javascript - Factorialize a Number - Stack Overflow

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I'm taking the freecodecamp course one of the exercises it's to create a Factorialize function, I know there is several ways to do it just not sure what this one keeps returning 5

function factorialize(num) {
    var myMax = num;
    var myCounter = 1;
    var myTotal = 0;

    for (i = 0; i>= myMax; i++) {
        num = myCounter * (myCounter + 1);
        myCounter++;
    }
    return num;
}

factorialize(5);

I'm taking the freecodecamp course one of the exercises it's to create a Factorialize function, I know there is several ways to do it just not sure what this one keeps returning 5

function factorialize(num) {
    var myMax = num;
    var myCounter = 1;
    var myTotal = 0;

    for (i = 0; i>= myMax; i++) {
        num = myCounter * (myCounter + 1);
        myCounter++;
    }
    return num;
}

factorialize(5);

• javascript
• math
• factorial

Share Improve this question Follow edited Mar 13, 2016 at 20:06 rene 42.5k7878 gold badges121121 silver badges165165 bronze badges asked Aug 21, 2015 at 19:09 VADOVADO 3311 gold badge11 silver badge66 bronze badges 3

• 5 We would want you to plete this challenge. Else whats the point of the codecamp ;) – Joy Biswas Commented Aug 21, 2015 at 19:11
• I know and I find another way to do it, just it´s bothering me this one is not working even if the Math to get the number is right, my problem is the return – VADO Commented Aug 21, 2015 at 19:16
• One hint Try recursion code gets nice. Best way to factorials !! – Joy Biswas Commented Aug 21, 2015 at 19:18

Add a ment  | 

9 Answers 9

Sorted by: Reset to default 6

This is a recursive solution of your problem:

function factorialize(num) {
    if(num <= 1) {
        return num
    } else {
        return num * factorialize(num-1) 
    }
}

factorialize(5)

This is the iterative solution:

function factorialize(num) {
    var cnt = 1;
    for (var i = 1; i <= num ; i++) {
        cnt *= i;
    }
    return cnt;
}

factorialize(5)

with argument 5, it will return the 5! or 120.

To answer your question, why your function is returning 5: Your function never reaches the inner part of the for-loop because your testing if i is greater than myMax instead of less than. So you are just returning your input parameter which is five.

But the loop does not calculate the factorial of num, it only multiplies (num+1) with (num+2);

My solution in pliance with convention for empty product

function factorializer(int) {
    if (int <= 1) {
        return 1;
    } else {
        return int * factorializer(int - 1);
    }
}

Here is another way to solve this challenge and I know it is neither the shortest nor the easiest but it is still a valid way.

function factorialiaze(num){
var myArr = []; //declaring an array.
if(num === 0 || num === 1){
return 1;
}
if (num < 0){ //for negative numbers.
return "N/A";
}
for (var i = 1; i <= num; i++){ // creating an array.
     myArr.push(i);
}
// Reducing myArr to a single value via .reduce:
num = myArr.reduce(function(a,b){
     return a * b;
 });
return num;
}
factorialiaze(5);

Maybe you consider another approach.

This solution features a very short - cut to show what is possible to get with an recursive style and a implicit type conversion:

function f(n) { return +!~-n || n * f(n - 1); }

• + convert to number
• ! not
• ~ not bitwise
• - negative

function f(n) { return +!~-n || n * f(n - 1); }

var i;
for (i = 1; i < 20; i++) {
    console.log(f(i));
}

.as-console-wrapper { max-height: 100% !important; top: 0; }

Try this function

const factorialize = (num) =>  num === 0 ? 1 : num * factorialize(num-1)

Use it like this:

factorialize(5) // returns 120

Try this :

function factorialize(num) {
  var value = 1;
    if(num === 1 || num ===0) {
      return value;
    } else {
      for(var i = 1; i<num; i++) {
        value *= i;
      }
      return num * value;
  }
}
factorialize(5);

// My solution

const factorialize = num => {
    let newNum = 1;
    for (let i = 1; i <= num; i++) {
        newNum *= i
    }
    return newNum;
}

I love syntactic sugar, so

let factorialize = num => num <= 1 ? num : num * factorialize(num -1)
factorialize(5) 

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