freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-47-distinct-primes-factors.md

id, challengeType, title, forumTopicId

id	challengeType	title	forumTopicId
5900f39c1000cf542c50feae	5	Problem 47: Distinct primes factors	302145

Description

The first two consecutive numbers to have two distinct prime factors are:

14 = 2 × 7
15 = 3 × 5

The first three consecutive numbers to have three distinct prime factors are:

644 = 2² × 7 × 23
645 = 3 × 5 × 43
646 = 2 × 17 × 19

Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?

Instructions

Tests

tests:
  - text: <code>distinctPrimeFactors(2, 2)</code> should return a number.
    testString: assert(typeof distinctPrimeFactors(2, 2) === 'number');
  - text: <code>distinctPrimeFactors(2, 2)</code> should return 14.
    testString: assert.strictEqual(distinctPrimeFactors(2, 2), 14);
  - text: <code>distinctPrimeFactors(3, 3)</code> should return 644.
    testString: assert.strictEqual(distinctPrimeFactors(3, 3), 644);
  - text: <code>distinctPrimeFactors(4, 4)</code> should return 134043.
    testString: assert.strictEqual(distinctPrimeFactors(4, 4), 134043);

Challenge Seed

function distinctPrimeFactors(targetNumPrimes, targetConsecutive) {

  return true;
}

distinctPrimeFactors(4, 4);

Solution

function distinctPrimeFactors(targetNumPrimes, targetConsecutive) {
  function numberOfPrimeFactors(n) {
    let factors = 0;

    //  Considering 2 as a special case
    let firstFactor = true;
    while (n % 2 == 0) {
      n = n / 2;
      if (firstFactor) {
        factors++;
        firstFactor = false;
      }
    }
    // Adding other factors
    for (let i = 3; i < Math.sqrt(n); i += 2) {
      firstFactor = true;
      while (n % i == 0) {
        n = n / i;
        if (firstFactor) {
          factors++;
          firstFactor = false;
        }
      }
    }

    if (n > 1) { factors++; }

    return factors;
  }

  let number = 0;
  let consecutive = 0;

  while (consecutive < targetConsecutive) {
    number++;
    if (numberOfPrimeFactors(number) >= targetNumPrimes) {
      consecutive++;
    } else {
      consecutive = 0;
    }
  }
  return number - targetConsecutive + 1;
}