2.5 KiB
2.5 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
59f40b17e79dbf1ab720ed7a | Department Numbers | 1 | 302249 | department-numbers |
--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]
--hints--
combinations
should be a function.
assert(typeof combinations === 'function');
combinations([1, 2, 3], 6)
should return an Array.
assert(Array.isArray(combinations([1, 2, 3], 6)));
combinations([1, 2, 3, 4, 5, 6, 7], 12)
should return an array of length 14.
assert(combinations(nums, total).length === len);
combinations([1, 2, 3, 4, 5, 6, 7], 12)
should return all valid combinations.
assert.deepEqual(combinations(nums, total), result);
--seed--
--after-user-code--
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]
];
--seed-contents--
function combinations(possibleNumbers, total) {
return true;
}
--solutions--
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;
}