Develop Reference

Javascript Happy Number detection error

My code works, but does not count correctly. For example, it is considering 22 and 208 as happy numbers, but they are not. The multiplication results are also wrong: after 42, it is not 20, but 45
function isHappy(x){
var b = x,
str = b.toString(),
strlen = str.length,
a = [],
dejavue = [],
sum = 0,
isOne = false,
result,
whilenumber = -1;
while (isOne == false){
whilenumber++;
for (var i=0;i<strlen;i++){
var ms = parseInt(str[i]);
a[i] = ms*ms;
//a[i] = Math.pow(ms, 2);
}
if (a.length>1){
for (var i=0;i<a.length;i++){
sum = sum + a[i];
}
} else { sum = a[0];}
var h = dejavue.indexOf(sum);
if (h==-1) {dejavue[whilenumber] = sum;} else {
result = " is not happy";
break;
}
if (sum!=1){
b = sum;
sum = 0;
str = b.toString();
strlen = str.length;
}else{
result = " is happy";
isOne = true;
}
}
return dejavue+"_"+x+result;
}

var l = 50;
a=n=>{if((r=[...(n+'')].map(t=>t*t).reduce((a,b)=>a+b))==1){alert('happy')}else{if(l){l--;a(r)}else{alert('unhappy')}}
}
a(22)
/* explanation */
// how many times we want our function to run ultil decide is a unhappy number. This is to avoid infinite loop
var l = 50;
// parameter n = positive integer
n=>{
if(
//convert integer to string and split each char into an array
(r=[...(n+'')]
// get each separated number its square
.map(t=>t*t)
// sum squares
.reduce((a,b)=>a+b))
//if result is equal to one
== 1
){
// we alert happy
alert('happy')
}else{
//otherwise we check if our loop counter (l) aint done looping
if(l){
// if it does, then we substract 1 from the counter and call this same function with the result of the above operation
l--;a(r)
}else{
// otherwise is a unhappy number
alert('unhappy')
}
}
}

You could change the code a bit by using less variables and use an exit by break.
basically you need not to iterate for getting the square and then another loop for getting the sum of the squares. The use of the array a with its former values was the reason for the wrong results.
function isHappy(x) {
var b = x,
str,
dejavue = [],
sum,
result;
while (true) {
str = b.toString();
sum = 0;
for (var i = 0; i < str.length; i++) {
sum += str[i] * str[i];
}
if (dejavue.indexOf(sum) !== -1) {
result = " is not happy";
break;
}
dejavue.push(sum);
if (sum === 1) {
result = " is happy";
break;
}
b = sum;
}
return dejavue.join('>') + ": " + x + result;
}
console.log(isHappy(22));
console.log(isHappy(7));

function sumDigits; How do I get this function to work with a negative number?

Question:
Write a function called sumDigits.
Given a number, sumDigits returns the sum of all its digits.
var output = sumDigits(1148);
console.log(output); // --> 14
If the number is negative, the first digit should count as negative.
var output = sumDigits(-316);
console.log(output); // --> 4
This is what I currently have coded and it works for positive values but I can't wrap my head around how to tackle the problem when given a negative value. When -316 is put into the function, NaN is returned and I understand that when I toString().split('') the number, this is what is returned: ['-', '3', '1', '6']. How do I deal with combining index 0 and 1?
function sumDigits(num) {
var total = 0;
var newString = num.toString().split('');
for (var i = 0; i < newString.length; i ++) {
var converted = parseInt(newString[i]);
total += converted;
}
return total;
}
sumDigits(1148);
Any hints on what methods I should be using? and is there a smarter way to even look at this?

