freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-78-coin-partitions.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f3ba1000cf542c50fecd	Problem 78: Coin partitions	5	302191	problem-78-coin-partitions

--description--

Let {p}(n) represent the number of different ways in which n coins can be separated into piles. For example, five coins can be separated into piles in exactly seven different ways, so {p}(5) = 7.

Coin piles
OOOOO
OOOO   O
OOO   OO
OOO   O   O
OO   OO   O
OO   O   O   O
O   O   O   O   O

Find the least value of n for which {p}(n) is divisible by divisor.

--hints--

coinPartitions(7) should return a number.

assert(typeof coinPartitions(7) === 'number');

coinPartitions(7) should return 5.

assert.strictEqual(coinPartitions(7), 5);

coinPartitions(10000) should return 599.

assert.strictEqual(coinPartitions(10000), 599);

coinPartitions(100000) should return 11224.

assert.strictEqual(coinPartitions(100000), 11224);

coinPartitions(1000000) should return 55374.

assert.strictEqual(coinPartitions(1000000), 55374);

--seed--

--seed-contents--

function coinPartitions(divisor) {

  return true;
}

coinPartitions(7);

--solutions--

function coinPartitions(divisor) {
  const partitions = [1];

  let n = 0;
  while (partitions[n] !== 0) {
    n++;
    partitions.push(0);

    let i = 0;
    let pentagonal = 1;
    while (pentagonal <= n) {
      const sign = i % 4 > 1 ? -1 : 1;
      partitions[n] += sign * partitions[n - pentagonal];
      partitions[n] = partitions[n] % divisor;

      i++;

      let k = Math.floor(i / 2) + 1;
      if (i % 2 !== 0) {
        k *= -1;
      }
      pentagonal = Math.floor((k * (3 * k - 1)) / 2);
    }
  }
  return n;
}