freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/rosetta-code/department-numbers.md

title	id	challengeType	forumTopicId
Department Numbers	59f40b17e79dbf1ab720ed7a	5	302249

Description

There is a highly organized city that has decided to assign a number to each of their departments:

• Police department
• Sanitation department
• Fire department

Each department can have a number between 1 and 7 (inclusive). The three department numbers are to be unique (different from each other) and must add up to the number 12. The Chief of the Police doesn't like odd numbers and wants to have an even number for his department.

Instructions

Write a program which outputs all valid combinations as an array.

[2, 3, 7] [2, 4, 6] [2, 6, 4]
[2, 7, 3] [4, 1, 7] [4, 2, 6]
[4, 3, 5] [4, 5, 3] [4, 6, 2]
[4, 7, 1] [6, 1, 5] [6, 2, 4]
[6, 4, 2] [6, 5, 1]

Tests

tests:
  - text: <code>combinations</code> should be a function.
    testString: assert(typeof combinations === 'function');
  - text: <code>combinations([1, 2, 3], 6)</code> should return an Array.
    testString: assert(Array.isArray(combinations([1, 2, 3], 6)));
  - text: <code>combinations([1, 2, 3, 4, 5, 6, 7], 12)</code> should return an array of length 14.
    testString: assert(combinations(nums, total).length === len);
  - text: <code>combinations([1, 2, 3, 4, 5, 6, 7], 12)</code> should return all valid combinations.
    testString: assert.deepEqual(combinations(nums, total), result);

Challenge Seed

function combinations(possibleNumbers, total) {

  return true;
}

After Test

const nums = [1, 2, 3, 4, 5, 6, 7];
const total = 12;
const len = 14;
const result = [
  [2, 3, 7],
  [2, 4, 6],
  [2, 6, 4],
  [2, 7, 3],
  [4, 1, 7],
  [4, 2, 6],
  [4, 3, 5],
  [4, 5, 3],
  [4, 6, 2],
  [4, 7, 1],
  [6, 1, 5],
  [6, 2, 4],
  [6, 4, 2],
  [6, 5, 1]
];

Solution

function combinations(possibleNumbers, total) {
  let firstNumber;
  let secondNumber;
  let thridNumber;
  const allCombinations = [];

  for (let i = 0; i < possibleNumbers.length; i += 1) {
    firstNumber = possibleNumbers[i];

    if (firstNumber % 2 === 0) {
      for (let j = 0; j < possibleNumbers.length; j += 1) {
        secondNumber = possibleNumbers[j];

        if (j !== i && firstNumber + secondNumber <= total) {
          thridNumber = total - firstNumber - secondNumber;

          if (thridNumber !== firstNumber && thridNumber !== secondNumber && possibleNumbers.includes(thridNumber)) {
            allCombinations.push([firstNumber, secondNumber, thridNumber]);
          }
        }
      }
    }
  }
  return allCombinations;
}