JavaScript - odd_man_out.txt

Take a string of numbers (like such: ("2 4 7 8 10”)), figure out if there is a single even number of a single odd number. return the ‘real’ index of that number (not 0 indexed)


- - - - - My Solution - - - - -

function iqTest(numbers){
    var numArr = numbers.split(' ');
    var evens = [];
    var odds = [];

    numbers.split(' ').map(function(num){
        num%2 === 0 ? evens.push(num) : odds.push(num);
    });
    return evens.length === 1 ? numArr.indexOf(evens[0])+1 : numArr.indexOf(odds[0])+1
}


iqTest("2 4 7 8 10");

- split the given number-string into an array
- create two dummy arrays, ‘evens’ and ‘odds’ to be pushed into
- .map over each number pushing it into either ‘evens’ or ‘odds’ arrays.
- if ‘evens’ has a single element, find the index+1 of it in the numArr
  - else find the index+1 of the odd number in the numArr.



- - - Alt - - -
whew. okay...
- m is being set to an odd number or an even number based on . . .
- . . .
- return the index+1 of ‘m’ in numbers (as split into a real array)



function iqTest(numbers){
  var m = numbers.match(/[02468]\b.*[02468]\b/.test(numbers) ? /\d*[13579]\b/ : /\d*[02468]\b/)[0];
  return numbers.split(' ').indexOf(m) + 1;
}


- all things told, I like my solution better. It is maybe not as sexy, but it is much more legible.
- but I still want to know regEx better.