· 4 Min read

My notes on Algorithms

Big 0 Notation

It is a metric for determining algorithm’s efficiency. It estimates how long it takes your code to run on different sets of input.

Where is the metric used

Time Complexity

How long does it take to execute in terms of computational steps.

Space Complexity

It measures the amount of memory your code uses.

Big-0 ⇒ It is the worst case scenario of a code (like maximum time or maximum memory )

How do we measure if a code is good or bad ?

Time ? Nah it can change from device to device and it is not consistent

Counting Operations ? Yes

Example ⇒ Find sum of n numbers

  1. Direct Formula
function sumOfNumbers(n){ 
	return n*(n+1)/2;
addition = 1 operation
multiplication = 1 operation
division = 1 operation
Total = 3 operations => O(1)
  1. For Loop
function sumOfNumbers(n){ 
	let total = 0;
	for (let i=0; i<=n ; i++){
		total = total + i;
	return total
total assignment to 0 = 1 operation
total + i = 1 operation * n times
total = sum = 1 operation * n times 
Total = 1 + n + n = 2n + 1 => O(n)
  • Constants doesn’t matter

$$ O(2n) ⇒ O(n) $$

$$ O(5n^2 + 3) => O(n^2) $$

  • Only care about the trends and not the detail

Big 0 Shorthands

  1. Arithmetic operations are constant
  2. Variable assignment is constant
  3. Accessing elements in an array(by index) or object(by key) is constant
  4. In a loop, the complexity is the length of the loop times the complexity of whatever happens inside of the loop


Big O Cheatsheet

Space Complexity

It refers to the total amount of memory space used by an algorithm/program, including the space of input values for execution.

Rules of thumb

  • Most primitives(booleans, numbers, undefined, null) are constant space
  • String requires O(n) space (where n is the string length)
  • Reference types are generally O(n), where n is the length(for arrays) or the number of keys (for objects)
function sumOfNumbers(n){ 
	let total = 0;
	for (let i=0; i<=n ; i++){
		total = total + i;
	return total
total declaration 1
for loop i declaration 1
Total = 2 => O(1)
function double(arr){
	let newArr = [];
	for(let i = 0 ; i < arr.length ; i++){
		newArr.push(2 * arr[i]);
	return newArr
newArr declaration O(n)
for loop i declaration 1
Total = O(n) + 1 => O(n)


It is inverse of exponent

$$ log_2(8) = 3 $$

$$ log_2(value) = e => 2^e = value

$$ log_2(n) === log_9(n) === log(n) $$

Analysing Performance


Insertion - O(1)

Removal - O(1)

Searching - O(n)

Access - O(1)

Object Methods

Object.keys - O(n)

Object.values - O(n)

Object.entries - O(n)

Object.hasOwnProperty - O(1)

When to use objects ?

  • When you don’t need order.
  • When you need fast access insertion and removal


Insertion - It depends

Removal - It depends

Searching - O(n)

Access - O(1)

Array Methods

Array.push & Array.pop - O(1)

Array.shift & Array.unshift - O(n) - We have to reindex every element

Array.concat - O(n) […arr1, …arr2] indexes of arr2 needs to be reindexed

Array.slice - O(n) Returns shallow copy of a portion of an array

Array.splice - O(n) Changes an array by removing existing elements and/or adding new elements

Array.sort - O(n x logn)

forEach/map/filter/reduce - O(n)

Problem Solving Approach

What is an Algorithm?

A process or set of steps to accomplish a certain task.

How do you improve?

  1. Devise a plan for solving problems
  2. Master common problem solving patterns

Problem Solving

  • Understand the problem
  • Explore concrete examples
  • Break it down
  • Solve/Simplify
  • Look back and refactor
  1. Understand the problem
    • Can i reinstate the problem in my own words
    • What are the inputs
    • What are the outputs that come from the solution
    • Can the outputs be determined from inputs. In other words do i have enough info to solve the problem?
  2. Exploring concrete examples
    • Start with some examples
    • Progress to more complex examples
    • Explore examples with Empty Inputs
    • Explore examples with Invalid inputs
  3. Break it down
    • Explicitly write out the steps you need to take
  4. Simplify
    • Find the core difficulty in what you’re trying to do
    • Temporarily ignore that difficulty
    • Write a simplified solution
    • Then incorporate that difficulty back in
  5. Look back and refactor
    • Can you check the result ?
    • Can you derive the result differently ?
    • Can you understand it at a glance?
    • Can you use the result or method for some other problem?
    • Can you improve the performance of your solution?
    • Can you think of other ways to refactor?
    • How have other people solved this problem

