expected '0th' to equal '0' - javascript

function numberToOrdinal(i) {
var j = i % 10,
k = i % 100;
if (j == 0 && k == 100) {
return '0th';
}
if (j == 1 && k != 11) {
return i + "st";
}
if (j == 2 && k != 12) {
return i + "nd";
}
if (j == 3 && k != 13) {
return i + "rd";
}
return i + "th";
}
It's not passing this test, what's wrong?
should handle single digits
expected '0th' to equal '0'

if you gonna devide a unknown input you wanna check if the input given might be zero. Else there might be an devisionbyzero exception thrown.
function numberToOrdinal(i) {
if(i === 0){ // since we devide i we wanna check if it might be zero.
return i ;
}
var j = i % 10,
k = i % 100;
//if (j == 0 && k == 100) { didnt seem to be doing anything
//return '0th';
//}
if (j == 1 && k != 11) {
return i + "st";
}
if (j == 2 && k != 12) {
return i + "nd";
}
if (j == 3 && k != 13) {
return i + "rd";
}
return i + "th";
}
console.log(numberToOrdinal(0));

you could just change these lines
if (j == 0 && k == 100) {
return '0th';
}
to be
if (i == 0) {
return i
}

If you want your function to return 0 when 0 is passed to the function, you will need to enter a condition to handle that, such as:
if (i === 0) {
return 0;
}

You have to enter another condition to handle it
function numberToOrdinal(i) {
var j = i % 10,
k = i % 100;
if (i === 0) {
return i;
}
if (j == 1 && k != 11) {
return `${i}st`;
}
if (j == 2 && k != 12) {
return `${i}nd`;
}
if (j == 3 && k != 13) {
return `${i}rd`;
}
return `${i}th`;
}

