Third if-clause can't be reached - javascript

This probably has an easy solution, but I simply don't see it at the moment.
I have three if-clauses that ashould be activated based on the length of an array. The first two ones seem to work fine, but for some odd reason I can't activate the third one (arr.length === 3). Right before the if clauses I have tried an alert to test whether it gives the right length of the array and it does.
function calculateDistances() {
var arr = [];
arr.push(posM, posL, posR);
alert(arr[1])
for (var i = 0; i < arr.length; i++) {
if (!arr[i]) {
arr.splice(i,1)
}
}
alert(arr.length)
if (arr.length === 0 || 1) {
return true;
}
else if (arr.length === 2 ) {
var diameter = calculateDiameter(arr[0], arr[1])
if (diameter > minDistance) {
return false;
}
else {
return true;
}
}
else if (arr.length === 3) {
alert("hello")
var diameter1 = calculateDiameter(arr[0], arr[1]);
var diameter2 = calculateDiameter(arr[0], arr[2]);
var diameter3 = calculateDiameter(arr[1], arr[3]);
if (diameter1 && diameter2 && diameter3 < minDistance) {
return true
}
else{
return false
}
}
}

Nor can you activate the second.
There's a bug here: if (arr.length === 0 || 1) {
The 1 casts to true.
Perhaps you meant: if (arr.length === 0 || arr.length === 1) {

You need this:
if (arr.length === 0 || arr.length === 1) {
The way you put it, it is equal to
if ((arr.length === 0) || true) {
which is always true.

I think what you are looking for is below condition in the first if condition
if (arr.length === 0 || arr.length === 1) {
return true;
}
this checks whether the length of the array is 1 or it's 0. Your first if condition is always true as it has 1 which is true.

(arr.length === 0 || 1)
is always true.
You could usethis instead
if (arr.length <= 1)
{
return true;
}

Related

can't pass challenge if code not optimized further

I can't pass this coding challenge: Code Challenge: https://www.codewars.com/kata/550f22f4d758534c1100025a/train/javascript
because my code is TOO SLOW. I'm not sure which part of my code is causing the problem. That's why I need help to optimize it.
function dirReduc(arr){
if (arr.length === 0 || arr.length === 1) return [];
let lengthTracker = arr.length;
for (let i = 0; i < arr.length; i++) {
if (lengthTracker > arr.length) {
lengthTracker = arr.length;
i = 0;
}
switch(arr[i]) {
case "NORTH":
arr[i-1] === "SOUTH"? arr.splice(i-1,2) :
arr[i+1] === "SOUTH"? arr.splice(i,2) : null
break;
case "SOUTH":
arr[i-1] === "NORTH"? arr.splice(i-1,2) :
arr[i+1] === "NORTH"? arr.splice(i,2) : null
break;
case "EAST":
arr[i-1] === "WEST"? arr.splice(i-1,2) :
arr[i+1] === "WEST"? arr.splice(i,2) : null
break;
case "WEST":
arr[i-1] === "EAST"? arr.splice(i-1,2) :
arr[i+1] === "EAST"? arr.splice(i,2) : null
break;
}
i===arr.length-1? i=0:null
}
return arr;
}
Splicing can be expensive. We can form a recurrence that assumes the function has already correctly reduced the next part of the list:
function matches(a, b){
return (a == "NORTH" && b == "SOUTH") ||
(b == "NORTH" && a == "SOUTH") ||
(a == "EAST" && b == "WEST") ||
(b == "EAST" && a == "WEST");
}
function f(A, i=0){
if (i == A.length)
return [];
const rest = f(A, i + 1);
const [head,...tail] = rest;
if (head){
if (matches(A[i], head))
return tail;
else
return [A[i]].concat(rest);
}
return [A[i]];
}
I see several problems with this. First, as I mentioned in the comments, splicing long arrays is costly and makes your algorithm O(n^2). Simple and faster would be to use a read-point and a write-point to copy the elements into itself one cell at a time, just skipping over the annihilations and then use splice once at the end to trim the uncopied cells off the end of the array. This would make it O(n).
Secondly, your code is looking both forward and backward for matches which is both unnecessary and can be confusing. Finally, there's no need for a switch (...) as all of the branches do the same thing.
Here is how I would use your code to accomplish this, changing the things mentioned above and noted in the comments.
function dirReduc(arr){
if (arr.length === 0 || arr.length === 1) return [];
let lengthTracker = 0; // the write-point
for(let i = 0; i < arr.length; i++) { // i is the read-point
if(lengthTracker == 0) {
// if no output, copy readpoint to write-point and advance
arr[lengthTracker++] = arr[i];
} else {
// replaces switch()
if (((arr[lengthTracker-1] === "NORTH") && (arr[i] === "SOUTH"))
|| ((arr[lengthTracker-1] === "SOUTH") && (arr[i] === "NORTH"))
|| ((arr[lengthTracker-1] === "EAST") && (arr[i] === "WEST"))
|| ((arr[lengthTracker-1] === "WEST") && (arr[i] === "EAST"))) {
lengthTracker--; // annihilate by decrementing the writepoint
} else {
// copy readpoint to writepoint and advance
arr[lengthTracker++] = arr[i];
}
}
}
//trim the array to only include what was written
arr.splice(lengthTracker);
return arr;
}

