LeetCode-in-All

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.

Solution

class MinStack {
    stack: number[]
    minStack: number[]

    constructor() {
        this.stack = []
        this.minStack = []
    }

    push(val: number): void {
        this.stack.push(val)
        if (this.minStack.length === 0 || this.minStack[this.minStack.length - 1] >= val) {
            this.minStack.push(val)
        }
    }

    pop(): void {
        const popValue = this.stack.pop()
        if (popValue === this.minStack[this.minStack.length - 1]) {
            this.minStack.pop()
        }
    }

    top(): number {
        return this.stack[this.stack.length - 1]
    }

    getMin(): number {
        return this.minStack[this.minStack.length - 1]
    }
}

/*
 * 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()
 */

export { MinStack }

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