4 ms·

Surely the spread operator is nicer here? [1, 2, 3, 4, 5].reduce((acc, n) => n % 2 === 1 ? [ ...acc, 2 * n ] : acc, [])

• v413 5y ago
  This is not efficient. Each iteration creates a new array instance due to the spread operator.
  • spiralx 5y ago
    `acc.concat()` also creates a new array instance, so I don't get your point.