Related

BarreCode EAN encoding

I try to generate the encoding of the EAN13 barcode in order to obtain for the code 6181100327649 the following line 6BSLLAA*dchgej+
This to be able to use the EAN font and generate in my creative visuals.
Here is my code but it does not work :
function EAN(chaine) {
var i = 0;
var first = 0;
var checksum = 0;
var CodeBarre = "";
var tableA = 0;
if(chaine.match("/^\d{12}$/")){
for (i = 1; i < 12; i += 2)
{
chaine.substr(i, 1);
checksum += parseInt(chaine.substr(i, 1));
}
checksum *= 3;
for (i = 0; i < 12; i += 2)
{
checksum += parseInt(chaine.substr(i, 1));
}
chaine += (10 - checksum % 10) % 10;
CodeBarre = chaine.substr(0, 1) + (chaine.fromCharCode(65 + parseInt(chaine.substr(1, 1)) ));
first = parseInt(chaine.substr(0, 1));
for (i = 2; i <= 6; i++)
{
tableA = false;
switch (i)
{
case 2: if (first >= 0 && first <= 3) tableA = true; break;
case 3: if (first == 0 || first == 4 || first == 7 || first == 8) tableA = true; break;
case 4: if (first == 0 || first == 1 || first == 4 || first == 5 || first == 9) tableA = true; break;
case 5: if (first == 0 || first == 2 || first == 5 || first == 6 || first == 7) tableA = true; break;
case 6: if (first == 0 || first == 3 || first == 6 || first == 8 || first == 9) tableA = true; break;
}
if (tableA)
CodeBarre += (chaine.fromCharCode(65 + parseInt(chaine.substr(i, 1)) ));
else
CodeBarre += (chaine.fromCharCode(75 + parseInt(chaine.substr(i, 1)) ));
}
CodeBarre += "*";
for (i = 7; i <= 12; i++)
{
CodeBarre += (chaine.fromCharCode(97 + parseInt(chaine.substr(i, 1)) ));
}
CodeBarre += "+";
}
return CodeBarre;
}
document.getElementById("demo").innerHTML = EAN("618110032764");
There was mutliple issues here, but here is a working example for you
function getCharCode(chaine, index, position) {
return String.fromCharCode(position + parseInt(chaine.substr(index, 1)));
}
function EAN(chaine) {
var i = 0;
var first = 0;
var checksum = 0;
var CodeBarre = "";
var tableA = 0;
if (chaine.match(/^\d{12}$/)) {
for (i = 1; i < 12; i += 2) {
chaine.substr(i, 1);
checksum += parseInt(chaine.substr(i, 1));
}
checksum *= 3;
for (i = 0; i < 12; i += 2) {
checksum += parseInt(chaine.substr(i, 1));
}
chaine += (10 - checksum % 10) % 10;
CodeBarre = chaine.substr(0, 1) + getCharCode(chaine, 1, 65);
first = parseInt(chaine.substr(0, 1));
for (i = 2; i <= 6; i++) {
tableA = false;
switch (i) {
case 2:
if (first >= 0 && first <= 3) tableA = true;
break;
case 3:
if (first == 0 || first == 4 || first == 7 || first == 8) tableA = true;
break;
case 4:
if (first == 0 || first == 1 || first == 4 || first == 5 || first == 9) tableA = true;
break;
case 5:
if (first == 0 || first == 2 || first == 5 || first == 6 || first == 7) tableA = true;
break;
case 6:
if (first == 0 || first == 3 || first == 6 || first == 8 || first == 9) tableA = true;
break;
}
if (tableA)
CodeBarre += getCharCode(chaine, i, 65);
else
CodeBarre += getCharCode(chaine, i, 75);
}
CodeBarre += "*";
for (i = 7; i <= 12; i++) {
CodeBarre += getCharCode(chaine, i, 97);
}
CodeBarre += "+";
}
return CodeBarre;
}
document.getElementById("demo").innerHTML = EAN("618110032764");
Among those issues :
Regex shouldn't be a string so no "" surrounding the expression
fromCharCode is a static method, should be used with String.fromCharCode
You should access characters from the string with the index on the loop
Here is the working demo: https://jsfiddle.net/t6r15qbd/1/
Sorry to come back to this topic, but can you give me another hand?
I can't get the encoding in txt_encoding
please, can you complete your help and help me?
function getCharCode(chaine, index, position) {return String.fromCharCode(position + parseInt(chaine.substr(index, 1)));}
function EAN(chaine) {
var code_ean13_sivop = document.getElementById("txt_ean13_sivop");
var txt_ean = document.getElementById("txt_ean");
var txt_encodage = document.getElementById("txt_encodage");
txt_ean.value = code_ean13_sivop.options[code_ean13_sivop.selectedIndex].text.substr(0, 12);
var i = 0;
var first = 0;
var checksum = 0;
var CodeBarre = "";
var tableA = 0;
if (chaine.match(/^\d{12}$/)) {
for (i = 1; i < 12; i += 2) {chaine.substr(i, 1); checksum += parseInt(chaine.substr(i, 1));}
checksum *= 3;
for (i = 0; i < 12; i += 2) {checksum += parseInt(chaine.substr(i, 1));}
chaine += (10 - checksum % 10) % 10;
CodeBarre = chaine.substr(0, 1) + getCharCode(chaine, 1, 65);
first = parseInt(chaine.substr(0, 1));
for (i = 2; i <= 6; i++) {
tableA = false;
switch (i) {
case 2: if (first >= 0 && first <= 3) tableA = true; break;
case 3: if (first == 0 || first == 4 || first == 7 || first == 8) tableA = true; break;
case 4: if (first == 0 || first == 1 || first == 4 || first == 5 || first == 9) tableA = true; break;
case 5: if (first == 0 || first == 2 || first == 5 || first == 6 || first == 7) tableA = true; break;
case 6: if (first == 0 || first == 3 || first == 6 || first == 8 || first == 9) tableA = true; break;
}
if (tableA) CodeBarre += getCharCode(chaine, i, 65); else CodeBarre += getCharCode(chaine, i, 75);
}
CodeBarre += "*";
for (i = 7; i <= 12; i++) {CodeBarre += getCharCode(chaine, i, 97); }
CodeBarre += "+";
}
return CodeBarre;
txt_encodage.value = EAN(txt_ean.value);
}
<select class="form-control form-control-sm text-monospace" id="txt_ean13_sivop" name="txt_ean13_sivop" onChange="EAN();">
<option value="" selected></option>
<option value="6181100327847">6181100327847</option>
<option value="6181100327854">6181100327854</option>
<option value="6181100328080">6181100328080</option>
<option value="6181100328097">6181100328097</option>
</select>
<input type="text" class="form-control form-control-sm text-monospace" id="txt_ean" name="txt_ean">
<input type="text" class="form-control form-control-sm text-monospace" id="txt_encodage" name="txt_encodage">

