I'm doing CS50: Introduction to Computer Science on edx.org. I find it's a great way to review what I learn by completing, rewriting and sharing some of my notes.
Note: Big O notation can be though of “on the order of” and it represents the running time of an algorithm. In the C examples, n
is equivalent to sizeof(arr)/sizeof(arr[0])
which translates in JavaScript to arr.length
.
Week 3 is about algorithms. 😺
Table Of Contents
Linear Search
To iterate across the array from left to right searching for a target element.
Pseudocode example #1:
Repeat, starting at the first element:
If the element is the target element, stop
Else, move to the next element
Pseudocode example #2:
For i from 0 to n–1
If i'th element is target_element
Return true
Return false
C example:
bool linearSearch(int arr[], int n, int target)
{
for (int i = 0; i < n; i++)
if (arr[i] == target) return true;
return false;
}
JavaScript example:
linearSearch = (arr, target) => {
for (let i = 0; i < arr.length; i++)
if (arr[i] === target) return true;
return false;
}
Linear Search algorithm's
Worst case scenario:
Having to look through the entire array ofn
elements in the case where the target element is the last one or it is not in the array.
In Big O notation, it translates to O(n).Best case scenario:
The target element is the 1st element.
In Big O notation, it translates to Ω(1).
Binary Search
To find the target element by reducing the search area by half each time. Make sure the array on which the binary search algorithm is used on is sorted, else it's impossible to make assumptions about its content.
Pseudocode example #1:
Repeat until the (sub)array is of size 0:
Calculate the middle point of the current (sub)array
If the target element is the middle element, stop
Else if it's less than the middle:
End point is now just to the left of the current middle, repeat
Else if it's greater than the middle:
Start point is now just to the right of the current middle, repeat
Pseudocode example #2:
If no items
Return false
If middle item is target_element
Return true
Else if target_element < middle item
Update end point
Search left half
Else if target_element > middle item
Update start point
Search right half
C example (recursive):
int binarySearch(int arr[], int target, int start, int end)
{
if (end >= start) {
// instead of (start+end)/2 to avoid overflow
int mid = start+(endstart)/2;
if (arr[mid] == target) return mid;
else if (arr[mid] > target) return binarySearch(arr, target, start, mid1);
else return binarySearch(arr, target, mid+1, end);
}
return 1;
}
JavaScript example (recursive):
binarySearch = (arr, target, start, end) => {
if (end >= start) {
let mid = Math.floor((start+end)/2);
if (arr[mid] === target) return mid;
else if(arr[mid] > target) return binarySearch(arr, target, start, mid1);
else return binarySearch(arr, target, mid+1, end);
}
return false;
}
Binary Search algorithm's
Worst case scenario:
Having to divide a list ofn
elements in half repeatedly to find the target element because the target is found at the end of the last division or it is not in the array.
In Big O notation, it translates to O(log n).Best case scenario:
The target element is at midpoint of the array, so we can stop searching immediately after we start.
In Big O notation, it translates to Ω(1).
Bubble Sort
To sort in a bubbling way: move higher values towards the right of the array and lower values, towards the left.
Pseudocode example #1:
Set swap counter to a nonzero value
Repeat until the swap counter is equal to 0:
Reset swap counter to 0
Look at each adjacent pair:
If two adjacent elements are not in order:
Swap them
Add one to the swap counter
Pseudocode example #2:
Repeat until no swaps
For i from 0 to n–2
If i'th and i+1'th elements out of order
Swap them
C example:
void bubbleSort(int arr[], int n)
{
for (int i = 0; i < n1; i++)
for (int j = 0; j < ni1; j++)
if (arr[j] > arr[j+1])
{
int temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
}
JavaScript example:
bubbleSort = arr => {
for (let i = 0; i < arr.length1; i++)
for (let j = 0; j < arr.lengthi1; j++)
if (arr[j] > arr[j+1]) {
let temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
return arr;
}
Because comparing the i
th and i+1
th element, the sorting only needs to go up to n2
for i
before swapping the two elements if they're out of order. Knowing the largest n1
elements will have bubbled to the right, the sorting can stop after n1
passes.
When regoing through the array, only consider the unsorted elements.
When the swap counter remains at 0
, there is nothing else to swap.
Bubble sort algorithm's
Worst case scenario:
Having to bubble each of the elements all the way across the array because the array is in reverse order. Since it's only possible to fully bubble one element into its sorted position per pass, the sorting must happenn
times.
In Big O notation, it translates to O(n²).Best case scenario:
The array is already perfectly sorted, resulting in no swapping on the first pass.
In Big O notation, it translates to Ω(n).
Selection Sort
To find the smallest unsorted element and add it to the end of the sorted list.
