For Interview JS translation

highestProductOf3 = function (arrayOfInts) {

if(arrayOfInts.length < 3){                 // if provided array has less than 3 values then return false
    console.log('Less than 3 items!');
    return false;
}


// we're going to start at the 3rd item (at index 2)
// so pre-populate highests and lowests based on the first 2 items.
// we could also start these as null and check below if they're set
// but this is arguably cleaner
var highest = Math.max(arrayOfInts[0], arrayOfInts[1]); 
var lowest = Math.min(arrayOfInts[0], arrayOfInts[1]);
var highestProductOf2 = arrayOfInts[0] * arrayOfInts[1];
var lowestProductOf2 = arrayOfInts[0] * arrayOfInts[1];


// except this one--we pre-populate it for the first *3* items.
// this means in our first pass it'll check against itself, which is fine.
var highestProductOf3 = arrayOfInts[0] * arrayOfInts[1] * arrayOfInts[2];

// walk through items, starting at index 2
for (var i = 2; i < arrayOfInts.length; i++) {
    var current = arrayOfInts[i];

    // do we have a new highest product of 3?
    // it's either the current highest,
    // or the current times the highest product of two
    // or the current times the lowest product of two
    highestProductOf3 = Math.max(Math.max(highestProductOf3, current * highestProductOf2), current * lowestProductOf2);

    // do we have a new highest product of two?
    highestProductOf2 = Math.max(Math.max(highestProductOf2, current * highest), current * lowest);

     // do we have a new lowest product of two?
    lowestProductOf2 = Math.min(Math.min(lowestProductOf2, current * highest), current * lowest);

    // do we have a new highest?
    highest = Math.max(highest, current);

    // do we have a new lowest?
    lowest = Math.min(lowest, current);
}
;
return highestProductOf3;
};

var arr = [1,25,50]; // a test !

var res = highestProductOf3(arr);

console.log(res); // console log result !