Sorted Union my solution

1

Continuing the discussion from freeCodeCamp Challenge Guide: Sorted Union:

i simply used spreed operator and a for loop. The solution is pretty basic though.

function uniteUnique(arr, …arrN) {
for (let i = 0; i < arrN.length; i++) {
for (let j = 0; j < arrN[i].length; j++) {
if (!arr.includes(arrN[i][j])) {
arr.push(arrN[i][j]);
}
}
}
return arr;
}
console.log(uniteUnique([1, 2, 3], [5, 2, 1, 4], [2, 1], [6, 7, 8]));

2

Hello there,

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Thank you.

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4

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6

Rest parameters + spread operator + array.flat() + filter = Easy and elegant solution.

function uniteUnique(...arr) {
  const array = [...arr].flat();
  const result = array.filter((a,b) => 
   array.indexOf(a) === b
   );
  return result;
}

uniteUnique([1, 3, 2], [5, 2, 1, 4], [2, 1]);
1 Like