Return the first divisible number from an array

I have a function:
function findDivisibleBy(array, num) {
for (i = 0; i < array.length; i++) {
if (array[i] % num == 0 && array[i] != 0) {
return array[i]
} else {
return ('No valid number found!')
}
}
}
I must return the first number in the array that is divisible by the num paramater, and that number can't be 0. The way i'm doing it is not working.
function findDivisibleBy(array, num) {
for (i = 0; i < array.length; i++) {
if (array[i] % num == 0 && array[i] != 0) {
return array[i]
}
}
return ('No valid number found!')
}
The else part is making your function return if the condition isn't met for the first item in the array.
function findDivisibleBy(array, num) {
for (let n of array) {
if (n % num == 0 && n != 0) {
return n
}
}
return ('No valid number found!')
}
You can try this simple code:
function findDivisibleBy(array, num) {
return array.find(x => x % num === 0 && x !== 0) || 'No valid number found!'
}
Use the find method and check for null before returning value.
const findDivisibleBy = (array, num) =>
array.find((val) => val % num === 0 && val !== 0) ?? "No valid number found!";
const input = [3, 0, 4, 5, 8];
console.log(findDivisibleBy(input, 2));
console.log(findDivisibleBy(input, 9));

Can't get the right result in condition

I can't get theright result, "Weird" on stdin, 18 and 20. Everything looks good to me, however something must be off.
if (N % 2 == 1) {
console.log("Weird");
}
else if ((N % 2 == 0) && (2 >= N <= 5)) {
console.log("Not Weird");
}
else if ((N % 2 == 0) && (5 <= N <= 20)) {
console.log("Weird");
}
else if ((N % 2 == 0) && (N > 20)) {
console.log("Not Weird");
}
else {
console.log("Weird");
}
'use strict';
process.stdin.resume();
process.stdin.setEncoding('utf-8');
let inputString = '';
let currentLine = 0;
process.stdin.on('data', inputStdin => {
inputString += inputStdin;
});
process.stdin.on('end', _ => {
inputString = inputString.replace(/\s*$/, '')
.split('\n')
.map(str => str.replace(/\s*$/, ''));
main();
});
function readLine() {
return inputString[currentLine++];
}
function main() {
const N = parseInt(readLine(), 10);
if (N%2==1) {
console.log("Weird");
}
else if ((N % 2 == 0) && (2 >= N <= 5)) {
console.log("Not Weird");
}
else if ((N % 2 == 0) && (5 <= N && N <= 20)) {
console.log("Weird");
}
else if ((N % 2 == 0) && (N > 20)) {
console.log("Not Weird");
}
else{
console.log("Weird");
}
}
I ve added the whole code. In the main function, in the second else if condition, there seems to be the problem. When n is given 18 or 20, I can not get the right output which should be "Weird"
You can't be doing two conditions in the same time
if (5<=N<=20) {}
Will evaluate 5<=N first which produces either true/false which are when compared to numbers will evaluate to (1/0) respectively. Then the second part ( <= 20) will be evaluated.
Combine two conditions only with AND / OR operators
if (5 <= N && N <= 20) {}
This will solve your problem.

