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Real world Javascript map/reduce, solving the Poker Hand problem

Mike Talbot ⭐ on June 16, 2020

You, like me, may have a play at solving some of the daily challenges here on DEV. Usually if the answer is more than one or two lines of some fan...

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Mike Talbot ⭐ • Jun 16 '20 • Edited

The original challenge specified no low straights from the Ace, but it jarred with me and this routine is easily adjustable to support it. We just have to worry about sort orders during straights and to recognise the straight in the first place. The following updated version of getHandDetails() manages both of these:

const order = "23456789TJQKA"
function getHandDetails(hand) {
    const cards = hand.split(" ")
    const faces = cards.map(a => String.fromCharCode([77 - order.indexOf(a[0])])).sort()
    const suits = cards.map(a => a[1]).sort()
    const counts = faces.reduce(count, {})
    const duplicates = Object.values(counts).reduce(count, {})
    const flush = suits[0] === suits[4]
    const first = faces[0].charCodeAt(0)
    //Also handle low straight
    const lowStraight = faces.join("") === "AJKLM"
    faces[0] = lowStraight ? "N" : faces[0]
    const straight = lowStraight || faces.every((f, index) => f.charCodeAt(0) - first === index)
    let rank =
        (flush && straight && 1) ||
        (duplicates[4] && 2) ||
        (duplicates[3] && duplicates[2] && 3) ||
        (flush && 4) ||
        (straight && 5) ||
        (duplicates[3] && 6) ||
        (duplicates[2] > 1 && 7) ||
        (duplicates[2] && 8) ||
        9

    return { rank, value: faces.sort(byCountFirst).join("") }

    function byCountFirst(a, b) {
        //Counts are in reverse order - bigger is better
        const countDiff = counts[b] - counts[a]
        if (countDiff) return countDiff // If counts don't match return
        return b > a ? -1 : b === a ? 0 : 1
    }
    function count(c, a) {
        c[a] = (c[a] || 0) + 1
        return c
    }
}

We perform a simple string check for the low straight and then switch the order of 'A' to 'N' for aces in this circumstance.

Mike Talbot ⭐ • Dec 12 '22

Yes, that's in the article list of hands picture and in the solution.

Макаров Валерий • Jun 19 '20

To take into account the "steel wheel" (straight 5432A) you need to add an imaginary face value of 1:

const order = "123456789TJQKA"

And if the input contains 2, 3, 4, 5 and A, then make the replacement of A by 1:

function SteelWheel(str) { if((str.indexOf('2') + 1) && (str.indexOf('3') + 1) && (str.indexOf('4') + 1) && str.indexOf('5') + 1) && (str.indexOf('A') + 1)) { return str.replace('A', '1') } else { return str } }

Mike Talbot ⭐ • Jun 19 '20 • Edited

As mentioned in my comment above, the initial challenge had no straight from the ace. My version in the comment above supports the "steel wheel"

Jeremy Gollehon • Jun 16 '20

Love this. Thanks! It's a really good problem to show off "modern" javascript.

Only piece of feedback... the nested ternary in byCountFirst is probably too terse.

Mike Talbot ⭐ • Jun 16 '20

Thanks! Yeah that's fair on the ternary. I'll adjust it.

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