Actually with the requirements specified as returning a string, you don't really need three test cases. Just put the i%3 test case before the i%5 test case and fizz then buzz will be added to the string resulting in fizzbuzz for the divisible by three and five test case. Use a flag to specify if either case gets triggered and if not then add the number to the string.

Why not let output be a string, then just build by concatenating with += after evaluating i? Could also demonstrate type coercion when i (integer) is coerced to string.

one of the test failing with result - secondTest ✘ expected:<[fizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzz]fizzbuzz> but was:<[12fizz4buzzfizz78fizzbuzz11fizz1314]fizzbuzz>

The expected string is strange, it doesn't have any numbers. Clearly the expected result is wrong.

[deleted]

My code is as follows
public static String fizzBuzz(Integer num) {
StringBuffer result = new StringBuffer();
for (int n=1; n<=num; n++) {
if (num % 3 == 0 && num % 5 == 0) result.append("fizzbuzz");
else if (num % 3 == 0) result.append("fizz");
else if (num % 5 == 0) result.append("buzz");
else result.append(String.valueOf(n));
}
return result.toString();
}

My bad. All tests pass now. Earlier, in the failed test case the expected string was a bit strange and the real issue got overlooked. Thanks.

"Expected space complexity: O(1)"

That seems hard to achieve, won't the resulting string itself always take up O(n) space?

Your suggested python solution is O(n) space complexity and the expected space complexity is O(1) . How can you claim it as final solution ? either change constraint on space complexity to O(n) or provide accurate solution..

For the JS implementation, you can just use an output string without an array so that you don't have to convert the array into a string at the end.

function fizzBuzz(n) {
  let output = '';

  for (let i = 1; i <= n; i++) {
    if (i % 3 === 0 && i % 5 === 0) {
      output += 'fizzbuzz';
    } else if (i % 3 === 0) {
      output += 'fizz';
    } else if (i % 5 === 0) {
      output += 'buzz';
    } else {
      output += i.toString();
    }
  }

  return output;
}