Want user to input number for for loop to run Javascript

Trying to have a user input a number and if they type in a string I would like to prompt them to enter a number. Seems like I got that part right but how can I get the second prompt to keep popping up until the user enters a an actual number. As of now, once the user enters a string again nothing runs after that. Would appreciate any kind of suggestions.
Here is the code:
function enterNumber(n) {
n = parseInt(prompt("Please enter a number: "));
if (isNaN(n)) {
n = parseInt(prompt("You did not enter a number. Please enter a number: "));
for(var i = 1; i <= n; i++) {
if (i % 15 === 0) {
document.write("Fizz Buzz" + "<br>");
continue;
}
else if (i % 3 === 0){
document.write("Fizz" + "<br>");
continue;
} else if (i % 5 === 0) {
document.write("Buzz" + "<br>");
continue;
}
document.write(i + "<br>");
}
}
};
enterNumber();
Use while loop until the entered is number.
function enterNumber(n) {
while (isNaN(parseInt(n))) {
n = parseInt(prompt("Please enter a number: "));
}
for (var i = 1; i <= n; i++) {
if (i % 15 === 0) {
document.write("Fizz Buzz" + "<br>");
continue;
} else if (i % 3 === 0) {
document.write("Fizz" + "<br>");
continue;
} else if (i % 5 === 0) {
document.write("Buzz" + "<br>");
continue;
}
document.write(i + "<br>");
}
};
enterNumber();
You can also shorten you code using nested Ternary operators as follow.
function enterNumber(n) {
while (isNaN(parseInt(n))) {
n = parseInt(prompt("Please enter a number: "));
}
for (var i = 1; i <= n; i++) {
var title = i % 15 === 0 ? 'Fizz Buzz' : i % 3 === 0 ? 'Fizz' : i % 5 === 0 ? 'Buzz' : i;
document.write(title + "<br>");
}
};
enterNumber();
Try like this
function enterNumber(n) {
while (isNaN(n))
n = parseInt(prompt("You did not enter a number. Please enter a number: "));
for (var i = 1; i <= n; i++) {
if (i % 15 === 0) {
document.write("Fizz Buzz" + "<br>");
continue;
} else if (i % 3 === 0) {
document.write("Fizz" + "<br>");
continue;
} else if (i % 5 === 0) {
document.write("Buzz" + "<br>");
continue;
}
document.write(i + "<br>");
}
};
enterNumber();

FizzBuzz program (details given) in Javascript [closed]

