isEven(n)

Overcomplicated recursion vs. the obvious one-liner — do they agree?

Enter any integer:

Overcomplicated

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Simple

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Quick-test numbers:

Overcomplicated isEven

function isEvenOvercomplicated(n) {
  // Handle negatives by working with absolute value
  if (n < 0) return isEvenOvercomplicated(-n);

  // Base cases
  if (n === 0) return true;   // 0 is even
  if (n === 1) return false;  // 1 is odd

  // Recursive step: subtract 2 until we hit 0 or 1
  return isEvenOvercomplicated(n - 2);
}

Simple isEven

function isEvenSimple(n) {
  return n % 2 === 0;
}