CODEWITHSUNDEEP
javascriptfactorial
Erick
Erick

Reputation: 369

Extra long factorial challenge giving me different result in JavaScript

I think that this code works but in the challenge they asking me to return "15511210043330985984000000" instead of "1.5511210043330984e+25" I don't understand what could be the problem. I've tried putting "BigInt" before my result but that gives me something like "15511210043330985984000000n" this "n" at the end of every number.

function extraLongFactorials(n) {
  let result = 1;
  for(let i = n; i >= 1; i--) {
    result *= i;
  }
  return result;
}

console.log(extraLongFactorials(25));

// should be 15511210043330985984000000

Upvotes: 2

Views: 2249

Answers (6)

Abhay Kumar Singh
Abhay Kumar Singh

Reputation: 51

let extraLongFactorials = (n) => {

    let mult = 1

    while (n >= 1) {

        mult = BigInt(mult) * BigInt(n)

        --n
    }

    console.log(BigInt(mult).toString())
}

Upvotes: -1

Enrique Arevalo
Enrique Arevalo

Reputation: 333

an easy and clean solution is:

 let memoization = [BigInt(0), BigInt(1)];

    const factorial = num => (typeof memoization[num] !== 'number')
        ? ((num - BigInt(1)) > 0
            ? (num * factorial(num - BigInt(1)))
            : BigInt(1)
        )
        : memoization[num]

    console.log(String(factorial(BigInt(n))));

Upvotes: 0

Masoumeh
Masoumeh

Reputation: 1

an easy solution for this challenge:

function extraLongFactorials(n) {
  let factorial = BigInt(1);
  for (let i = BigInt(2); i <= n; i++) {
    factorial *= i;
  }
  return factorial.toString();
}

Upvotes: 0

oguzhanslmtemiz
oguzhanslmtemiz

Reputation: 11

Javascript recursive solution

function extraLongFactorials(n) {
  const factorial = (int) => {
    const bigInt = BigInt(int)
    return bigInt === 0n ? 1n : bigInt * factorial(bigInt - 1n)
  }
  console.log(factorial(n).toString())
}
extraLongFactorials(25)

Upvotes: 1

Erick
Erick

Reputation: 369

Well, I found this solution. It is quite close to what @undg was saying but a little different that is the answer I was expecting.

function extraLongFactorials(n) {
  n = BigInt(n)
  let result = BigInt(1);
  
  
  for(let i = n; i >= 1; i--) {
    result *= i;
  }
  return result.toString();
}

console.log(extraLongFactorials(25)); // 15511210043330985984000000

Upvotes: 3

undg
undg

Reputation: 825

Use BigInt() followed by toString()

BigInt(Number.MAX_VALUE).toString()
//'179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368'

Upvotes: 0

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