Factorialize a Number challenge

Factorialize a Number challenge
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#1

function factorialize(num) {
var result = 1;
if (num > 0) {
for (var i = 1; i <= num; i++) {
result = result * i;
}
}
num = 1;

return result;
}

factorialize(-7);

I know that there is other more simple solution but I only wanted to focus on my way of thinking .
I am doing these challenges for a week to get used to the “logical thinking” and it is really hard. I hope 1 day I get better at writing because the are a lot of amazing terms, methods, function, prototypes to learn =)


#2

OPS, LET ME CORRECT AND PUT SOME COMMENTS

function factorialize(num) {
var result = 1; // result is assigned with 1
if (num > 0) { // if this condition is true, then result will be assigned with other value.
for (var i = 1; i <= num; i++) {
result = result * i;
}
}

return result;
}

factorialize(-7);


#3
  1. The factorial of 1 is 1, you can change your if condition so that your for loop is not executed in that case (like you did for for num = 0).
  2. Instead of result = result * i you can use the equivalent result *= i, which means "multiply result by i".
  3. With your function, the factorial of any negative number would be 1, which is wrong. You should return another value (for example undefined) if someone tries to find the factorial of a negative number.
  4. You can format your code putting it between backtick characters:

my code

This results in something nicer to read, so it is easier for other to help you.

function factorialize(num) {
  var result = 1; // result is assigned with 1
  if (num > 0) { // if this condition is true, then result will be assigned with other value.
    for (var i = 1; i <= num; i++) {
      result = result * i;
    }
  }

  return result;
}


#5

You are right and I did not find anyone writing about negative values. So what should I do?


#6
function factorialize(num) {
  var result = 1; // result is assigned with 1
  if (num > 0) { // if this condition is true, then result will be assigned with other value.
    for (var i = 1; i <= num; i++) {
      result = result * i;
    }
  }
  if  (num < 0) {
    result = undefined;
  }

  return result;
}

factorialize(-7);


#7

I inserted another if, since else would not be logical to put. Is it alright now?


#8

Generally you should check error cases first, so your program doesn’t do unnecessary work.
Put your second if at the beginning. And you don’t have to do result = undefined - if you return without value it defaults to undefined. So the first line could be:

function factorialize(num) {
  if  (num < 0) return; // number is invalid, return undefined
...
...

#9

Tks! Did not know that just putting return it could work!


#10

a more primitive solution from a non-professional///
function factorialize(num) {

var k=num;
if (num>0) {
for (var i=num;i>2;i–){
k=k*(num-1); num= num-1;
} }
if (num===0){
k=1;}
return k;
}

factorialize(5);
///