DEV Community: Loveena Saran

Pascal’s Triangle

Loveena Saran — Mon, 31 Jan 2022 13:42:38 +0000

Deduction made from how the Pascal's triangle is constructed:

The boxes at the start and end of the list are always set to '1'

Initially the first 2 levels do not have any adding up of elements and the process of adding starts from the 3rd level

Adding up the previous level answer leads to the new numbers at the center of the triangle

How to approach this problem?

We can keep a pointer to here 'i' for each level of the triangle
Also if we notice there are center boxes that grow in triangular pattern.

Notice when level is '1' we have '0' elements to be added from previous levels.
At level 1 --> 0 elements are formed by adding
At level 2 --> 1 elements are formed by adding
At level 3 --> 2 elements are formed by adding
and so on...

we can have a 'j' variable that we initialize when we are at 1st level of 'i'.
Here we are initializing 'j' to be 'zero' and less than 'i'.

'i' and 'j' values Serve multiple purposes in this problem

'i' indicates each level and 'j' indicates the elements formed by adding up previous elements.
'j' runs at a 1 less than 'i' as it starts from '0' and creates boxes at each level.
As the answer is stored in List i.e (List of Lists). We can to retrieve the previous level data and add this to form the current value.

Code:

   for(int i=1; i<=numRows; i++){
            for(int j=0; j<i; j++){

For the first and the last box we want to add '1'

for(int i=1;i<=numRows;i++){      
    for(int j=0; j<i; j++){
      if(j==0 || j==i-1){  // add '1' to first box and last box
          curr.add(1);  // Add to list curr
               }

-For boxes at the center ranging from (1:i-1)

for(int i=1;i<=numRows;i++){
           List<Integer> curr = new ArrayList<>();       
           for(int j=0;j<i;j++){
               if(j==0 || j==i-1){
                   curr.add(1);
               }
               else{
curr.add(levels.get(i-1).get(j)+ levels.get(i-1).get(j-1)); // to form new elements by  adding values from previous levels
           }
        }

Dry Run:

Do like if you found this helpful. This would help me learn and create new articles like this.
Thanks!

Knights Tour using Backtracking

Loveena Saran — Tue, 25 Jan 2022 16:31:42 +0000

Let's discuss Knight's Tour

Motive of the problem is to print the various paths that the knight can take in a N*N board covering all the boxes given that we place the knight at a specific row and column.

Constraints if we look at a chess board that is of 3*3 size it and place the knight at position (1R-1C) the knight cannot move in the "L" shaped manner in this board .

So ideally we want the board to be 5*5 that is the minimum we should expect for our problem to work.
(Note: There are other Algorithms which handle N*M size of pattern but we are sticking to N*N for our problem)

If we place the Knight at the center (2R-2C) we can say the movements will be as follows:

Our basic requirement is to initiate and put '1' from the first row column that user gives us.
which represents the first step taken by the knight and keep incrementing it till covers all the rows*columns.

Let's create a function to take the row column with our first step by setting 'move' variable to '1'

Once we place the knight at a given place we have a choice of placing our knight at the 8 open positions that are valid according to it's movement.

And this is when Recursion enters !!✨
Why recursion?
we want to give the 2nd step the same 8 possible valid
options we gave the 1st step to place the next step of
the knight.

Code:

public static void knightsTour(int[][] chess, int r, int c, int move) 
{
        chess[r][c]=move;
        knightsTour(chess,r-2,c+1,move+1);
        knightsTour(chess,r-1,c+2,move+1);
        knightsTour(chess,r+1,c+2,move+1);
        knightsTour(chess,r+2,c+1,move+1);
        knightsTour(chess,r+2,c-1,move+1);
        knightsTour(chess,r+1,c-2,move+1);
        knightsTour(chess,r-1,c-2,move+1);
        knightsTour(chess,r-2,c-1,move+1);


    }

knightsTour(chess,row,col,1);  //function call

In order to keep the moves valid and inside the N*N board

limits also be careful to not overwrite a previous step we add our conditions.

public static void knightsTour(int[][] chess, int r, int c, int move) 
{
  if(r<0 || c<0 || r>=chess.length || c >=chess.length || chess[r][c]>0)
        {
         return;   
        }
  ....[code]
 }

As the Recursion call proceeds it is sure to hit a case where the border is less than 0 ,equal/greater to chess length or the path is already set in the box.