This should do it:
function sumDigits(num) {
var total = 0;
var newString = num.toString().split('');
for (var i = 0; i < newString.length; i ++) {
if(newString[i]==='-') { //check to see if the first char is -
i++; //if it is, lets move to the negative number
var converted = parseInt(newString[i]); // parse negative number
total -= converted; // subtract value from total
continue; // move to the next item in the loop
}
var converted = parseInt(newString[i]);
total += converted;
}
return total;
}
console.log(sumDigits(-316));

You could always use String#replace with a function as a parameter:
function sumDigits (n) {
var total = 0
n.toFixed().replace(/-?\d/g, function (d) {
total += +d
})
return total
}
console.log(sumDigits(-1148)) //=> 14

One way to do this, is to do a split that will keep the minus and the first digit together, not split.
You can do that with a regular expression, and use match instead of split:
var newString = num.toString().match(/-?\d/g);
function sumDigits(num) {
var total = 0;
var newString = num.toString().match(/-?\d/g);
for (var i = 0; i < newString.length; i++) {
var converted = parseInt(newString[i]);
total += converted;
}
return total;
}
var result = sumDigits(-316);
console.log(result);
In a bit shorter version, you could use map and reduce, like this:
function sumDigits(num) {
return String(num).match(/-?\d/g).map(Number).reduce( (a, b) => a+b );
}
console.log(sumDigits(-316));

Is there a smarter way to even look at this?
You can avoid the conversion from number to string and back by using the modulo operator to extract the last digit. Repeat this step until you got all digits:
function sumDigits(num) {
let total = 0, digit = 0;
while (num != 0) {
total += digit = num % 10;
num = (num - digit) * 0.1;
}
return total < 0 ? digit + digit - total : total;
}
console.log(sumDigits(-316)); // 4
console.log(sumDigits(1148)); // 14
console.log(sumDigits(Number.MAX_SAFE_INTEGER)); // 76

function sumDigits(num) {
let string = num.toString();
let zero = 0;
let total = 0;
for (var i = 0; i < string.length; i++) {
if (Math.sign(num) === 1) {
total = zero += Number(string[i]);
} else {
for (var i = 2; i < string.length; i++) {
total = (zero += Number(string[i])) - Number(string[1]);
}
}
}
return total;
}

A while loop to add the digits of a multi-digit number together? (Javascript)

I need to add the digits of a number together (e.g. 21 is 2+1) so that the number is reduced to only one digit (3). I figured out how to do that part.
However,
1) I may need to call the function more than once on the same variable (e.g. 99 is 9+9 = 18, which is still >= 10) and
2) I need to exclude the numbers 11 and 22 from this function's ambit.
Where am I going wrong below?
var x = 123;
var y = 456;
var z = 789;
var numberMagic = function (num) {
var proc = num.toString().split("");
var total = 0;
for (var i=0; i<proc.length; i++) {
total += +proc[i];
};
};
while(x > 9 && x != 11 && x != 22) {
numberMagic(x);
};
} else {
xResult = x;
};
console.log(xResult);
//repeat while loop for y and z

Here are the problems with your code
var x = 123;
var y = 456;
var z = 789;
var numberMagic = function (num) {
var proc = num.toString().split("");
var total = 0;
for (var i=0; i<proc.length; i++) {
total += +proc[i]; // indentation want awry
}; // don't need this ; - not a show stopper
// you're not returning anything!!!!
};
while(x > 9 && x != 11 && x != 22) {
numberMagic(x);
}; // ; not needed
// because x never changes, the above while loop would go on forever
} else { // this else has no if
xResult = x; // even if code was right, x remains unchanged
};
console.log(xResult);
Hope that helps in some way
Now - here's a solution that works
var x = 123;
var y = 456;
var z = 789;
var numberMagic = function (num) {
while (num > 9) {
if (num == 11 || num == 22) {
return num;
}
var proc = num.toString().split("");
num = proc.reduce(function(previousInt, thisValueString) {
return previousInt + parseInt(thisValueString);
}, 0);
}
return num;
}
console.log(numberMagic(x));
console.log(numberMagic(y));
console.log(numberMagic(z));

I'm not sure to understand what you want..
with this function you reduce any number to one single digit
while(num > 9){
if(num == 11 || num == 22) return;
var proc = num.toString();
var sum = 0;
for(var i=0; i<proc.length; i++) {
sum += parseInt(proc[i]);
}
num = sum;
}
is it what you are looking at?