Problem solving patterns

  1. Frequency Counters

    This patterns uses objects or sets to collect values/frequencies of values

    This can of tern avoid the need for nested loops or O(N^2) operations with arrays / strings

    • Squared Array comparison problem

      A naive solution - Time Complexity O(n^2)

      function same(arr1, arr2){
      	if(arr1.length !== arr2.length){
      		return false
      	for (let i = 0;i<arr1.length;i++){
      		let correctIndex = arr2.indexOf(arr1[i] ** 2)
      		if(correctIndex === -1){
      			return false;

      A slight better solution - Time Complexity O(n*logn)

      function same(arr1, arr2) {
        if (arr1.length !== arr2.length) {
          return false;
        arr1 = arr1.map(val => val ** 2).sort();
        arr2 = arr2.sort();
        for (let i = 0; i < arr1.length; i++) {
          if (arr1[i] !== arr2[i]) {
            return false;
        return true;

      Frequency Counter Pattern - Time Complexity O(n)

      function same(arr1, arr2) {
        if (arr1.length !== arr2.length) {
          return false;
        let frequencyCounter1 = {};
        let frequencyCounter2 = {};
        for (let val of arr1) {
          frequencyCounter1[val] = (frequencyCounter1[val] || 0) + 1;
        for (let val of arr2) {
          frequencyCounter2[val] = (frequencyCounter2[val] || 0) + 1;
        for (let key in frequencyCounter1) {
          if (!(key ** 2 in frequencyCounter2)) {
            return false;
          if (frequencyCounter2[key ** 2] !== frequencyCounter1[key]) {
            return false;
        return true;
  2. Multiple Pointers

    Create pointers or values that correspond to an index or position and move towards the beginning, end or middle based on a certain condition. You can use it on some sort of linear structure, like an array, string or a linked list, to search for a pair of values, or searching for something that meets a condition. You use two reference points in the linear structure, and you work toward the middle.

    • Sum Zero problem

      Naive Solution - Time Complexity - O(n^2)

      function sumZero(arr){
      	for(let i=0;i<arr.length;i++){
      		for(let j=i+1;j<arr.length;j++){
      			if(arr[i] === -arr[j] ){
      				return [arr[i], arr[j]]

      Pointer Solution - Time Complexity - O(n)

      function sumZero(arr){
      	let left = 0
      	let right = arr.length - 1
      		let sum = left + right
      			return [arr[left], arr[right]]
      		} else if(sum>0){
      		} else {
  3. Sliding Window

    This pattern involves creating a window which can either be an array or number from one position to another. Depending on certain condition window either increases or closes. Very useful for keeping track of a subset of data in an array/string

    • Max Subarray sum problem

      Naive Solution

      function maxSubArraySum(arr,n){
      	if (num < arr.length){
      		return null
      	let total;
      	for (let i=0;i<arr.length - n + 1;i++){
      		let temp = 0;
      		for(let j = 0 ; j<n ;j++){
      				temp+= arr[i+j]
      		if(temp > total){
      				total = temp;
      	return total

      Sliding Window Solution

      function maxSubarraySum(arr, num){
        let maxSum = 0;
        let tempSum = 0;
        if (arr.length < num) return null;
        for (let i = 0; i < num; i++) {
          maxSum += arr[i];
        tempSum = maxSum;
        for (let i = num; i < arr.length; i++) {
          tempSum = tempSum - arr[i - num] + arr[i];
          maxSum = Math.max(maxSum, tempSum);
        return maxSum;
  4. Divide and Conquer

    This pattern involves dividing a data set into smaller chunks and then repeating a process with a subset of data.

