## DEV Community 👩‍💻👨‍💻 Posted on • Updated on

# Algorithms in JavaScript with visual examples.

Hello Programmers,

Most of us are scared of algorithms, and don't ever start to learn it. But we shouldn't be scared of it. An algorithm is just steps to solve a problem.

Today let's cover major Algorithms in an easy and illustrative manner.

Don't try to memorize them, algorithm is more about problem solving. So, sit with a paper and pen.

The terms in table of content might seem very scary, but just be with me I promise to explain everything in the simplest manner possible.

## Tabel of content:

Before we begin.

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## Understanding Big O Notation

Big O Notation is a way to represent the time and space complexity of an algorithm.

• Time Complexity: Time taken by the algorithm to complete the execution.
• Space Complexity: The memory occupied by the algorithm. There are few expressions (notations) which represents the time complexity of an algorithm.

• O(1): Constant time complexity. This is the ideal case.
• O(log n): Logarithmic time complexity. If `log(n) = x` then it is same as `10^x`
• O(n): Linear time complexity. Time increases with the number of input in a linear manner. For ex, If one input takes 1 ms, 4 inputs will take 4 ms to execute the algo.
• O(n^2): Quadratic time complexity. This mostly happens in case of nested loops.
• O(n!): Factorial time complexity. This is the worst case senario, which should be avoided.

You should try to write your algorithm such that it can be represented by the first 3 notations. And last two should be avoided as often as possible. You want to keep your complexity as low and straight as possible, ideally avoiding anything above O(n).

In further sections of this article you will see examples of each notation. For now this is all you need to know.

## Algorithm

### What is algorithm and why to care?

The way to solve a problem or we can say the steps, procedure, or set of rules to solve a problem is known as Algorithm.

ex: Search Engine algorithm to find out data related to a search string.

As a programmer you will come across many problems that needs to be solved with these algorithms. So, it's better if you already know them.

### Recursion

A function calling itself is recursion. Think of it as an alternative to loop.

``````function recursiveFn() {
console.log("This is a recursive function");
recursiveFn();
}

recursiveFn();
``````

In the above snippet look at line 3 recursiveFn is called in recursiveFn itself. As I mentioned earlier recursion is an alternative to loop.

So, how many times this function is exactly going to run?

Well, this will create an infinite loop, because there is nothing to stop it at any point. Let's say we need to run the loop only 10 times. On the 11th iteration function should return. That will stop the loop.

``````let count = 1;
function recursiveFn() {
console.log(`Recursive \${count}`);
if (count === 10) return;
count++;
recursiveFn();
}

recursiveFn();
``````

In the above snippet line 4 returns and stops the loop at count 10.

Now let's see a more realistic example. Our task is to return an array of odd numbers from a given array. This can be achieved in a number of ways including for-loop, Array.filter method, e.t.c

But to showcase the use of recursion I will use a helperRecursive function.

``````function oddArray(arr) {
let result = [];
function helperRecursiveFn(arr) {
if(arr.length === 0) {
return; // 1
} else if(arr % 2 !== 0) {
result.push(arr); // 2
}
helperRecursiveFn(arr.slice(1)); // 3
}
helperRecursiveFn(arr);
return result;
}

oddArray([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]);
// OutPut -> [1, 3, 5, 7, 9]
``````

Here the recursive function is helperRecursiveFn.

1. Return if the array length is 0.
2. Push the element to result array if the element is odd.
3. Call helperRecursiveFn with first element of array sliced. Every time the first element of array will be sliced, because we have already cheked it for odd or even.

For ex: First time helperRecursiveFn will be called with `[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]`. Next time it will be called with `[2, 3, 4, 5, 6, 7, 8, 9, 10]` and so on untill the array length is 0.

### Linear Search Algorithm

Linear search algorithm is pretty simple. Say that you need to find if a number exists in a given array or not.

