elp1987 elp1987 - 4 years ago 81
Javascript Question

Any Suggestions for Improving my Code on (Negative) Factorial (in JavaScript)?

I was doing a basic problem on coding a factorial \$n!\$:

function factorialize(num) {
if (num < 0) {
return undefined;
} else if (num === 0) {
return 1;
}
else {
return num * factorialize(num - 1);
}
}


The site I'm learning from accepted this solution, but they're only testing for nonnegative integers, i.e for \$n \ge 0\$, not negative integers.

I got curious on how to compute a negative factorial \$(-n)!\$. Many pages say it's undefined, but I found two saying it could be defined.



The gist I got is that:

\$|n!| = |(-n)!|\$

Their absolute values are the same but the negative factorials change signs.

Examples:

\$4! = (-4)! = 24\$

\$5! = 120\$ but \$(-5)! = -120\$

The formula that I gathered from the two linked pages is:

\$(-n)! = |n|! * (-1)^n\$

And this reflects my code. From test cases, I think I've nailed it. I just want to ask if there's a better way of coding it. Someone from here
remarked that using recursion is memory-inefficient.

function factorialize(num) {
if (num === 0) {
return 1;
} else if (num > 0) {
return num * factorialize(num - 1);
} else {
return Math.pow(-1, num) * Math.abs(num) * factorialize(Math.abs(num) - 1);
}
}

// test cases below
console.log(factorialize(-1)); // -1
console.log(factorialize(1)); // 1
console.log(factorialize(0)); // 1
console.log(factorialize(-2)); // 2
console.log(factorialize(2)); // 2
console.log(factorialize(-3)); // -6
console.log(factorialize(3)); // 6
console.log(factorialize(-4)); // 24
console.log(factorialize(4)); // 24
console.log(factorialize(-5)); // -120
console.log(factorialize(5)); // 120

Answer Source

You could also do this using iteration (normally everything that can be done recursively can also be done iteratively). There is also no need to raise -1 to the power of your number. It would be much more efficient, if your just checked if it was odd or even.

function factorialize(n){
  var absNum = Math.abs(n);
  var i = 1;
  var factorial = 1;
  while(i <= absNum){
    factorial *= i;
    i += 1;
  }
  if(absNum % 2 === 1 && n < 0){
    return -factorial
  }
  return factorial;
}
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