    • Binary Search Problem

      Naive solution

      const search = (arr,val) => {
      	arr.map((item, index) => {
      		if(item === val){
      			return index
      	return -1

      Divide and conquer

      const search = (arr, val) => {
      	let min = 0
      	let max = arr.length - 1
      	while(min <= max){
      		let middle = Math.floor((min+max)/2);
      		let currentElement = array[middle];
      		if(array[middle] < val){
      			min = middle + 1
      		else if (array[middle] > val){
      			max= middle - 1
      		} else {
      				return middle;


Recursion is a programming technique where a function calls itself to solve smaller instance of the same problem until it reaches a base case, which is a condition which stops recursion.

  • The Dragon and the Nested Caves Story

    Recursion is a programming technique where a function calls itself to solve smaller instances of the same problem until it reaches a base case, which is a condition that stops the recursion. To explain recursion, let's use a story involving a dragon.

    The Dragon and the Nested Caves

    Imagine a dragon that has hidden its treasure in the deepest chamber of a series of nested caves. Each cave leads to another smaller cave, and this sequence continues until the smallest, innermost cave, which contains the treasure.

    To find the treasure, you must enter the first cave, then the next cave within that, and so on, until you reach the last cave with the treasure. However, the path to the treasure is the same process repeated at each cave: enter the cave, look for the treasure or another cave, and if there's another cave, enter it and repeat the process.

    Relating to Recursion

    In this story:

    • Each cave represents a recursive call. Just like entering a new cave is the same action repeated, a recursive function calls itself with a smaller portion of the problem.
    • The treasure represents the base case. The base case in recursion is what stops the recursion from continuing indefinitely. Finding the treasure means there are no more caves to enter, just as reaching the base case means no more recursive calls are needed.
    • The process of checking for treasure or another cave is the recursive condition. In a recursive function, there's usually a condition that checks whether it should make another recursive call (enter another cave) or return a value (find the treasure).
    function findTreasure(cave) {
        if (cave.containsTreasure()) {
            // Base case: Treasure found, stop recursion
            return "Treasure"
        } else {
            // Recursive case: Enter the next cave
            return findTreasure(cave.nextCave())

    Understanding Recursion Through the Story

    The dragon story helps illustrate how recursion breaks down a problem (finding the treasure) into smaller instances of the same problem (entering the next cave to look for the treasure) until a stopping condition is met (the treasure is found). Each step brings you closer to the goal, just as each recursive call brings the function closer to the base case.

Simple Count Down

function countDown(number){
	for(let i=number; i > 0 ; i++){
	console.log('Coundown completed')

Recursive Count Down

function countDown(number){
		if(number <= 0){
			console.log('Count Down Completed')
  • Factorial


    function factorial(num){
    	let total = 1
    		total = total * num;
    	return total;


    function factorial(num){
    	let total = 1
    	if(num === 1){
    		return 1;
    	total = num * factorial(num - 1)
    	return total

Avoiding These Pitfalls:

To avoid these common pitfalls, consider the following best practices when writing recursive functions:

  • Clearly Define the Base Case: Ensure that your recursive function has a well-defined base case that is reached by every possible path through the function.
  • Ensure Progress Toward the Base Case: Make sure each recursive call changes the state or input in a way that progresses toward the base case.
  • Limit Recursion Depth: Be mindful of the potential depth of recursion and consider iterative alternatives or algorithms that minimise recursion depth, such as bottom-up dynamic programming, for deep recursion scenarios.
  • Use Memoization: For problems with overlapping subproblems, use memoization or other dynamic programming techniques to store and reuse the results of subproblems, avoiding redundant computations.
  • Be Cautious with Global Variables and Side Effects: Minimise the use of global variables and be aware of side effects within your recursive functions to ensure predictable and correct behavior.
  • Test Thoroughly: Due to the nature of recursion and the variety of paths through the recursive function, thorough testing is crucial to uncover and fix potential issues related to base cases, state management, and edge cases.

Searching Algorithms

Searching algorithms are a fundamental aspect of computer science, designed to retrieve information stored within some data structure. Their primary purpose is to find an item or group of items with specific properties within a dataset. These algorithms vary widely in their mechanisms, efficiency, and the types of data structures they are best suited for.

