Developing a JavaScript program for ATMs that can efficiently handle and dispense money in the fewest number of notes possible

Question

Developing a JavaScript program for ATMs that can efficiently handle and dispense money in the fewest number of notes possible

When a certain amount is entered, the code should be capable of handling figures up to 20000. For instance, if the input amount is 2600 with a card balance of 3000, the output will be as follows: New Balance - 400 Notes: 2000 * 1 500 * 1 100 * 1

Only three banknotes are allowed: 2000, 500, and 100, and there is a cash limit of 20000.

I am relatively new to the world of JavaScript and struggling to write the necessary code. Can someone please assist me?

var h = 5;
var f = 2;
var t = 1;

// More code snippets here...

Apologies for the simple program, but I would greatly appreciate if someone could provide a solution in TypeScript code, returning the required denomination in an array. Thank you!

javascript typescript

Answer 1

Answer №1

function separateNotes(amount) {
  let hundredCount = 0;
  let fiftyCount = 0;
  let twentyCount = 0;

  while (amount >= 20) {
    if (amount >= 100 && ((amount % 100) % 50 === 0 || (amount % 100) % 20 === 0)) {
      amount -= 100;
      hundredCount++;
    } else if (amount >= 50 && ((amount % 50) % 100 === 0 || (amount % 50) % 20 === 0)) {
      amount -= 50;
      fiftyCount++;
    } else {
      amount -= 20;
      twentyCount++;
    }
  }

  return [hundredCount, fiftyCount, twentyCount];
}

console.log(separateNotes(230));
console.log(separateNotes(250));

Answer 2

function separateNotes(amount) {
  let hundredCount = 0;
  let fiftyCount = 0;
  let twentyCount = 0;

  while (amount >= 20) {
    if (amount >= 100 && ((amount % 100) % 50 === 0 || (amount % 100) % 20 === 0)) {
      amount -= 100;
      hundredCount++;
    } else if (amount >= 50 && ((amount % 50) % 100 === 0 || (amount % 50) % 20 === 0)) {
      amount -= 50;
      fiftyCount++;
    } else {
      amount -= 20;
      twentyCount++;
    }
  }

  return [hundredCount, fiftyCount, twentyCount];
}

console.log(separateNotes(230));
console.log(separateNotes(250));

Answer 3

Answer №2

amtArray = [2000, 500, 100];  // array containing denominations to find

 for (let i = 0; i < this.amtArray.length; i++) {
            this.resultArray.push(Math.floor(total / this.amtArray[i]));
            // Calculate new total
            total = total % this.amtArray[i];
        }
        var twothousands_notes = this.resultArray[0];
        var fivehundred_notes = this.resultArray[1];
        var hundred_notes = this.resultArray[2];

        console.log('Calculated amounts: ' + '100 : ' +
            hundred_notes + '   500 :  ' +
            fivehundred_notes + '  2000 :  ' +
            twothousands_notes);

You can customize the logic based on the given amount.

Hope this explanation is helpful! :)

Answer 4

amtArray = [2000, 500, 100];  // array containing denominations to find

 for (let i = 0; i < this.amtArray.length; i++) {
            this.resultArray.push(Math.floor(total / this.amtArray[i]));
            // Calculate new total
            total = total % this.amtArray[i];
        }
        var twothousands_notes = this.resultArray[0];
        var fivehundred_notes = this.resultArray[1];
        var hundred_notes = this.resultArray[2];

        console.log('Calculated amounts: ' + '100 : ' +
            hundred_notes + '   500 :  ' +
            fivehundred_notes + '  2000 :  ' +
            twothousands_notes);

You can customize the logic based on the given amount.

Hope this explanation is helpful! :)

Answer 5

Answer №3

Here is a convenient solution for all your scenarios:

function dispenseCase (inputAmount) {
 var notes = [];
 var balance = 3000;

  if(inputAmount !== 0 && inputAmount % 100 == 0 && inputAmount <= balance) {
    var notes2000 = Math.round(inputAmount / 2000);
    var notes500 = Math.round((inputAmount - (notes2000 * 2000)) / 500 );
    var notes100 = Math.round((inputAmount - ((notes2000 * 2000) + (notes500 * 500))) / 100);

   notes.push(notes2000);
   notes.push(notes500);
   notes.push(notes100);

   console.log("Your account balance = ", balance - inputAmount);
   console.log(notes);
  }
  else if (inputAmount > balance) {
   console.log("Insufficient funds in your account");
  }
  else if ( inputAmount % 100 != 0 || inputAmount < 100 ) {
   console.log( "Invalid amount entered, amount must be multiples of 100");
}
}

dispenseCase(2600);