You will run a simple for loop and check every element untill you find the one you are looking for.

``````const array = [3, 8, 12, 6, 10, 2];

// Find 10 in the given array.
function checkForN(arr, n) {
for(let i = 0; i < array.length; i++) {
if (n === array[i]) {
return `\${true} \${n} exists at index \${i}`;
}
}

return `\${false} \${n} does not exist in the given array.`;
}

checkForN(array, 10);
``````

That's linear search algorithm. You search for each element in the array one by one in a linear manner. #### Time Complexity of Linear Search Algorithm

There is only one for loop which will run n times. Where n (in worst case) is the length of the given array. Here number of iterations (in worst case) is directly propotional to the input (array of length).

Hence the time complexity for linear search algorithm is Linear Time Complexity: O(n).

### Binary Search Algorithm

In linear search you can eliminate one element at a time. But with binary search algorithm you can eliminate multiple elements at once. That's why binary search is faster than linear search.

The point to be noted here is that binary search works only on sorted array.

This algorithm follows divide and conquer approach. Let find the index of 8 in [2, 3, 6, 8, 10, 12].

Step 1:
Find the middleIndex of the array.

``````const array = [2, 3, 6, 8, 10, 12];
let firstIndex = 0;
let lastIndex = array.length - 1;
let middleIndex = Math.floor((firstIndex + lastIndex) / 2); // middleIndex -> 2
``````

Step 2:
Check if middleIndex element > 8. If so, that means 8 is on the left of middleIndex. Hence, change lastIndex to (middleIndex - 1).

Step 3:
Else if middleIndex element < 8. That means 8 is on the right of middleIndex. Hence, change firstIndex to (middleIndex + 1);

``````if (array[middleIndex] > 8) {
lastIndex = middleIndex - 1;
} else {
firstIndex = middleIndex + 1;
}
``````

Step 4:
With every iteration middleIndex is again set as per the new firstIndex or lastIndex.

Let's see all those steps together in code format.

``````function binarySearch(array, element) {
let firstIndex = 0;
let lastIndex = array.length - 1;
let middleIndex = Math.floor((firstIndex + lastIndex) / 2);

while (array[middleIndex] !== element && firstIndex <= lastIndex) {
if(array[middleIndex] > element) {
lastIndex = middleIndex - 1;
}else {
firstIndex = middleIndex + 1;
}
middleIndex = Math.floor((firstIndex + lastIndex) / 2);
}
return array[middleIndex] === element ? middleIndex : -1;
}

const array = [2, 3, 6, 8, 10, 12];
binarySearch(array, 8); // OutPut -> 3
``````

Here is visual representation of the above code.

``````firstIndex = middleIndex + 1;
``````
``````lastIndex = middleIndex - 1;
``````
``````array[middleIndex] === 8 // Found It
``````

#### Time Complexity of Binary Search

There is only one while loop which will run n times. But here number of iterations does not depend on the input (array length).

Hence the time complexity for binary search algorithm is Logarithmic Time Complexity: O(log n). And you can check the O-notation graph. O(log n) is faster than O(n).

### Naive Search Algorithm

Naive search algorithm is used to find if a string contains a given substring. For example, check if "helloworld" contains the substring "owo".

1. First loop on the main string ("helloworld").
2. Run a nested loop on the substring ("owo").
3. If character does not match then break inner loop else keep looping.
4. If inner loop is completed and got a match, then return true else keep the outer loop going.

Here is a visual representation. Here is the implementation in code.

``````function naiveSearch(mainStr, subStr) {
if (subStr.length > mainStr.length) return false;

for(let i = 0; i < mainStr.length; i++) {
for(let j = 0; j < subStr.length; j++) {
if(mainStr[i + j] !== subStr[j]) break;
if(j === subStr.length - 1) return true;
}
}
return false;
}
``````

Now, let's try to understand the code above.

• At line 2, return false if subString length is greater than mainString length.
• At line 4, start looping on mainString.
• At line 5, start nested loop on subString.
• At line 6, break inner loop if no match is found, And move on to next iteration for outer loop.
• At line 7, return true at last iteration of inner loop.

#### Time Complexity of Naive Search

There is a loop inside a loop (Nested Loop). Both loop run n times. Hence time complexity for naive search algo is (n * n) Quadratic Time Complexity: O(n^2).