Pseudocode example #1:
Repeat until there is no unsorted elements remaining:
Search unsorted part of data to find the smallest value
Swap the found value with the first element of the unsorted part
Pseudocode example #2:
For i from 0 to n–1
Find smallest item between i'th item and last item
Swap smallest item with i'th item
C example:
void selectionSort(int arr[], int n)
{
for (int i = 0; i < n1; i++)
{
int min = i;
for (int j = i+1; j < n; j++)
if (arr[j] < arr[min]) min = j;
int temp = arr[min];
arr[min] = arr[i];
arr[i] = temp;
}
}
JavaScript example:
selectionSort = arr => {
for (let i = 0; i < arr.length1; i++) {
let min = i;
for (let j = i+1; j < arr.length; j++)
if (arr[j] < arr[min]) min = j;
let temp = arr[min];
arr[min] = arr[i];
arr[i] = temp;
}
return arr;
}
Selection sort algorithm's
Worst case scenario:
Having to repeat the sorting processn
times to iterate each of then
elements of the array to find the smallest unsorted element and sort it. Only one element gets sorted on each pass.
In Big O notation, it translates to O(n²).Best case scenario:
The same as the worst case scenario as there is no way to guarantee the array is sorted until the sorting process iterates over all of the elements of the array.
In Big O notation, it translates to Ω(n²).
Insertion Sort
To build a sorted array in place; shifting elements out of the way to make room if necessary as the array is being built.
Pseudocode example #1:
Call the first element of the array sorted
Repeat until all elements are sorted:
Insert next unsorted item into sorted part shifting the required number of items
Pseudocode example #2:
For i from 1 to n–1
Insert next unsorted item into sorted part shifting i items
C example:
void insertionSort(int arr[], int n)
{
for (int i = 1; i < n; i++) {
int key = arr[i];
int j = i1;
while (j >= 0 && arr[j] > key) {
arr[j+1] = arr[j];
j = j1;
}
arr[j+1] = key;
}
}
JavaScript example:
insertionSort = arr => {
for (let i = 1; i < arr.length; i++) {
let key = arr[i];
let j = i1;
while (j >= 0 && arr[j] > key) {
arr[j+1] = arr[j];
j = j1;
}
arr[j+1] = key;
}
return arr;
}
Insertion sort algorithm's
Worst case scenario:
Having to shift each of then
elements fromn
positions each time to make an insertion because the array is in reverse order.
In Big O notation, it translates to O(n²).Best case scenario:
The array is already sorted. Only gotta keep moving between unsorted and sorted elements as we iterate over each of them.
In Big O notation, it translates to Ω(n).
Recursion
To code elegantly. 🌹
Recursion is related to how an algorithm or a function is implemented, it isn't an algorithm itself.
A recursive function invokes itself as part of its execution.
Detailed example using the factorial function:
 n! is defined over all positive integers
 n! equals all of the positive integers less than or equal to n, multiplied together

n! as
fact(n)
:
Pseudocode example #1:
fact(1) = 1
fact(2) = 2 * 1
fact(3) = 3 * 2 * 1
…
Pseudocode example #2:
fact(1) = 1
fact(2) = 2 * fact(1)
fact(3) = 3 * fact(2)
…
The basis for a recursive definition of the factorial function:
fact(n) = n * fact(n1)
Recursive function has two cases that can apply given any input:
 Base case: terminates the recursive process when triggered
 Recursive case: where the recursion happens
int fact(int n)
{
// base case
if (n == 1)
return 1;
// recursive case
else
return n * fact(n1);
}
There can be multiple base cases.
Example the fibonacci number sequence where:
 1st element is
0
 2nd element is
1

n
th element is the sum of(n1)+(n2)
There can be multiple recursive cases.
Example the collatz conjecture.
The next C and JavaScript examples define a collatz
function that calculates how many steps it takes to get "back to 1":
C example:
int collatz(int steps)
{
// base case
if (steps == 1) return 0;
// recursive case: even numbers
else if ((steps % 2) == 0) return 1+collatz(steps/2);
// recursive case: odd numbers
else return 1+collatz(3*steps+1);
}
JavaScript example:
collatz = steps => {
// base case
if (steps == 1) return 0;
// recursive case: even numbers
else if ((steps % 2) == 0) return 1+collatz(steps/2);
// recursive case: odd numbers
else return 1+collatz(3*steps+1);
}
Merge Sort
To divide an array into smaller arrays to sort and then, combine those sorted arrays back together in sorted order.
