# Interview Prep Day 3

Dec 05, 2022
#dsa#stacks

## Day 3

These stack problems were actually difficult for me to comprehend first try. Below is my implementation of NeetCode's solution.

## Stacks

The first solution of each problem is my attempt at solving with just intuition. It's likely not the most optimal solution. The second solution is what I learned from NeetCode's solution and then implementing it myself.

### LeetCode 20. Valid Parentheses

#### Stack + Hashmap Solution --> O(n) time (for loop)

//  @param {string} s
//  @return {boolean}
var isValid = function (s) {
if (s.length % 2 !== 0) return false;
let stack = [];
let map = {
")": "(",
"]": "[",
"}": "{",
};
for (const c of s) {
if (c in map) {
if (stack.length > 0 && stack.at(-1) === map[c]) {
stack.pop();
} else {
return false;
}
} else {
stack.push(c);
}
}
return stack.length == 0;
};
• First we check to see if the length of the string is odd. If it's odd, we return false because there can't be closed pairs if the length is odd.
• Create our stack and a map for each closing parentheses and their corresponding open parenthesis.
• Our stack will hold all of the open parentheses.
• Loop through each character of the string.
• If the character is in map, it means it is a closing parenthesis
• Size of stack must be > 0 because it would be ), which isn't valid.
• We also need to check that the top value in the stack (most recent) is a corresponding open parenthesis.
• Ex: c = ), top value of the stack must be (, which is map[c] --> map["("].
• If both conditions are true, then we can remove the open parenthesis from the stack.
• If both are false, then there won't be matching pairs --> return false
• If the character is not in map, it is an open parenthesis. We push this to the top of the stack.
• At the end, if the stack size is equal to 0, this means all the open parentheses had corresponding closing parentheses, we can return true. Otherwise, there weren't enough pairs, so the expression would return false.

This was a tough one 😅! Took me rereading a bunch of times and rewatching the solution video to really understand the algorithm.

### LeetCode 155. Min Stack

#### Intuitive Solution --> O(n) time (spread operator (...))

let MinStack = class {
constructor() {
this.stack = [];
}
};

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

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

//  @return {number}
MinStack.prototype.top = function () {
return this.stack.at(-1);
};

//  @return {number}
MinStack.prototype.getMin = function () {
return Math.min(...this.stack);
};
• Initialized a stack in the constructor
• Used the built in functions for push() and pop(), which should be O(1) time complexity.
• at(-1) gives the last element of the stack
• Math.min() returns the minimum value of a bunch of provided values. Using the ... spread operator is actually O(n), which isn't optimal. Each function is supposed to be O(1), which I did not accomplish. Let's see what NeetCode's solution is!

#### Two Stacks Solution --> O(1) time (normal stack operations)

let MinStack = class {
constructor() {
this.stack = [];
this.minStack = [];
}
};

//  @param {number} val
//  @return {void}
MinStack.prototype.push = function (val) {
this.stack.push(val);
if (this.minStack.length == 0 || this.minStack.at(-1) >= val) {
this.minStack.push(val);
}
};

//  @return {void}
MinStack.prototype.pop = function () {
let topOfStack = this.stack.pop();
let minVal = this.minStack.at(-1);
if (topOfStack == minVal) {
this.minStack.pop();
}
};

//  @return {number}
MinStack.prototype.top = function () {
return this.stack.at(-1);
};

//  @return {number}
MinStack.prototype.getMin = function () {
return this.minStack.at(-1);
};
• Initialize two stacks. First stack will be the general stack we return. The second stack, minStack, will keep track of the minimum values that are pushed to the stack.
• The top value of minStack will always be the minimum value.
• When we push, we check two conditions for minStack. If either are true, we will also push val to minStack.
• If minStack has no elements yet, we can push val to be the first minimum value.
• If the top element of minStack is greater than or equal to val, we can push val because it's less than or equal to.
• We push val even if it may equal because there may be two of the same values, and one of them might be removed.
• Example: We push two 3's into stack but only 3 once in minStack. If we pop() a 3, minStack is empty, even though the minimum value is still 3.
• We can keep track of what element we popped from stack. If this element is equal to the element at the top of minStack, then we also remove the element from minStack.
• For our getMin function, we can return the element at the top of minStack because the topmost element must be the minimum value.