Answer 6

Here is a convenient solution for all your scenarios:

function dispenseCase (inputAmount) {
 var notes = [];
 var balance = 3000;

  if(inputAmount !== 0 && inputAmount % 100 == 0 && inputAmount <= balance) {
    var notes2000 = Math.round(inputAmount / 2000);
    var notes500 = Math.round((inputAmount - (notes2000 * 2000)) / 500 );
    var notes100 = Math.round((inputAmount - ((notes2000 * 2000) + (notes500 * 500))) / 100);

   notes.push(notes2000);
   notes.push(notes500);
   notes.push(notes100);

   console.log("Your account balance = ", balance - inputAmount);
   console.log(notes);
  }
  else if (inputAmount > balance) {
   console.log("Insufficient funds in your account");
  }
  else if ( inputAmount % 100 != 0 || inputAmount < 100 ) {
   console.log( "Invalid amount entered, amount must be multiples of 100");
}
}

dispenseCase(2600);

Answer 7

Answer №4

Creating a program in JavaScript to calculate ATM denominations.

This program will determine the minimum number of notes needed to make up a specified amount, starting from the highest denomination to the lowest.

function countCurrency(amount) {
  var notes = [2000, 500, 200, 100];
  var noteCounter = [0, 0, 0, 0];

  for (var i = 0; i < 4; i++) {
    if (amount >= notes[i]) {
      noteCounter[i] = Math.floor(amount / notes[i]);
      amount = amount - noteCounter[i] * notes[i];
    }
  }
  // Display notes denomination
  console.log("Denomination Count:");
  for (var j = 0; j < 4; j++) {
    if (noteCounter[j] !== 0) {
      console.log(notes[j] + " : " + noteCounter[j]);
    }
  }
}

countCurrency(3300);

For a live demo of the program, click here

Answer 8

Creating a program in JavaScript to calculate ATM denominations.

This program will determine the minimum number of notes needed to make up a specified amount, starting from the highest denomination to the lowest.

function countCurrency(amount) {
  var notes = [2000, 500, 200, 100];
  var noteCounter = [0, 0, 0, 0];

  for (var i = 0; i < 4; i++) {
    if (amount >= notes[i]) {
      noteCounter[i] = Math.floor(amount / notes[i]);
      amount = amount - noteCounter[i] * notes[i];
    }
  }
  // Display notes denomination
  console.log("Denomination Count:");
  for (var j = 0; j < 4; j++) {
    if (noteCounter[j] !== 0) {
      console.log(notes[j] + " : " + noteCounter[j]);
    }
  }
}

countCurrency(3300);

For a live demo of the program, click here

Answer 9

Answer №5

If you input an amount, this code will calculate the number of notes in denominations of 2000, 500, and 100 needed to dispense that amount.

function calculateNotes (enteredAmount) {
 var notes = [];

  if(enteredAmount !== 0) {
    var notes2000 = Math.round(enteredAmount / 2000);
    var notes500 = Math.round((enteredAmount - (notes2000 * 2000)) / 500 );
    var notes100 = Math.round((enteredAmount - ((notes2000 * 2000) + (notes500 * 500))) / 100);

   notes.push(notes2000);
   notes.push(notes500);
   notes.push(notes100);

   console.log(notes);
  }
}

calculateNotes(2600);

Hopefully, this explanation clarifies the process for you.

Answer 10

If you input an amount, this code will calculate the number of notes in denominations of 2000, 500, and 100 needed to dispense that amount.

function calculateNotes (enteredAmount) {
 var notes = [];

  if(enteredAmount !== 0) {
    var notes2000 = Math.round(enteredAmount / 2000);
    var notes500 = Math.round((enteredAmount - (notes2000 * 2000)) / 500 );
    var notes100 = Math.round((enteredAmount - ((notes2000 * 2000) + (notes500 * 500))) / 100);

   notes.push(notes2000);
   notes.push(notes500);
   notes.push(notes100);

   console.log(notes);
  }
}

calculateNotes(2600);

Hopefully, this explanation clarifies the process for you.

Answer 11

Answer №6

//ATM Cash Denominations //Cash Input Value Already been Provided in this method // You may use a input stream method to input a user input value

public class ATMWithdrawal
{
 public static void main(String args[])//throws IOException
 {

 int notes[]={200,100,50,20,10}; //storing all the denominations in an array
 int amount = 500;

 int copy=amount; //Making a copy of the amount
 int totalNotes=0,count=0;

 System.out.println("\nATM CASH DENOMINATIONS: \n");

 for(int i=0;i<5;i++) //Since there are 5 different types of notes, hence we check for each note.
 {
 count=amount/notes[i]; // counting number of notes[i] notes
 if(count!=0) //printing that denomination if the count is not zero
 {
 System.out.println(notes[i]+"\tx\t"+count+"\t= "+notes[i]*count);
 }
 totalNotes=totalNotes+count; //finding the total number of notes
 amount=amount%notes[i]; //finding the remaining amount whose denomination is to be found
 }