And as discussed at top, any time complexity above O(n) should be avoided if possible. We will see a better approach with less time complexity in next algo.

### KMP Algorithm

KMP algo is a pattern recongnition algorithm, and it is a bit tough to understand. Ok, let's try to find if the string "abcabcabspl" contains the sub string "abcabs".

If we try to solve this with Naive Search Algo, it will match for first 5 characters but not for 6th character. And we will have to start over again with next iteration, we will lost all the progress in previous iteration. So, in order to save our progress and use it, we must use something called LPS table. Now in our matched string "abcab" we will find the longest same prefix and suffix.

Here, in our string "abcab" "ab" is the longest same prefix and suffix. Now, we will begin the next search iteration from index 5 (for main string). We saved two characters from our previous iteration. In order to figure out the prefix, suffix, and where to start next iteration from we use LPS table.

LPS for our substring ("abcabs") is "0 0 0 1 2 0". Here is how to calculate LPS table.

``````function calculateLpsTable(subStr) {
let i = 1;
let j = 0;
let lps = new Array(subStr.length).fill(0);

while(i < subStr.length) {
if(subStr[i] === subStr[j]) {
lps[i] = j + 1;
i += 1;
j += 1;
} else {
if(j !== 0) {
j = lps[j - 1];
} else {
i += 1;
}
}
}
return lps;
}
``````

Here is the implementation in code using LPS table.

``````function searchSubString(string, subString) {
let strLength = string.length;
let subStrLength = subString.length;
const lps = calculateLpsTable(subString);

let i = 0;
let j = 0;

while(i < strLength) {
if (string[i] === subString[j]) {
i += 1;
j += 1;
} else {
if (j !== 0) {
j = lps[j - 1];
} else {
i += 1;
}
}
if (j === subStrLength) return true;
}

return false;
}
``````

#### Time Complexity of KMP Algorithm

There is only one loop which run n times. Hence time complexity for KMP algo is Linear Time Complexity: O(n).

Notice how time complexity is improved as compared to that of Naive search algo.

### Bubble Sort Algorithm

Sorting means rearranging data in ascending or descending order. Bubble sort is one of many sorting algorithms.

In bubble sort algo, we swap the larger number to the end by comparing each number with the previous number. Here is a visual representation. Bubble sort code implementation.

``````function bubbleSort(array) {
let isSwapped;

for(let i = array.length; i > 0; i--) {
isSwapped = false;

for(let j = 0; j < i - 1; j++) {
if(array[j] > array[j + 1]) {
[array[j], array[j+1]] = [array[j+1], array[j]];
isSwapped = true;
}
}

if(!isSwapped) {
break;
}
}
return array;
}
``````

Let's try to understand the above code.

• Looping from end of array with variable i towards beginning.
• Start inner loop with variable j until (i - 1).
• If array[j] > array[j + 1] swap them.
• return sorted array.

#### Time Complexity of Bubble Sort Algorithm

There is a nested loop and both loop runs n times, so time complexity for this algo is (n * n) that is Quadratic Time Complexity O(n^2).

### Merge Sort Algorithm

Merge sort algorithm follows divide and conquer approach. It's a combination of two things - merge and sort.

In this algo first we divide main array into multiple indivisual sorted arrays. Then we merge the indivisual sorted elements together into final array.

Let's look at the implementation in code.

Merge Sorted Array

``````function mergeSortedArray(array1, array2) {
let result = [];
let i = 0;
let j = 0;

while(i < array1.length && j < array2.length) {
if(array1[i] < array2[j]) {
result.push(array1[i]);
i++;
} else {
result.push(array2[j]);
j++;
}
}

while (i < array1.length) {
result.push(array1[i]);
i++;
}

while (j < array2.length) {
result.push(array2[j]);
j++;
}

return result;
}
``````

The above code merges two sorted array into a new sorted array.