Once it does that particular Recursion Stack Path will return to its the Parent Recursion call.

So Ideally we went down a given path as far as we could but could not cover all the boxes. So we backtrack to our previous decision (basically time travel 😅) to the last mistake that we did and erase it .
to do that we Simply remove the previous step that was added in the parent recursion call and go to the parent parent call (bear with me😂) and look for other steps from our 8-1 options and make a new Recursion call to that step...

public static void knightsTour(int[][] chess, int r, int c, int move) {

 ...[code]
        knightsTour(chess,r-1,c-2,move+1);
        knightsTour(chess,r-2,c-1,move+1);
        chess[r][c]=0;   <-- this erases our mistakes

    }

Now we are nearing the end... I can see the light👩‍🚀
We just Print the chessboard as soon as the N*N move is set in the board.
But There is one little tiny thing which is still left..
At the end of the Recursion when our move is 25 for the case (5*5) we have reached our limit and there is no need to run a Recursion call for this case as this will return no possible new steps... and not handling this will
reset the value of that given box to '0' as chess[r][c]=0; runs after RC call.

Also Be sure to remove the 25th element after printing the chessboard as we want the chessboard to be reset with empty values for the next possible paths...

public static void knightsTour(int[][] chess, int r, int c, int move) {

        if(r<0 || c<0 || r>=chess.length || c >=chess.length || chess[r][c]>0)
        {
         return;   
        }
      else if(move==chess.length*chess.length){      
            chess[r][c]=move;
            displayBoard(chess);   
/*displayBoard can be a function to iterate over the i*j values of chessboard and print them */
            chess[r][c]=0;
            return;
            }

THe eND!!

Avatar API support in Appwrite

Loveena Saran — Wed, 13 Oct 2021 06:46:17 +0000

Appwrite provides an AMAZING solution that offers YOU an array of REST APIs , tools , Management Console UI for YOUR core backend needs.

If you haven't already try it out for yourself AppWrite
Go Ahead don't be shy...

Are YOU Looking to make your Service Standout?

Are YOU Tired of importing multiple packages for Icon support?

Are YOU Hoping to make your Life a little bit easier?

Enhance your Appwrite App with Images, Icons, and Avatars...

Adding payment methods logos , fetching the desired country flags and Generating QR codes is all that you desire.

Be it Food Delivery Services ,Language Learning App or a Fintech App these small additions that will make it a lot easier.

How can you use Appwrite's Avatar API ?

Create a Service Class to fetch the endpoint. In our case avatars and create a constructor to access it within the method.

class AppwriteService {
constructor() {   
appwrite.setEndpoint(Server.endpoint).setProject(Server.project);
    this.avatars= appwrite.avatars;
  }

2.Access your desired method getCreditCard present under Avatar API and pass in the mandatory attributes cardname for the method to work.
Optional Attributes Include: width,height,quality

 createCreditAvatar =(cardname) =>{
    return this.avatars.getCreditCard(cardname);
  }

3.Call the method Created in your file that you desire to add your avatar. Don't forget to pass the value to the attributes that were Set earlier.
Example: amex, argencard, cabal, censosud, diners, discover, elo, hipercard, jcb, mastercard, naranja, targeta-shopping, union-china-pay, visa, mir, maestro

  let creditcard = appwrite.createCreditAvatar('amex');
  console.log(creditcard);

4.In the return section of your Project call the component inside the image tag to view the icon

   return(
    <div >
      <img src={creditcard} />
    </div>
   );

5.And Voilà!! You have your icon displayed on your UI

Few Other Examples To get you started:

createFlagAvatar =(flagname) =>{
    return this.avatars.getFlag(flagname);
  }
  createQrAvatar =(text,size) =>{
    return this.avatars.getQR(text,size);
  }

let flagname = appwrite.createFlagAvatar('IN');
let  qrcode  = appwrite.createQrAvatar('FinTechApp',100);

console.log(flagname);
console.log(qrcode);
   return(
    <div >
      <img src={flagname} />
      <br/>
      <img src={qrcode} />
      <br/>
    </div>
   );

Start Appwriting Away!!!

Read for any Doubts:
Getting Started with Appwrite
Appwrite React Series