First Unique Character in a String Leetcode - using pointers (javascript)

Given a string, find the first non-repeating character in it and return its index. If it doesn't exist, return -1.
leetcode question
"cc" // -1
"ccdd" // -1
"leetcode" // 1
"loveleetcode" // 2
"abcabd" // 2
"thedailybyte" // 1
"developer" // 0
My approach passed all the test cases except the 2nd test case "ccdd". I am expecting -1 but receiving 4. Not sure why.
var firstUniqChar = function(s) {
if(!s || s.length === 0 ) return -1
else if(s.length === 1) return 0
let pointer1 = 0
let pointer2 = pointer1 + 1
if(s.length > 2){
while(pointer2 <= s.length - 1){
if(s[pointer1] !== s[pointer2])
pointer2++
else if(s[pointer1] === s[pointer2])
pointer1++
}
return pointer1
}
return -1
};
this is too old, but still can be too much shorter:
const firstUniqChar = (_str) => {
for (let i= 0; i < _str.length; i+= 1) {
if (_str.indexOf(_str[i]) === _str.lastIndexOf(_str[i])) return i+1;
}
return -1;
}

Finding if a given set of parentheses is valid or not

Below is my code, it works for some strings but not for all.
Ex: "()()()()()((" expected is false, my code returns true.
function validParentheses(parens){
var stack = [];
parens.split('').map((cur, index) =>{
if(stack.length === 0 || stack[index-1] === cur) stack.push(cur);
else stack.pop();
});
return stack.length > 0 ? false : true;
}
stack[index - 1] will be valid so long as you push every iteration. In the case that you pop an element, the incrementing index will always be out of bounds.
Change it to stack.length - 1 to always get the last element, regardless of what is pushed or popped.
For every '(' there must be a exactly one ')'. So you need a counter to see that there is an exact match
function validParentheses(parens){
const chars = parens.split('');
const numChars = chars.length;
let ii;
let numOpenParens = 0;
for (ii = 0; ii < numChars; ii += 1) {
curChar = chars[ii];
numOpenParens += curChar == '(' ? 1 : -1;
// return false if there is one too many closed parens
if (numOpenParens < 0) {
return false;
}
}
// return true only if all parens have been closed
return numOpenParens === 0;
}
For case when stack's length is greater than 0:
if top of the stack is equal to current iterated parenthesis, push that to stack
else pop the stack
function validParentheses(parens) {
var stack = []
parens.split("").forEach((cur) => {
if (stack.length > 0) {
if (stack[stack.length - 1] === cur) {
stack.push(cur)
} else {
stack.pop()
}
} else {
stack.push(cur)
}
})
return stack.length > 0 ? false : true
}
console.log(validParentheses("()()()()()(("))
console.log(validParentheses("()()()()()()"))
console.log(validParentheses("((()))"))
console.log(validParentheses("((())))"))
in stack[index-1] === cur
you are comparing if the char isn't the same like the one stored in the stack, so )( opposite parens will be valid
you can try do something like this
function validParentheses(parens) {
if (parens % 2 == 1) return false;
for (let i = 0; i < parens.length; i++) {
const char = parens[i];
if (char == "(") {
if (parens[i + 1] == ")") {
i++;
} else {
return false
}
} else {
return false
}
}
return true;
}
You need to check the last added value as well, because an unresolves closing bracket should remain in he stack.
BTW, Array#forEach is the method of choice, because Array#map returns a new array, which is not used here.
function validParentheses(parens) {
var stack = [];
parens.split('').forEach((cur, index) => {
if (cur === ')' && stack[stack.length - 1] === '(') stack.pop();
else stack.push(cur);
});
return !stack.length;
}
console.log(validParentheses("(())()"));
console.log(validParentheses("()()()()()(("));
console.log(validParentheses("))(())"));

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));

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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