JavaScript Translated Code

gistfile1.md

JavaScript

function highestProductOf3(arrayOfInts) {

    if (arrayOfInts.length < 3) {
        throw new Error('Less than 3 items!');
    }

    // 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
    let highest = Math.max(arrayOfInts[0], arrayOfInts[1]);
    let lowest  = Math.min(arrayOfInts[0], arrayOfInts[1]);

    let highestProductOf2 = arrayOfInts[0] * arrayOfInts[1];
    let 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.
    let highestProductOf3 = arrayOfInts[0] * arrayOfInts[1] * arrayOfInts[2];

    // walk through items, starting at index 2
    for (let i = 2; i < arrayOfInts.length; i++) {
        let 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(
            highestProductOf3,
            current * highestProductOf2,
            current * lowestProductOf2);

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

        // do we have a new lowest product of two?
        lowestProductOf2 = 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;
}