    next?: SListNode;
}

class MinStack {
    // Use singly linked list
    private dummyHead?: SListNode;

    constructor() {
        // Initialize with undefined - no dummy head yet
    }

    push(val: number): void {
        if (!this.dummyHead) {
            this.dummyHead = {
                key: val,
                value: val
            }
        } else {
            this.dummyHead = {
                key: val,
                value: Math.min(val, this.dummyHead.value),
                next: this.dummyHead
            }
        }
    }

    pop(): void {
        this.dummyHead = this.dummyHead!.next
    }

    top(): number {
        return this.dummyHead!.key
    }

    getMin(): number {
        return this.dummyHead!.value
    }
}

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