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Can someone please correct this code of mine for FizzBuzz? There seems to be a small mistake. This code below prints all the numbers instead of printing only numbers that are not divisible by 3 or 5.
Write a program that prints the numbers from 1 to 100. But for multiples of three, print "Fizz" instead of the number, and for the multiples of five, print "Buzz". For numbers which are multiples of both three and five, print "FizzBuzz".
function isDivisible(numa, num) {
if (numa % num == 0) {
return true;
} else {
return false;
}
};
function by3(num) {
if (isDivisible(num, 3)) {
console.log("Fizz");
} else {
return false;
}
};
function by5(num) {
if (isDivisible(num, 5)) {
console.log("Buzz");
} else {
return false;
}
};
for (var a=1; a<=100; a++) {
if (by3(a)) {
by3(a);
if (by5(a)) {
by5(a);
console.log("\n");
} else {
console.log("\n");
}
} else if (by5(a)) {
by5(a);
console.log("\n");
} else {
console.log(a+"\n")
}
}
for (let i = 1; i <= 100; i++) {
let out = '';
if (i % 3 === 0) out += 'Fizz';
if (i % 5 === 0) out += 'Buzz';
console.log(out || i);
}
/*Write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz”*/
var str="",x,y,a;
for (a=1;a<=100;a++)
{
x = a%3 ==0;
y = a%5 ==0;
if(x)
{
str+="fizz"
}
if (y)
{
str+="buzz"
}
if (!(x||y))
{
str+=a;
}
str+="\n"
}
console.log(str);
Your functions return falsy values no matter what, but will print anyway. No need to make this overly complicated.
fiddle: http://jsfiddle.net/ben336/7c9KN/
Was fooling around with FizzBuzz and JavaScript as comparison to C#.
Here's my version, heavily influenced by more rigid languages:
function FizzBuzz(aTarget) {
for (var i = 1; i <= aTarget; i++) {
var result = "";
if (i%3 === 0) result += "Fizz";
if (i%5 === 0) result += "Buzz";
if (result.length ===0) result = i;
console.log(result);
}
}
I like the structure and ease of read.
Now, what Trevor Dixon cleverly did is relay on the false-y values of the language (false , null , undefined , '' (the empty string) , 0 and NaN (Not a Number)) to shorten the code.
Now, the if (result.length ===0) result = i; line is redundant and the code will look like:
function FizzBuzz(aTarget) {
for (var i = 1; i <= aTarget; i++) {
var result = "";
if (i%3 === 0) result += "Fizz";
if (i%5 === 0) result += "Buzz";
console.log(result || i);
}
}
Here we relay on the || operator to say : "if result is false, print the iteration value (i)". Cool trick, and I guess I need to play more with JavaScript in order to assimilate this logic.
You can see other examples (from GitHub) that will range from things like :
for (var i=1; i <= 20; i++)
{
if (i % 15 == 0)
console.log("FizzBuzz");
else if (i % 3 == 0)
console.log("Fizz");
else if (i % 5 == 0)
console.log("Buzz");
else
console.log(i);
}
No variables here, and just check for division by 15,3 & 5 (my above one only divides by 3 & 5, but has an extra variable, so I guess it's down to microbenchmarking for those who care, or style preferences).
To:
for(i=0;i<100;)console.log((++i%3?'':'Fizz')+(i%5?'':'Buzz')||i)
Which does it all in on line, relaying on the fact that 0 is a false value, so you can use that for the if-else shorthanded version (? :), in addition to the || trick we've seen before.
Here's a more readable version of the above, with some variables:
for (var i = 1; i <= 100; i++) {
var f = i % 3 == 0, b = i % 5 == 0;
console.log(f ? b ? "FizzBuzz" : "Fizz" : b ? "Buzz" : i);
}
All in all, you can do it in different ways, and I hope you picked up some nifty tips for use in JavaScript :)
.fizz and .buzz could be CSS classes, no? In which case:
var n = 0;
var b = document.querySelector("output");
window.setInterval(function () {
n++;
b.classList[n%3 ? "remove" : "add"]("fizz");
b.classList[n%5 ? "remove" : "add"]("buzz");
b.textContent = n;
}, 500);
output.fizz:after {
content: " fizz";
color:red;
}
output.buzz:after {
content: " buzz";
color:blue;
}
output.fizz.buzz:after {
