Smallest Common Multiple - fCC Solution 4 - Ternary operator

[vietan00892b]

Smallest Common Multiple

I’m having trouble understanding one of their solution:

Solution 4 - Prime factorization (Click to Show/Hide)

// Find the SCM of a range of numbers
function smallestCommons(arr) {
  let primeFactors = {};
  const [min, max] = arr.sort((a, b) => a - b);
  for (let i = min; i <= max; i++) {
    // Factorize number in range
    let primes = getPrimeFactors(i);
    for (let j in primes) {
      // Add factor to set or update number of occurrences
      if (!primeFactors[j] || primes[j] > primeFactors[j]) {
        primeFactors[j] = primes[j]
      }
    }
  }
  // Build SCM from factorization
  let multiple = 1;
  for (let i in primeFactors) {
    multiple *= i ** primeFactors[i]
  }
  return multiple;
}

// Compute prime factors of a number
function getPrimeFactors(num) {
  const factors = {};
  for (let prime = 2; prime <= num; prime++) {
    // Count occurances of factor
    // Note that composite values will not divide num
    while ((num % prime) === 0) {
      factors[prime] = (factors[prime]) ? factors[prime] + 1 : 1;
      num /= prime;
    }
  }
  return factors;
}

smallestCommons([1, 5]);

Specifically, the ternary operator inside the while loop in getPrimeFactors function:

 while ((num % prime) === 0) {
      factors[prime] = (factors[prime]) ? factors[prime] + 1 : 1; // confused
      num /= prime;
    }

For example if I pass in getPrimeFactors(20), I get out an object { 2: 2, 5: 1 }.

This is me trying to understand it by rewriting in conventional if-else statement:

while (num % prime === 0) {
	if (factors[prime]) {
		factors[prime] = factors[prime] + 1;
	} else factors[prime] = 1;

    num /= prime;
}

So what they’re trying to do is:

• If the object factors doesn’t have property prime (prime is 2):
  • Create a new prop with value 1 (1 being occurence count).
  • Then num /= prime will make change num from 20 to 10 for the next loop.
• Next loop, because now factors does have property { 2: 1 }
  • The condition if (factors[prime]) is true,
  • we increase the occurence count, hence { 2: 1 } becomes { 2: 2 }.
• We repeat this until num % prime != 0 (num is 5 now) and end the while loop.

I can’t help but feel like I’m missing something. Any insight is appreciated.

I was actually able to pass the challenge without looking up solutions, using the same approach. Then I found this solution and the code is like 3 times shorter so I’m scratching my head trying to understand it.

[kevinSmith]

This is me trying to understand it by rewriting in conventional if-else statement:

Yes. Saying:

x = condition ? a : b

is the same as:

if (condition)
  x = a
else
  x = b

That is a very common usage for a ternary operator - when doing an assignment that is conditional.

So what they’re trying to do is…

That sounds basically right. The only nitpicky thing I would add is:

The condition if (factors[prime]) is true ,

That shouldn’t say “true”, but “truthy”.

I can’t help but feel like I’m missing something. Any insight is appreciated.

What do you think you are missing? It sounds like you understand it.

When I am confused, I try to log things out to see what is happening. Consider this:

function getPrimeFactors(num) {
  const factors = {};
  for (let prime = 2; prime <= num; prime++) {
    console.log('\n* prime =', prime, '... and factors =', factors)
    while ((num % prime) === 0) {
      console.log('\n*** num =', num)
      console.log('*** factors[prime]', factors[prime])
      console.log('***', factors[prime] ? 'this prime already exists, will just use that' : 'this prime does not exist, must create')
      console.log('*** before factors', factors)
      factors[prime] = (factors[prime]) ? factors[prime] + 1 : 1;
      console.log('*** after factors', factors)
      num /= prime;
    }
  }
  return factors;
}

console.log('final =', getPrimeFactors(20));

[vietan00892b]

Thank you. I guess this is one of those times where you figure out the answer while typing out the question.

[kevinSmith]

It happens more times than you’d think. That happens to me all the time at work. That’s one of the big reasons why we always try to get people to write explicit questions in the forum. Keep on truckin’.