127 lines
2.5 KiB
Markdown
127 lines
2.5 KiB
Markdown
---
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id: 59f40b17e79dbf1ab720ed7a
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title: Department Numbers
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challengeType: 5
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forumTopicId: 302249
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dashedName: department-numbers
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---
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# --description--
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There is a highly organized city that has decided to assign a number to each of their departments:
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<ul>
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<li>Police department</li>
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<li>Sanitation department</li>
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<li>Fire department</li>
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</ul>
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Each department can have a number between 1 and 7 (inclusive).
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The three department numbers are to be unique (different from each other) and must add up to the number 12.
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The Chief of the Police doesn't like odd numbers and wants to have an even number for his department.
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# --instructions--
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Write a program which outputs all valid combinations as an array.
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```js
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[2, 3, 7] [2, 4, 6] [2, 6, 4]
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[2, 7, 3] [4, 1, 7] [4, 2, 6]
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[4, 3, 5] [4, 5, 3] [4, 6, 2]
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[4, 7, 1] [6, 1, 5] [6, 2, 4]
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[6, 4, 2] [6, 5, 1]
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```
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# --hints--
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`combinations` should be a function.
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```js
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assert(typeof combinations === 'function');
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```
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`combinations([1, 2, 3], 6)` should return an Array.
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```js
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assert(Array.isArray(combinations([1, 2, 3], 6)));
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```
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`combinations([1, 2, 3, 4, 5, 6, 7], 12)` should return an array of length 14.
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```js
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assert(combinations(nums, total).length === len);
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```
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`combinations([1, 2, 3, 4, 5, 6, 7], 12)` should return all valid combinations.
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```js
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assert.deepEqual(combinations(nums, total), result);
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```
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# --seed--
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## --after-user-code--
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```js
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const nums = [1, 2, 3, 4, 5, 6, 7];
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const total = 12;
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const len = 14;
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const result = [
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[2, 3, 7],
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[2, 4, 6],
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[2, 6, 4],
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[2, 7, 3],
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[4, 1, 7],
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[4, 2, 6],
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[4, 3, 5],
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[4, 5, 3],
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[4, 6, 2],
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[4, 7, 1],
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[6, 1, 5],
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[6, 2, 4],
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[6, 4, 2],
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[6, 5, 1]
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];
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```
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## --seed-contents--
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```js
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function combinations(possibleNumbers, total) {
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return true;
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}
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```
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# --solutions--
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```js
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function combinations(possibleNumbers, total) {
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let firstNumber;
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let secondNumber;
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let thridNumber;
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const allCombinations = [];
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for (let i = 0; i < possibleNumbers.length; i += 1) {
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firstNumber = possibleNumbers[i];
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if (firstNumber % 2 === 0) {
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for (let j = 0; j < possibleNumbers.length; j += 1) {
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secondNumber = possibleNumbers[j];
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if (j !== i && firstNumber + secondNumber <= total) {
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thridNumber = total - firstNumber - secondNumber;
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if (thridNumber !== firstNumber && thridNumber !== secondNumber && possibleNumbers.includes(thridNumber)) {
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allCombinations.push([firstNumber, secondNumber, thridNumber]);
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}
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}
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}
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}
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}
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return allCombinations;
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}
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```
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