content: " fizzbuzz";
color:magenta;
}
<output>0</output>
With ternary operator it is much simple:
for (var i = 0; i <= 100; i++) {
str = (i % 5 == 0 && i % 3 == 0) ? "FizzBuzz" : (i % 3 == 0 ? "Fizz" : (i % 5 == 0) ? "Buzz" : i);
console.log(str);
}
for(i = 1; i < 101; i++) {
if(i % 3 === 0) {
if(i % 5 === 0) {
console.log("FizzBuzz");
}
else {
console.log("Fizz");
}
}
else if(i % 5 === 0) {
console.log("Buzz");
}
else {
console.log(i)
}
}
In your by3 and by5 functions, you implicitly return undefined if it is applicable and false if it's not applicable, but your if statement is testing as if it returned true or false. Return true explicitly if it is applicable so your if statement picks it up.
As an ES6 generator: http://www.es6fiddle.net/i9lhnt2v/
function* FizzBuzz() {
let index = 0;
while (true) {
let value = ''; index++;
if (index % 3 === 0) value += 'Fizz';
if (index % 5 === 0) value += 'Buzz';
yield value || index;
}
}
let fb = FizzBuzz();
for (let index = 0; index < 100; index++) {
console.log(fb.next().value);
}
Codeacademy sprang a FizzBuzz on me tonight. I had a vague memory that it was "a thing" so I did this. Not the best way, perhaps, but different from the above:
var data = {
Fizz:3,
Buzz:5
};
for (var i=1;i<=100;i++) {
var value = '';
for (var k in data) {
value += i%data[k]?'':k;
}
console.log(value?value:i);
}
It relies on data rather than code. I think that if there is an advantage to this approach, it is that you can go FizzBuzzBing 3 5 7 or further without adding additional logic, provided that you assign the object elements in the order your rules specify. For example:
var data = {
Fizz:3,
Buzz:5,
Bing:7,
Boom:11,
Zing:13
};
for (var i=1;i<=1000;i++) {
var value = '';
for (var k in data) {
value += i%data[k]?'':k;
}
console.log(value?value:i);
}
This is what I wrote:
for (var num = 1; num<101; num = num + 1) {
if (num % 5 == 0 && num % 3 == 0) {
console.log("FizzBuzz");
}
else if (num % 5 == 0) {
console.log("Buzz");
}
else if (num % 3 == 0) {
console.log("Fizz");
}
else {
console.log(num);
}
}
for (var i = 1; i <= 100; i++) {
if (i % 3 === 0 && i % 5 === 0) console.log("FizzBuzz");
else if (i%3 === 0) console.log("Fizz");
else if (i%5 === 0) console.log("Buzz");
else console.log(i);
}
One of the easiest way to FizzBuzz.
Multiple of 3 and 5, at the same time, means multiple of 15.
Second version:
for (var i = 1; i <= 100; i++) {
if (i % 15 === 0) console.log("FizzBuzz");
else if (i%3 === 0) console.log("Fizz");
else if (i%5 === 0) console.log("Buzz");
else console.log(i);
}
In case someone is looking for other solutions: This one is a pure, recursive, and reusable function with optionally customizable parameter values:
const fizzBuzz = (from = 1, till = 100, ruleMap = {
3: "Fizz",
5: "Buzz",
}) => from > till || console.log(
Object.keys(ruleMap)
.filter(number => from % number === 0)
.map(number => ruleMap[number]).join("") || from
) || fizzBuzz(from + 1, till, ruleMap);
// Usage:
fizzBuzz(/*Default values*/);
The from > till is the anchor to break the recursion. Since it returns false until from is higher than till, it goes to the next statement (console.log):
Object.keys returns an array of object properties in the given ruleMap which are 3 and 5 by default in our case.
Then, it iterates through the numbers and returns only those which are divisible by the from (0 as rest).
Then, it iterates through the filtered numbers and outputs the saying according to the rule.
If, however, the filter method returned an empty array ([], no results found), it outputs just the current from value because the join method at the end finally returns just an empty string ("") which is a falsy value.
Since console.log always returns undefined, it goes to the next statement and calls itself again incrementing the from value by 1.
A Functional version of FizzBuzz
const dot = (a,b) => x => a(b(x));
const id = x => x;
function fizzbuzz(n){
const f = (N, m) => n % N ? id : x => _ => m + x('');
