Journal Archive

2022


var MinStack = function() {
    this.stack = [];
    this.min = null;
};

/** 
 * @param {number} val
 * @return {void}
 */
MinStack.prototype.push = function(val) {
    this.stack[this.stack.length] = val;
};

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

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

/**
 * @return {number}
 */
MinStack.prototype.getMin = function() {
    return Math.min(...this.stack)
};

/** 
 * 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()
 */

Day 62: Solving one of LeetCode problems

155. Min Stack Difficulty - 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