Linear search is a straightforward method where each element of the array is checked sequentially from the beginning to the end until the target element is found or the array ends. It's simple to implement but not the most efficient, especially for large arrays.

function linearSearch(arr, target) {
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] === target) {
      return i; // Return the index of the target element
  return -1; // Element not found

Best O(1) Worst O(n) Average O (n)

2. Binary Search (sorted*)

Binary search is a much more efficient algorithm but requires the array to be sorted. It works by repeatedly dividing the search interval in half. If the value of the search key is less than the item in the middle of the interval, the search continues in the lower half, or if the value is greater, it continues in the upper half. This process repeats until the target value is found or the interval is empty.

function binarySearch(sortedArr, target) {
  let left = 0;
  let right = sortedArr.length - 1;
  while (left <= right) {
    let mid = Math.floor((left + right) / 2);
    if (sortedArr[mid] === target) {
      return mid; // Target found
    } else if (sortedArr[mid] < target) {
      left = mid + 1; // Continue search in the right half
    } else {
      right = mid - 1; // Continue search in the left half
  return -1; // Target not found

Best O(1) Worst O(log n) Average O(log n)

Sorting Algorithms

Sorting algorithms are methods used in computer science to rearrange a sequence of elements in a given array or list into a specific order, typically in numerical or lexicographical (alphabetical) ascending or descending order.

1. Bubble Sort

Bubble Sort is one of the simplest sorting algorithms. It repeatedly steps through the list, compares adjacent elements, and swaps them if they are in the wrong order. The pass through the list is repeated until the list is sorted. Despite its simplicity, Bubble Sort is highly inefficient for large datasets, with an average and worst-case time complexity of

$$ O(n^2) $$

n is the number of elements. Best case is O(n)

function bubbleSort(arr){
	let sorted = false
		let isSwapped = false
		for (i=0;i<arr.length - 1;i++){
				if(arr[i] > arr[i+1]){
					isSwapped = true;
					[arr[i], arr[i+1]] = [arr[i+1], arr[i]];
		if(!isSwapped) {
			sorted = true
     return arr

2. Selection Sort

Selection Sort divides the input list into two parts: a sorted sublist of items that is built up from left to right at the front of the list, and an unsorted sublist that occupies the rest of the list. The algorithm proceeds by finding the smallest (or largest, depending on sorting order) element in the unsorted sublist, swapping it with the leftmost unsorted element, and moving the sublist boundary one element to the right. Like Bubble Sort, it has a time complexity of

$$ O(n^2) $$

making it inefficient for large lists.

function selectionSort(arr){
	let left = 0
	while(left !== arr.length - 1){
		let min = left
		for (i=left;i<arr.length;i++){
				if(arr[min] > arr[i]){
				    min = i
				[arr[left], arr[min]] = [arr[min], arr[left]]
	return arr

3. Insertion Sort

Insertion Sort builds the final sorted array (or list) one item at a time. It is much less efficient on large lists than more advanced algorithms like quicksort, heapsort, or merge sort. However, insertion sort provides several advantages, such as simple implementation, efficiency for small data sets, and more efficient in practice than most other simple quadratic algorithms like bubble sort or selection sort. It has an average and worst-case time complexity of

$$ O(n^2) $$

function insertionSort(arr) {
    for (let i = 1; i < arr.length; i++) {
        let current = arr[i];
        let j = i - 1;
        // Move elements of arr[0..i-1], that are greater than current,
        // to one position ahead of their current position
        while (j >= 0 && arr[j] > current) {
            arr[j + 1] = arr[j];
        arr[j + 1] = current;
    return arr;

Comparing Bubble Sort, Selection Sort and Insertion Sort

| Algorithm | Time complexity (Best) | Time complexity (Average) | Time complexity (Worst) | Space complexity | | --- | --- | --- | --- | --- | | Bubble Sort | O(n) | O(nn) | O(nn) | O(1) | | Selection Sort | O(n) | O(nn) | O(nn) | O(1) | | Insertion Sort | O(nn) | O(nn) | O(n*n) | O(1) |