Merge Sort Algorithm

``````function mergeSortedAlgo(array) {
if(array.length <= 1) return array;

let midPoint = Math.floor(array.length / 2);
let leftArray = mergeSortedAlgo(array.slice(0, midPoint));
let rightArray = mergeSortedAlgo(array.slice(midPoint));

return mergeSortedArray(leftArray, rightArray);
}
``````

The above algo uses recursion to divide the array into multiple single element array.

#### Time Complexity of Merge Sort Algorithm

Let's try to calculate time complexity of merge sort algorithm. So, taking our previous example ([6, 3, 5, 2]), it took 2 steps to divide it into multiple single element array.

**

`It took 2 steps to divide an array of length 4 - (2^2)`

**.

Now if we double the length of array (8), it will take 3 steps to divide - (2^3). Means doubling the array length didn't doubled the steps.

Hence time complexity of merge sort algorithm is Logarithmic Time Complexity O(log n).

### Quick Sort Algorithm

Quick sort is one of the fastest sorting algorithm. In quick sort we select a single element known as pivot and we will move all element (smaller than pivot) to the left of pivot.

A visual representation. We will repeat this process for array to left and right of pivot until the array is sorted.

Code implementation

Pivot Utility

``````function pivotUtility(array, start=0, end=array.length - 1) {
let pivotIndex = start;
let pivot = array[start];

for(let i = start + 1; i < array.length; i++) {
if(pivot > array[i]) {
pivotIndex++;
[array[pivotIndex], array[i]] = [array[i], array[pivotIndex]];
}
}

[array[pivotIndex], array[start]] = [array[start], array[pivotIndex]];
return pivotIndex;
}
``````

The above code identifies the correct position of pivot and returns that position index.

``````function quickSort(array, left=0, right=array.length-1) {
if (left < right) {
let pivotIndex = pivotUtility(array, left, right);
quickSort(array, left, pivotIndex - 1);
quickSort(array, pivotIndex + 1, right);
}

return array;
}
``````

The above code uses recursion to keep moving pivot to it's correct position for left and right array of pivot.

#### Time Complexity of Quick Sort Algorithm

BEST CASE: Logarithmic Time Complexity - O(n log n)

AVERAGE CASE: Logarithmic Time Complexity - O(n log n)

WORST CASE: O(n^2) Radix sort is also know as Bucket sort algorithm.

Here first we build 10 buckets of index from 0 to 9. Then we take last character in each number and push the number to the corresponding bucket. Retrive the new order and repeat for the second last character of each number.

Keep repeating the above process till the array is sorted.

Implementation in code.

// Count Digits: The code below counts the number of digits the given element have.

``````function countDigits(number) {
if(number === 0) return 1;

return Math.floor(Math.log10(Math.abs(number))) + 1;
}
``````

// Get Digit: The code below gives the digit at index i from right.

``````function getDigit(number, index) {
const stringNumber = Math.abs(number).toString();
const currentIndex = stringNumber.length - 1 - index;

return stringNumber[currentIndex] ? parseInt(stringNumber[currentIndex]) : 0;
}
``````

// MaxDigit: The below snippet finds the number with maximum digits.

``````function maxDigit(array) {
let maxNumber = 0;

for(let i = 0; i < array.length; i++) {
maxNumber = Math.max(maxNumber, countDigits(array[i]));
}

return maxNumber;
}
``````

// The Radix Algo: Utilises all the above snippets to sort the array.

``````function radixSort(array) {
let maxDigitCount = maxDigits(array);

for(let i = 0; i < maxDigitCount; i++) {
let digitBucket = Array.from({length: 10}, () => []);

for(let j = 0; j < array.length; j++) {
let lastDigit = getDigit(array[j], i);
digitBucket[lastDigit].push(array[j]);
}

array = [].concat(...digitBucket);
}

return array;
}
``````

#### Time complexity of Radix Sort Algorithm

There is a nested for loop, and we know that time complexity for a nested for loop is O(n^2). But in this case both for loop does not run n times.

The outer loop runs k (maxDigitCount) times and inner loop runs m (length of array) times. Hence time complexity of Radix Sort is O(k x m) - (where k x m = n) Linear time complexity O(n)

Allright, with that we are at the end of this post. It is OK if some algos didn't clicked instantly, go through them multiple times.