I wrote an example at Jsfiddle that you can turn any given number into a single digit:
Example input: 551
array of [5, 5, 1] - add last 2 digits
array of [5, 6] - add last 2 digits
array of [1, 1] - add last 2 digits
array of [2] - output
Here is the actual code:
var number = 1768;
var newNumber = convertToOneDigit(number);
console.log("New Number: " + newNumber);
function convertToOneDigit(number) {
var stringNumber = number.toString();
var stringNumberArray = stringNumber.split("");
var stringNumberLength = stringNumberArray.length;
var tmp;
var tmp2;
var tmp3;
console.log("Array: " + stringNumberArray);
console.log("Array Length: " + stringNumberLength);
while (stringNumberLength > 1) {
tmp = parseInt(stringNumberArray[stringNumberLength - 1]) + parseInt(stringNumberArray[stringNumberLength - 2]);
stringNumberArray.pop();
stringNumberArray.pop();
tmp2 = tmp.toString();
if (tmp2.length > 1) {
tmp3 = tmp2.split("");
for (var i = 0; i < tmp3.length; i++) {
stringNumberArray.push(tmp3[i]);
}
} else {
stringNumberArray.push(tmp2);
}
stringNumberLength = stringNumberArray.length;
console.log("Array: " + stringNumberArray);
console.log("Array Length: " + stringNumberLength);
}
return stringNumberArray[0];
}

function addDigits(n) {
let str = n.toString().split('');
let len = str.length;
let add,
acc = 0;
for (i=0; i<=len-1; i++) {
acc += Number(str[i]);
}
return acc;
}
console.log( addDigits(123456789) ); //Output: 45
Just make it a While loop, remember a While loops it's just the same as a For loop, only you add the counter variable at the end of the code, the same way you can do with a Do{code}while(condition) Only need to add a counter variable at the end and its gonna be the same. Only that the variable its global to the loop, I mean comes from the outside.
Ej.
let i = 0; //it's global to the loop, ( wider scope )
while (i<=x) {
//Code line;
//Code line;
//Code line;
//Code line;
i++
}
Now this is working with an outside variable and it's NOT recommended.. unless that var its local to a Function.

Please look at the this solution also
var x = 123;
var y = 456;
var z = 789;
var numberMagic = function (num) {
var total = 0;
while (num != 0) {
total += num % 10;
num = parseInt(num / 10);
}
console.log(total);
if (total > 9)
numberMagic(total);
else
return total;
}
//Call first time function
numberMagic(z);

JavaScript Function Arrays

How would I use a function that returns the sum of a given array while getting the sum of the even numbers and sum the odd numbers? I'm not understanding how that is done. Can someone please explain a little more in depth?
Here is my entire code:
function main()
{
var evenNum = 0;
//need a total Even count
var oddNum = 0;
//need a total Odd count
var counter = 1;
var num = 0;
function isOdd(x) {
if ((num % 2) == 0)
{
return false;
}
else
{
return true;
}
}
function isEven(x) {
if ((num % 2) == 0)
{
return false;
}
else
{
return true;
}
}
for (counter = 1; counter <= 100; counter++)
{
num = Math.floor(1 + Math.random() * (100-1));
var total = 0;
for(var j = 0; j < length; j++)
total += a[j];//Array?
console.log(num);
console.log("The count of even number is " + evenNum);
console.log("The count of odd number is " + oddNum);
return 0;
}
main()

If I understand your question correctly, you need a function that returns two values, one for the sum of even numbers and one for the sum of odd numbers. It's not clear if you use even/odd referring to the index of the array or the values in array.
In both cases you can return an object that contains both values:
function sum(array) {
var evenSum = 0;
var oddSum = 0;
...calculate...
var res = {};
res.evenSum = evenSum;
res.oddSum = oddSum;
return res;
}
Hope this will help

Find the largest prime factor with Javascript

Thanks for reading. Pretty new to Javascript and programming in general.
I'm looking for a way to return the largest prime factor of a given number. My first instinct was to work with a while loop that counts up and finds prime factors of the number, storing the factors in an array and resetting each time it finds one. This way the last item in the array should be the largest prime factor.
var primerizer = function(input){
var factors = [];
var numStorage = input
for (x=2; numStorage != 1; x++){ // counter stops when the divisor is equal to the last number in the
// array, meaning the input has been fully factorized
if (result === 0) { // check if the number is prime; if it is not prime
factors.push(x); // add the divisor to the array of prime numbers
numStorage = numStorage/x // divide the number being calculated by the divisor
x=2 // reset the divisor to 2 and continue
};
};
primeFactor = factors.pop();
return primeFactor;
}
document.write(primerizer(50))
This only returned 2, undefined, or nothing. My concern was that the stop condition for the for loop must be defined in terms of the same variable as the start condition, so I tried it with a while loop instead.
var primerizer = function(input){
var factors = [];
var numStorage = input
x=2
while (numStorage != 1){
var result = numStorage%x;
if (result === 0) {
factors.push(x);
numStorage = numStorage/x
x=2
}
else {
x = x+1
}
}
return factors.pop();
}
document.write(primerizer(50)
Same problem. Maybe there's a problem with my syntax that I'm overlooking? Any input is much appreciated.
Thank you.