 System.out.println("--------------------------------");
 System.out.println("TOTAL\t\t\t= "+copy); //printing the total amount
 System.out.println("--------------------------------");
 System.out.println("Total Number of Notes\t= "+totalNotes); //printing the total number of notes
 }
}

Answer 12

//ATM Cash Denominations //Cash Input Value Already been Provided in this method // You may use a input stream method to input a user input value

public class ATMWithdrawal
{
 public static void main(String args[])//throws IOException
 {

 int notes[]={200,100,50,20,10}; //storing all the denominations in an array
 int amount = 500;

 int copy=amount; //Making a copy of the amount
 int totalNotes=0,count=0;

 System.out.println("\nATM CASH DENOMINATIONS: \n");

 for(int i=0;i<5;i++) //Since there are 5 different types of notes, hence we check for each note.
 {
 count=amount/notes[i]; // counting number of notes[i] notes
 if(count!=0) //printing that denomination if the count is not zero
 {
 System.out.println(notes[i]+"\tx\t"+count+"\t= "+notes[i]*count);
 }
 totalNotes=totalNotes+count; //finding the total number of notes
 amount=amount%notes[i]; //finding the remaining amount whose denomination is to be found
 }

 System.out.println("--------------------------------");
 System.out.println("TOTAL\t\t\t= "+copy); //printing the total amount
 System.out.println("--------------------------------");
 System.out.println("Total Number of Notes\t= "+totalNotes); //printing the total number of notes
 }
}

Answer 13

Answer №7

let sumToDenominate=Math.floor(Math.random() * 100);
let billsValues = [100, 50, 20, 10, 5,1];

let restAfterDenomination = [];
let billsNumber = [];



function denomination(sum, billsValues) {



printInitialValue( sumToDenominate, billsValues);

initializeArray( sumToDenominate, billsValues);


for (let i = 1; i <= billsValues.length; i++) {

    if (restAfterDenomination[i - 1] > 0 || restAfterDenomination < billsNumber[i]) {

        billsNumber.push(Math.floor(restAfterDenomination[i - 1] / billsValues[i]));

        console.log(`bill's number of `, billsValues[i], "=>", billsNumber[i]);

        restAfterDenomination.push(restAfterDenomination[i - 1] - (billsNumber[i] * billsValues[i]));


    } else {
        console.log(`rest is less than smallest bill or equal to 0`);
        billsNumber.push(0);
        // break;
    }
}



}

function printInitialValue(amount, billsValue) {
console.log("Denomination sumToDenominate: ", amount);

console.log("____________");
for (const logEntry of billsValue) {
    console.log(logEntry);
}
console.log("__________");
}

function initializeArray(amount, billsValues) {


billsNumber.push(Math.floor(amount / billsValues[0]));

console.log(`bill's number of`, billsValues[0], "=>", billsNumber[0]);

restAfterDenomination.push(amount - (billsNumber[0] * billsValues[0]));


denomination(sumToDenominate,billsValues);

Answer 14

let sumToDenominate=Math.floor(Math.random() * 100);
let billsValues = [100, 50, 20, 10, 5,1];

let restAfterDenomination = [];
let billsNumber = [];



function denomination(sum, billsValues) {



printInitialValue( sumToDenominate, billsValues);

initializeArray( sumToDenominate, billsValues);


for (let i = 1; i <= billsValues.length; i++) {

    if (restAfterDenomination[i - 1] > 0 || restAfterDenomination < billsNumber[i]) {

        billsNumber.push(Math.floor(restAfterDenomination[i - 1] / billsValues[i]));

        console.log(`bill's number of `, billsValues[i], "=>", billsNumber[i]);

        restAfterDenomination.push(restAfterDenomination[i - 1] - (billsNumber[i] * billsValues[i]));


    } else {
        console.log(`rest is less than smallest bill or equal to 0`);
        billsNumber.push(0);
        // break;
    }
}



}

function printInitialValue(amount, billsValue) {
console.log("Denomination sumToDenominate: ", amount);

console.log("____________");
for (const logEntry of billsValue) {
    console.log(logEntry);
}
console.log("__________");
}

function initializeArray(amount, billsValues) {


billsNumber.push(Math.floor(amount / billsValues[0]));

console.log(`bill's number of`, billsValues[0], "=>", billsNumber[0]);

restAfterDenomination.push(amount - (billsNumber[0] * billsValues[0]));


denomination(sumToDenominate,billsValues);

Developing a JavaScript program for ATMs that can efficiently handle and dispense money in the fewest number of notes possible

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

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