Else and Return

I know how to solve the following question prompt, but I have a question on why an alternate solution won't work. The code challenge is this: Around line 17, create a function named 'arrayCounter' that takes in a parameter which is an array. The function must return the length of an array passed in or 0 if a 'string', 'number' or 'undefined' value is passed in.

If I just write "return array.length;" after I have used an if-statement to return a value of 0 for 'string', 'number' or 'undefined', I can pass the solution. But I don't understand why it is logically incorrect to write "return array.length" as an else-statement.

In the first case, an if-statement is used, and then if the code doesn't satisfy it, the program will execute with the return array.length. In the second case, an if-statement is used, and if it doesn't satisfy it, the program doesn't execute with the else-statement. Why not? It seems that both would exhaust all logical options. Either the input is a 'string', 'number' or 'undefined' or it isn't. If it is, return 0, if it isn't, return array.length. I don't understand what distinguishes the two, that is, why else won't execute identically to the return.