return dot(f(3, 'fizz'), f(5, 'buzz')) (id) (n);
}
for more options in the above replace dot with dots as below
const dots = (...a) => f0 => a.reduceRight((acc, f) => f(acc), f0);
function fizzbuzz(n){
const f = (N, m) => n % N ? id : x => _ => m + x('');
return dots(f(3, 'fizz'), f(5, 'buzz'), f(7, 'bam')) (id) (n);
}
Reference: FizzBuzz in Haskell by Embedding a Domain-Specific Language
by Maciej Piro ́g
for (i=1; i<=100; i++) {
output = "";
if (i%5==0) output = "buzz";
if (i%3==0) output = "fizz" + output;
if (output=="") output = i;
console.log(output);
}
Functional style! JSBin Demo
// create a iterable array with a length of 100
// and map every value to a random number from 1 to a 100
var series = Array.apply(null, Array(100)).map(function() {
return Math.round(Math.random() * 100) + 1;
});
// define the fizzbuzz function which takes an interger as input
// it evaluates the case expressions similar to Haskell's guards
var fizzbuzz = function (item) {
switch (true) {
case item % 15 === 0:
console.log('fizzbuzz');
break;
case item % 3 === 0:
console.log('fizz');
break;
case item % 5 === 0:
console.log('buzz');
break;
default:
console.log(item);
break;
}
};
// map the series values to the fizzbuzz function
series.map(fizzbuzz);
Another solution, avoiding excess divisions and eliminating excess spaces between "Fizz" and "Buzz":
var num = 1;
var FIZZ = 3; // why not make this easily modded?
var BUZZ = 5; // ditto
var UPTO = 100; // ditto
// and easily extended to other effervescent sounds
while (num < UPTO)
{
var flag = false;
if (num % FIZZ == 0) { document.write ("Fizz"); flag = true; }
if (num % BUZZ == 0) { document.write ("Buzz"); flag = true; }
if (flag == false) { document.write (num); }
document.write ("<br>");
num += 1;
}
If you're using using jscript/jsc/.net, use Console.Write(). If you're using using Node.js, use process.stdout.write(). Unfortunately, console.log() appends newlines and ignores backspaces, so it's unusable for this purpose. You could also probably append to a string and print it. (I'm a complete n00b, but I think (ok, hope) I've been reasonably thorough.)
"Whaddya think, sirs?"
check this out!
function fizzBuzz(){
for(var i=1; i<=100; i++){
if(i % 3 ===0 && i % 5===0){
console.log(i+' fizzBuzz');
} else if(i % 3 ===0){
console.log(i+' fizz');
} else if(i % 5 ===0){
console.log(i+' buzz');
} else {
console.log(i);
}
}
}fizzBuzz();
Slightly different implementation.
You can put your own argument into the function. Can be non-sequential numbers like [0, 3, 10, 1, 4]. The default set is only from 1-15.
function fizzbuzz (set) {
var set = set ? set : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
var isValidSet = set.map((element) => {if (typeof element !== 'number') {return false} else return true}).indexOf(false) === -1 ? true : false
var gotFizz = (n) => {if (n % 3 === 0) {return true} else return false}
var gotBuzz = (n) => {if (n % 5 === 0) {return true} else return false}
if (!Array.isArray(set)) return new Error('First argument must an array with "Number" elements')
if (!isValidSet) return new Error('The elements of the first argument must all be "Numbers"')
set.forEach((n) => {
if (gotFizz(n) && gotBuzz(n)) return console.log('fizzbuzz')
if (gotFizz(n)) return console.log('fizz')
if (gotBuzz(n)) return console.log('buzz')
else return console.log(n)
})
}
var num = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
var runLoop = function() {
for (var i = 1; i<=num.length; i++) {
if (i % 5 === 0 && i % 3 === 0) {
console.log("FizzBuzz");
}
else if (i % 5 === 0) {
console.log("Buzz");
}
else if (i % 3 === 0) {
console.log("Fizz");
}
else {
console.log(i);
}
}
};
runLoop();
Just want to share my way to solve this
for (i = 1; i <= 100; i++){
if (i % 3 === 0 && i % 5 === 0) {
console.log('fizzBuzz');
} else if (i % 3 === 0) {
console.log('fizz');
} else if (i % 5 === 0){
console.log('buzz');
} else {
console.log(i);
}
}
