ds-algo-study/DS-n-Algos/Misc/whiteboarding/whiteboarding-solutions.js

Just experimenting with publishing a package

// PROBLEM 1 - Two Sum
// https://leetcode.com/problems/two-sum/
// ============================================================================
// Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
// You may assume that each input would have exactly one solution, and you may not use the same element twice.
// You can return the answer in any order.

// Example 1:
// Input: nums = [2,7,11,15], target = 9
// Output: [0,1]
// Output: Because nums[0] + nums[1] == 9, we return [0, 1].

// Example 2:
// Input: nums = [3,2,4], target = 6
// Output: [1,2]

// Example 3:
// Input: nums = [3,3], target = 6
// Output: [0,1]

// Constraints:
// 2 <= nums.length <= 105
// -109 <= nums[i] <= 109
// -109 <= target <= 109
// Only one valid answer exists.

// Time: O(n)
// // We are only passing through the input array one time at most
const twoSum = function(nums, target) {
	const map = {};
	for (let i = 0; i < nums.length; i++) {
		// The complement is the value that would add up with this element to our target
		const complement = target - nums[i];
		// If we already encountered our complement, return the index of that value
		// and this one, ie the pair that sums to the target.
		if (map[complement] !== undefined) {
			return [ map[complement], i ];
		}
		// If we didn't already encounter the complement, add our value and index to
		// the hash. This will allow us to find the pair if we encounter it later on
		map[nums[i]] = i;
	}
	// If we made it all the way through the for loop, the function returns undefined
};

// PROBLEM 2 - Reverse a linked list
// ============================================================================
// Amazon and Microsoft ask you to show an algorithm that will reverse a singly-linked
// list, that is, a list that is made of nodes between which there is a unidirectional
// association as in the following image.

function reverseList(list) {
	// A list of length <= 1 is the same reversed
	if (list.length <= 1) return list;

	// Our currentNode starts off at the head. As we iterate we want to reverse the
	// order, so the node that is next in the original order will have to be swapped.
	// We'll do that by assigning that nextNode's `next` to be our currentNode.
	let currentNode = list.head;
	let nextNode = currentNode.next;

	// We reverse our pointers for our `tail` and `head` properties.
	list.head = list.tail;
	list.tail = currentNode;

	// Since our currentNode is now the tail, its `next` should be null.
	currentNode.next = null;

	// We loop until we no longer have a nextNode. This will happen after assigning
	// our original tail node to point to what was originally before it in the list.
	// When we reassign our nextNode at the end of the loop to tempNode, it will be
	// null, exiting out of our loop.
	while (nextNode) {
		let tempNode = nextNode.next;
		nextNode.next = currentNode;
		currentNode = nextNode;
		nextNode = tempNode;
	}

	// After our loop, all of our `next` pointers have been reversed. Our `head` and
	// `tail` properties have already been reassigned and our new tail's `next` already
	// points to null. We're ready to return the list, which has been reversed.
	return list;
}
// Our function runs in O(n) time, since we only have to iterate through the list once.
// We also take up O(1) space. We only create three new letiables no matter the
// size of the list and do all of our reversing in place.

// PROBLEM 3 - The missing value
// ============================================================================
// Amazon and Microsoft ask you to derive an algorithm that will inspect an
// array of of numbers that contains the values between 0 and the length of the
// list, inclusive, and find the missing value. For example, you may be given an
// array that of length 6 that contains
// [0, 2, 3, 4, 5, 6]
// It is your job to determine that the missing value from the array is 1.

// O(n)
function findNum(arr) {
	let hash = {};

	// Loop through our input array and create a mapping of the value to true
	// This will cost us n iterations up front, but allow us to access the numbers
	// in constant time for future lookup.
	// O(n)
	for (let i = 0; i < arr.length; i++) {
		hash[arr[i]] = true;
	}

	let j = 0;
	// Increment from 0 to arr.length, which is all of the potential values that we
	// would expect in our array.
	// When we get to a value that was not found in our first loop, we return that
	// missing number.
	// O(n)
	while (j < arr.length) {
		if (!hash[j]) {
			return j;
		}
		j++;
	}
	// If we made it beyond the while loop, we found all numbers 0 to length - 1,
	// meaning we are missing the last value of arr.length
	// (This also could have been accomplished by looping up to and including the
	// length instead of being exclusive in the conditional.)
	return arr.length;
}

// PROBLEM 4 - Stack min
// ============================================================================
// Google and Apple ask you to design a stack that, in addition to the push and
// pop functions, has a function min that returns the minimum element in the stack
// without removing it. All three functions push, pop, and min should operate in
// O(1) time.

class minStack {
	// We keep track of a separate array that will only get elements added to it if
	// they represent a new minimum value (or tie for the current value)
	constructor() {
		this.elements = [];
		this.mins = [];
	}

	push(val) {
		this.elements.push(val);
		const curMin = this.mins[this.mins.length - 1];
		// If we don't have a minimum yet (first element pushed), or if this value is
		// a new minimum (or equal value), we also push it into the mins array.
		if (this.mins.length === 0 || val <= curMin) {
			this.mins.push(val);
		}
	}

	pop() {
		const popped = this.elements.pop();
		const curMin = this.mins[this.mins.length - 1];
		// If the element we just removed was also the current minimum, we remove it
		// from our mins array. The element at the end of our mins array will always
		// represent our current minimum.
		if (popped === curMin) this.mins.pop();
		return popped;
	}

	min() {
		// Since the element at the end of mins is always our minimum, we can key
		// directly in to see that value no matter where it is in the stack.
		return this.mins[this.mins.length - 1];
	}
}

// PROBLEM 5 - Test a retractable ballpoint pen
// ============================================================================

// Facebook asks you to write the tests cases for testing a ballpoint pen.
// What would you consider to be good tests for the pen?
// Try to be as exhaustive as possible.