The shortest answer I've found is this:
function largestPrimeFactor(n){
var i=2;
while (i<=n){
if (n%i == 0){
n/=i;
}else{
i++;
}
}
console.log(i);
}
var a = **TYPE YOUR NUMBER HERE**;
largestPrimeFactor(a)

You can try with this
var x = 1, div = 0, primes = [];
while(primes.length != 10001) {
x++;
for(var i = 2; i < x && !div; i++) if(!(x % i)) div++;
if(!div) primes.push(x); else div = 0;
}
console.log(primes[primes.length-1]);
or this: (This solution uses more of your memory)
var dont = [], max = 2000000, primes = [];
for (var i = 2; i <= max; i++) {
if (!dont[i]) {
primes.push(i);
for (var j = i; j <= max; j += i) dont[j] = true;
}
}
console.log(primes);

here is my own solution.
//function
function largestPrimeFactor (num) {
//initialize the variable that will represent the divider
let i = 2;
//initialize the variable that will represent the quotient
let numQuot = num;
//array that will keep all the dividers
let primeFactors = [];
//execute until the quotient is equal to 1
while(numQuot != 1) {
/*check if the division between the number and the divider has no reminder, if yes then do the division keeping the quotient in numQuot, the divider in primeFactors and proceed to restart the divider to 2, if not then increment i by one an check again the condition.*/
if(numQuot % i == 0){
numQuot /= i;
primeFactors.push(i);
i = 2;
} else {
i++;
}
}
/*initialize the variable that will represent the biggest prime factor. biggest is equal to the last position of the array, that is the biggest prime factor (we have to subtract 1 of .length in order to obtain the index of the last item)*/
let biggest = primeFactors[primeFactors.length - 1];
//write the resutl
console.log(biggest);
}
//calling the function
largestPrimeFactor(100);

<script>
function LPrimeFactor() {
var x = function (input) {
var factors = [];
var numStorage = input;
x = 2;
while (numStorage != 1) {
var result = numStorage % x;
if (result === 0) {
factors.push(x);
numStorage = numStorage / x;
x = 2;
}
else {
x = x + 1;
}
}
return factors.pop();
}
document.write(x(50));
}
</script>
<input type="button" onclick="LPrimeFactor();" />
Here is an example i tried with your code

Here is the solution I used that should work in theory... except for one small problem. At a certain size number (which you can change in the code) it crashes the browser due to making it too busy.
https://github.com/gordondavidescu/project-euler/blob/master/problem%203%20(Javascript)
Adding the code inline:
<p id="demo">
</p>
<script>
function isPrime(value) {
for(var i = 2; i < value; i++) {
if(value % i === 0) {
return false;
}
}
return value > 1;
}
function biggestPrime(){
var biggest = 1;
for(var i = 600851470000; i < 600851475143; i++){
if (isPrime(i) != false)
{
biggest = i;
}
document.getElementById("demo").innerHTML = biggest;
}
}
biggestPrime();
</script>
</p>

<script>
//Finds largest prime factor
find = 2165415 ; // Number to test!
var prime = 0;
loop1:
for (i = 2; i < find; i++){
prime = 0;
if (find%i == 0){
document.write(find/i);
for (j = 2; j < (find / i); j++){
if ((find / i )%j == 0){
document.write(" divides by "+j+"<br>");
prime = prime + 1;
break;
}
}
if (prime == 0){
document.write("<br>",find/i, "- Largest Prime Factor")
prime = 1;
break;
}
}
}
if (prime==0)
document.write("No prime factors ",find," is prime!")