var limit = prompt("Enter the number limit");
var n = parseInt(limit);
var series = 0;
for(i=1;i<n;i++){
series = series+" " +check();
}
function check() {
var result;
if (i%3==0 && i%5==0) { // check whether the number is divisible by both 3 and 5
result = "fizzbuzz "; // if so, return fizzbuzz
return result;
}
else if (i%3==0) { // check whether the number is divisible by 3
result = "fizz "; // if so, return fizz
return result;
}
else if (i%5==0) { // check whether the number is divisible by 5
result = "buzz "; // if so, return buzz
return result;
}
else return i; // if all the above conditions fail, then return the number as it is
}
alert(series);
Thats How i did it :
Not the best code but that did the trick
var numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
for(var i = 0 ; i <= 19 ; i++){
var fizz = numbers[i] % 3 === 0;
var buzz = numbers[i] % 5 === 0;
var fizzBuzz = numbers[i] % 5 === 0 && numbers[i] % 3 === 0;
if(fizzBuzz){
console.log("FizzBuzz");
} else if(fizz){
console.log("Fizz");
} else if(buzz){
console.log("Buzz");
} else {
console.log(numbers[i]);
}
}
As much as this is easy logic it can be a daunting task for beginners. Below is my solution to the FizzBuzz problem:
let i = 1;
while(i<=100){
if(i % 3 ==0 && i % 5 == 0){
console.log('FizzBuzz');
}
else if(i % 3 == 0){
console.log('Fizz');
}
else if(i % 5 == 0){
console.log('Buzz');
}
else{
console.log(i);
}
i++;
}
considering performance and readability, please find my take on this problem
way 1: instead of doing a math modules operation in an if loop, which results in performing 3 times taking it a step above reduces the overhead
function fizzBuzz(n) {
let count =0;
let x = 0;
let y = 0;
while(n!==count)
{
count++;
x = count%3;
y = count%5;
if(x === 0 && y ===0)
{
console.log("fizzbuzz");
}
else if(x === 0)
{
console.log("fizz");
}
else if(y === 0)
{
console.log("buzz");
}
else
{
console.log(count);
}
}
}
fizzBuzz(15);
way 2: condensing the solution
function fizzBuzz(n) {
let x = 0;
let y = 0;
for (var i = 1; i <= n; i++) {
var result = "";
x = i%3;
y = i%5;
if (x === 0 && y === 0) result += "fizzbuzz";
else if (x === 0) result += "fizz";
else if (y === 0) result += "buzz";
console.log(result || i);
}
}
fizzBuzz(5)
Here's my favorite solution. Succinct, functional & fast.
const oneToOneHundred = Array.from({ length: 100 }, (_, i) => i + 1);
const fizzBuzz = (n) => {
if (n % 15 === 0) return 'FizzBuzz';
if (n % 3 === 0) return 'Fizz';
if (n % 5 === 0) return 'Buzz';
return n;
};
console.log(oneToOneHundred.map((i) => fizzBuzz(i)).join('\n'));
function fizzBuzz(n) {
for (let i = 1; i < n + 1; i++) {
if (i % 15 == 0) {
console.log("fizzbuzz");
} else if (i % 3 == 0) {
console.log("fizz");
} else if (i % 5 == 0) {
console.log("buzz");
} else {
console.log(i);
}
}
}
fizzBuzz(15);
Different functional style -- naive
fbRule = function(x,y,f,b,z){return function(z){return (z % (x*y) == 0 ? f+b: (z % x == 0 ? f : (z % y == 0 ? b: z))) }}
range = function(n){return Array.apply(null, Array(n)).map(function (_, i) {return i+1;});}
range(100).map(fbRule(3,5, "fizz", "buzz"))
or, to incorporate structures as in above example: ie [[3, "fizz"],[5, "buzz"], ...]
fbRule = function(fbArr,z){
return function(z){
var ed = fbArr.reduce(function(sum, unit){return z%unit[0] === 0 ? sum.concat(unit[1]) : sum }, [] )
return ed.length>0 ? ed.join("") : z
}
}
range = function(n){return Array.apply(null, Array(n)).map(function (_, i) {return i+1;});}
range(100).map(fbRule([[3, "fizz"],[5, "buzz"]]))
OR, use ramda [from https://codereview.stackexchange.com/questions/108449/fizzbuzz-in-javascript-using-ramda ]
var divisibleBy = R.curry(R.compose(R.equals(0), R.flip(R.modulo)))
var fizzbuzz = R.map(R.cond([
[R.both(divisibleBy(3), divisibleBy(5)), R.always('FizzBuzz')],
[divisibleBy(3), R.aklways('Fizz')],
[divisibleBy(5), R.always('Buzz')],
[R.T, R.identity]
]));
console.log(fizzbuzz(R.range(1,101)))

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