// Basic Functionality
// The pen has a clicking/retracting functionality
// Clicking the pen from a retracted position extends the pen
// Clicking the pen from an extended position retracts the pen
// You are able to write from an extended position
// You are not able to write from a retracted position
// Additional Desired Functionality
// Writing large amounts results in smooth lettering (no missing letters, lines, etc.)
// Ink does not fade while writing (coloring is consisitent, compare beginning to end)
// Ink does not smudge (test interacting with lettering and compare after interaction)
// Ink is waterproof
// The pen can write on desired surfaces (glossy, smooth, rough, etc.)
// The pen lasts for at least X pages (test a new pen and make sure output is possible for X pages)
// etc.

// PROBLEM 6 - OOParking Lot
// ============================================================================

// Amazon and Microsoft ask you to specify the classes that it would take to write
// software to manage a paid parking lot. It should know where cars are parked,
// be able to identify the cars, know where the keys are hanging, how many cars
// are in the lot, what time the cars come and go, and how much it costs someone
// when they leave the parking lot based on the following schedule:

// Time	       -> Rate per hour
// 8pm - 6am	    $3
// 6am - noon	    $10
// noon - 6pm	    $8
// 6pm - 8pm	    $6

// This is mostly pseudocode
// The important aspects are being able to define the roles that you have
// Possible changes:
// // Using a Car class instead of a plain object to fill spaces
// // Account for minimum charges or a max daily rate when calculating fees
class ParkingLot {
	constructor(numSpaces) {
		// track the spaces in the lot
		// track the number of vacancies
		this.spaces = new Array(numSpaces).fill(null);
		this.vacancies = numSpaces;
		this.nextAvailableSpace = 0;
	}

	fillSpot(plateNumber, keyLocation) {
		// return false if !this.spotAvailable()
		// decrement vacancies
		// Add in to this.spaces[this.nextAvailableSpace] { entryTime: Date.now(), plateNumber, keyLocation }
		// check this.spaces at this.nextAvailableSpace+1, incrementing until a vacancy is found. Reassign this.nextAvailableSpace
	}

	spotAvailable() {
		// return if vacancies > 0
	}

	calculateFee(spaceNumber) {
		// Get vehicle info with this.spaces[spaceNumber]
		// Take time of entry (vehicle.entryTime) and time of exit (Date.now()) on object
		// calculate based on time in each fee window
		const { entryTime } = this.spaces[spaceNumber];
		const exitTime = Date.now();
		let totalFee = 0;
		if (entryTime.getHours() < 6) {
			const hours = 6 - entryTime.getHours() - 1;
			const fraction = (60 - getMinutes()) / 60; // add fraction of hour
			totalFee += (hours + fraction) * 3; // multiply by rate
		} // etc.
	}

	removeVehicle(spaceNumber) {
		// calculateFee
		// increment vacancies
		// reassign this.spaces[spaceNumber] = null
	}

	inspectSpace(spaceNumber) {
		// return the vehicle at this.spaces[spaceNumber];
	}

	findByPlate(plateNumber) {
		// iterate over this.spaces
		// if a vehicle exists, compare the plateNumber argument to the object
		// if it is a match, return the spaceNumber (index of this.spaces where match exists)
	}
}

// PROBLEM 7 - Valid Parentheses
// https://leetcode.com/problems/valid-parentheses/
// ============================================================================

// Given a string s containing just the characters '(', ')', '{', '}', '[' and ']',
// determine if the input string is valid.

// An input string is valid if:

// Open brackets must be closed by the same type of brackets.
// Open brackets must be closed in the correct order.

const isValid = function(s) {
	const stack = [];
	const openings = { '(': ')', '{': '}', '[': ']' };
	const closings = { ')': '(', '}': '{', ']': '[' };
	for (let i = 0; i < s.length; i++) {
		// If the character is an opening, we push it on to the stack
		if (openings[s[i]]) {
			stack.push(s[i]);
			// If it is a closing, it has to match the top opening on the stack
			// If it does, pop the opening off of the stack
		} else if (closings[s[i]] === stack[stack.length - 1]) {
			stack.pop();
			// The only way we get to the else is if it is a closing that does not match
			// In this case, we return false immediately
		} else {
			return false;
		}
	}
	// If we made it out of the for loop without invalid closings, we just have to
	// make sure all of our openings were closed.
	// This occurs when our stack is empty, indicating no unmatched characters.
	return stack.length === 0;
};