freeCodeCamp/curriculum/challenges/chinese-traditional/10-coding-interview-prep/project-euler/problem-43-sub-string-divisibility.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f3971000cf542c50feaa	Problem 43: Sub-string divisibility	5	302100	problem-43-sub-string-divisibility

--description--

The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.

Let d_1 be the 1^{st} digit, d_2 be the 2^{nd} digit, and so on. In this way, we note the following:

• {d_2}{d_3}{d_4} = 406 is divisible by 2
• {d_3}{d_4}{d_5} = 063 is divisible by 3
• {d_4}{d_5}{d_6} = 635 is divisible by 5
• {d_5}{d_6}{d_7} = 357 is divisible by 7
• {d_6}{d_7}{d_8} = 572 is divisible by 11
• {d_7}{d_8}{d_9} = 728 is divisible by 13
• {d_8}{d_9}{d_{10}} = 289 is divisible by 17

Find the sum of all 0 to n pandigital numbers with sub-strings fulfilling n - 2 of these divisibility properties.

Note: Pandigital numbers starting with 0 are to be considered in the result.

--hints--

substringDivisibility(5) should return a number.

assert(typeof substringDivisibility(5) === 'number');

substringDivisibility(5) should return 12444480.

assert.strictEqual(substringDivisibility(5), 12444480)

substringDivisibility(7) should return 1099210170.

assert.strictEqual(substringDivisibility(7), 1099210170)

substringDivisibility(8) should return 1113342912.

assert.strictEqual(substringDivisibility(8), 1113342912)

substringDivisibility(9) should return 16695334890.

assert.strictEqual(substringDivisibility(9), 16695334890)

--seed--

--seed-contents--

function substringDivisibility(n) {

  return true;
}

substringDivisibility(5);

--solutions--

function substringDivisibility(n) {
  function isSubDivisable(digits) {
    const factors = [2, 3, 5, 7, 11, 13, 17];

    for (let i = 1; i < digits.length - 2; i++) {
      const subNumber = digits[i] * 100 + digits[i + 1] * 10 + digits[i + 2];
      if (subNumber % factors[i - 1] !== 0) {
        return false;
      }
    }
    return true;
  }

  function heapsPermutations(k, digits, conditionCheck, results) {
    if (k === 1) {
      if (conditionCheck(digits)) {
        const number = parseInt(digits.join(''), 10);
        results.push(number);
      }
      return;
    }

    heapsPermutations(k - 1, digits, conditionCheck, results);

    for (let i = 0; i < k - 1; i++) {
      if (k % 2 === 0) {
        [digits[i], digits[k - 1]] = [digits[k - 1], digits[i]];
      } else {
        [digits[0], digits[k - 1]] = [digits[k - 1], digits[0]];
      }
      heapsPermutations(k - 1, digits, conditionCheck, results);
    }
    return;
  }

  const allowedDigits = [...new Array(n + 1).keys()];
  const divisablePandigitals = [];
  heapsPermutations(
    allowedDigits.length,
    allowedDigits,
    isSubDivisable,
    divisablePandigitals
  );

  let sum = 0;
  for (let i = 0; i < divisablePandigitals.length; i++) {
    sum += divisablePandigitals[i];
  }

  return sum;
}