What is the most efficient way to check for a sequence of numbers in a JavaScript array?

Question

Background

For a technical interview I was tasked with implementing a missing algorithm in JavaScript. The interviewer supplied me with some code and 18 failing unit tests that once the algorithm was successfully implemented would pass. I'm sure there is a more efficient way to have solved this problem as I tried a few different approaches during my allotted time. This way is the first way that I got to work, which for the technical test was enough but I'd like to know a better way of solving the problem.

Problem

Work out if the cards in a poker hand form a straight. (I've already ordered the hand in ascending order.)

My Solution

PokerHand.prototype._check_straight_function = function(arr) {
    var isStraight = false;
    for (var j = i = 4; i >= 0 && j > 1; i--)
        if (arr[i].value() - 1 == arr[--j].value()) {
            isStraight = true;
        } else {
            isStraight = false;
        }
    };
    return isStraight;
};

Other approaches

Things I didn't get working that I think might work faster, I'd really appreciate if someone could talk me through a working version of the below approach(es) and help me understand which is the fastest to evaluate.

• recursive use of arr.pop().value - 1 == arr.pop().value()
• filter the array to create a new array that contains only values where the next index (arr[++i])is the current index + 1 and then see if the new array is the same length.
• a for loop with a break / continue to short circuit as soon as the straight ends.

Answer 1

There's no need to assign a variable isStraight at all.

PokerHand.prototype._check_straight_function = function(arr) {
    for (var i = i = 4; i++) {        
        if (arr[i].value() - 1 != arr[i-1].value()) {
            return false;
        }
    };

    return true;
};

Answer 2

Just throwing it out there, I think this one does it with the least amount of lines (3), whether or not that is "better" is subjective as it is probably less clear

    var last = arr.pop().value(), popped;
    while ((popped = arr.pop().value()) === --last);
    return popped === undefined;

Answer 3

here is another way

var isOrdered=true;
[5,6,8,9].sort(function(a, b) {
  (b-a != 1)?isOrdered=false:true;
});
alert(isOrdered);

DEMO