Pseudocode example #1:
If only one element
Return
Else
Sort left half of elements
Sort right half of elements
Merge sorted halves
Pseudocode example #2:
Sort the left half of the array (assuming n > 1)
Sort right half of the array (assuming n > 1)
Merge the two halves together
C example (recursive):
// merges two subarrays of arr[]
void merge(int arr[], int leftIndex, int mid, int rightIndex)
{
int n1 = midleftIndex+1;
int n2 = rightIndexmid;
// temp arrays
int Left[n1], Right[n2];
// copy data to temp arrays
for (int i = 0; i < n1; i++)
Left[i] = arr[leftIndex+i];
for (int j = 0; j < n2; j++)
Right[j] = arr[mid+1+j];
// merge the temp arrays back into arr[]
int i = 0; // init index of 1st subarray
int j = 0; // init index of 2nd subarray
int k = leftIndex; // init index of merged subarray
while (i < n1 && j < n2)
{
if (Left[i] <= Right[j])
{
arr[k] = Left[i];
i++;
}
else
{
arr[k] = Right[j];
j++;
}
k++;
}
// copy the remaining elements of Left[], if any
while (i < n1)
{
arr[k] = Left[i];
i++;
k++;
}
// copy the remaining elements of Right[], if any
while (j < n2)
{
arr[k] = Right[j];
j++;
k++;
}
}
void mergeSort(int arr[], int leftIndex, int rightIndex)
{
if (leftIndex < rightIndex)
{
// instead of (l+r)/2 to avoid overflow
int mid = leftIndex+(rightIndexleftIndex)/2;
// sort first and second halves
mergeSort(arr, leftIndex, mid);
mergeSort(arr, mid+1, rightIndex);
// merge them back together
merge(arr, leftIndex, mid, rightIndex);
}
}
JavaScript example (recursive):
// to merge left subarray and right subarray
merge = (left, right) => {
let resultArray = [], leftIndex = 0, rightIndex = 0;
// concat values into the resultArray in order
while (leftIndex < left.length && rightIndex < right.length) {
if (left[leftIndex] < right[rightIndex]) {
resultArray.push(left[leftIndex]);
leftIndex++;
} else {
resultArray.push(right[rightIndex]);
rightIndex++;
}
}
// concat remaining element from either left OR right
return resultArray
.concat(left.slice(leftIndex))
.concat(right.slice(rightIndex));
}
mergeSort = arr => {
// if array has one element or is empty, no need to sort
if (arr.length <= 1) return arr;
const mid = Math.floor(arr.length/2);
// divide the array into left and right
const left = arr.slice(0, mid);
const right = arr.slice(mid);
// merge back together using recursion
return merge(mergeSort(left), mergeSort(right));
}
Merge sort algorithm's
Worst case scenario:
Having to splitn
elements up before recombining them effectively, doubling the sorted subarrays as they are built.
In Big O notation, it translates to O(n log n).Best case scenario:
The array is already sorted, but still gotta be split and recombined back together to know it is sorted.
In Big O notation, it translates to Ω(n log n).
Resources:
Top comments (12)
I enjoyed the video immensely, by the way!
You may also consider looking into two more, both of which are among the fastest known sorting algorithms in existence.
Thank you, I'll look into them! 😊👍
It's encouraging to see you summarize what learn from courses, at the same time others can learn from you as well. Keep doing it 👍🏻
The best thing to test whatever you have learnt/studied is to share and explain others about it. It boosts confidence as well as help understand it better.
Came across these algorithms a year ago while studying C/C++. Thanks for refreshing these algorithms. Great notes for revision. ✌
I agree, but find it hard to find the time to do so in an interesting way sometimes. I guess it's an habit to take up as part of a learning & sharing routine. Like it should be done right after and/or before starting to learn something else.
Thanks to you for the positive comment. =)
Yeah that true. Its difficult to take out time. I am a beginner so trying my best to implement this habit from the beginning. The good thing about doing this is that the work is done more systematically and with focus. No one would work lazily or just for formality if he/she has to deliver that to a bunch of people afterwards. Also the quality will be maintained. And the best thing is that it would act as a revision /summary for him/her and others who would come across it. Its a win win situation for everyone. 🙂
Thanks for these notes. Hope to read more in future and gain knowledge. Great job and all the best.✌
Totally agree once again. 🙂
Thanks to you and all the best as well! 💮
I didn't get avoid overflow part
To prevent integer overflow for large numbers which could end up being too big to fit the integer's storage space.
en.wikipedia.org/wiki/Integer_over...
fresh2refresh.com/cprogramming/c...
Aha, I see
So we get the difference first and divide it by 2 then add it to the start.
instead of adding the start and the end.
But why we didn't do the same with the js example ?!
To highlight both possibilities? You don't always have to worry about overflow; When you do, it's nice to know there's a simple solution. These are just notes.
:)