That's how I understood them.

Also, I run a weekly Newsletter where I share epic content on web-dev and programming. Subscribe to build your skillset. Jeff Chavez

Wow. I'll learn this.  You actually have a wrong LPS algorithm. Here's the right one:

``````function calculateLpsTable(subStr) {
let i = 1;
let j = 0;
let lps = new Array(subStr.length).fill(0);

while(i < subStr.length) {
if(subStr[i] === subStr[j]) {
lps[i] = j + 1;
i += 1;
j += 1;
} else {
if(j !== 0) {
j = lps[j - 1];
} else {
i += 1;
}
}
}
return lps;
}
`````` There was a small typo at line 8. Wrote "i" instead of "1".

Thanks for pointing. LPS value is actually correct. It is [0, 0, 0, 1, 2, 0]

Thanks for pointing. Those code snippets are not added via any website. I wrote the article in markdown along with the code, and the snippets' design is native to dev.to platform itself. You got the O(n^2) complexity wrong. O(n^2) is quadratic complexity, a special case of polynomial complexity O(n^k).
Exponential complexity is O(2^n) and none of your example algorithms have exponential complexity. Yes, you are right. Thaks for this. I will fix it soon. Lazar Eric • Edited on

Wonderful post, thank you.

Radix sort improvement to handle negative integers, in my own writing:

``````// Including the sign (+,-)
function maxDigits(array) {
return Math.max(...array.map(item => String(item).length));
}

function getDigit(value, i) {
let response = String(value);

const index = response.length - 1 - i;

response = index >= 0 ? response[index] : undefined;

return ['+', '-'].includes(response) ? -1 : !response ? 0 : +response;
}

let array = value;
const digits = maxDigits(array);

// Digits
for (let i = 0; i < digits; i++) {
// sign, and 10 number digits 0-9, 11
const buckets = Array.from({ length: 11 }, () => []);

// Array
for (let j = 0; j < array.length; j++) {
// -1 for sign, 0 for 0 or no value, and 0-9 for a value
const index = getDigit(array[j], i);

buckets[index + 1].push(array[j]);
}

array = buckets.flatMap(item => item);
}

return array;
}
`````` Toni Stojev

In binary search, you have an infinite loop for non-existent member search.
Correct while loop exit term should be:

``````  while (array[middleIndex] !== element && firstIndex !== middleIndex && middleIndex!==firstIndex)
`````` Andrew Baisden

Super post! So much great content shared here. Algorithms are such a pain to learn and really hold some developers back from applying for jobs. Thank you Andrew. 😊

This motivates me to write more. Bruno Panassi

This is gold, thank you for sharing this! Thank you Bruno. 🙏

I am glad that you found this useful. 😊 Isn’t it a bit cheating to say that in the radix sort algorithm k x m = n?

I mean, it is true but if the largest number in the array is at least propotional to the size of the array (which is very likely) then k becomes Ω(log m) (lower bound) and in this case for being more fair with the other algorithms, if we define n to be the length of the array we will get something close to O(n log n).

Just putting it here for those who would confuse and think that radix sort is absolutely better then quick sort or merge sort. It depends on the data and the free space you have.

• Quick sort is an in place sorting algorithm while radix uses additional O(m) space.
• Sorting one million people by their age (as an int) is a classic radix sort example because of the small range numbers can be in, but sorting one million people by the amount of money they have in their bank account will probably do better with quick sort. Friday candour

Hmm, you just let me know I don't have to buy a course to learn algorithms, I love your content, eyes on you bro, thanks for making me happy. This is the best comment I have ever got.

Thanks for being so kind, brother. 🙏

You literally made my day. 😊 Thanks for your kind words Alejandro. It means a lot. 😊 Have you thought about making a website with algorithms explained in javascript? Hey Amr,

Thank you so much for your interest. I haven't thought about it, but it seems to be a really great idea.

I will definitely think about it. Thank you for the suggestion man.  