Sorting algorithm in Javascript

Question

Write a JavaScript callback for the jQuery function $("#sort").click. Allow the user to enter three numbers in any order. Output the numbers in order from lowest to highest.

$(document).ready(function() {
  $("#sort").click(function() {
    var a = Number($("#a").val());
    var b = Number($("#b").val());
    var c = Number($("#c").val());
    var message = "";
   if (b > c) {
     if ((b + c) > (a + c)) {
       message = c + " " + a + " " + b;
     } else {
       message = c + " " + b + " " + a;
     }
   } else {
     message = b + " " + a + " " + c;
   }
    if (b > a) {
      if ((a + b) > (a + c)) {
        message = a + " " + c + " " + b;
      } else {
        message = a + " " + b + " " + c;
      }
    } else {
      message = b + " " + c + " " + a;
    }
  $("#output").html(message)
  });
});

Would anyone mind looking at this code and saying what's wrong?

Sorry I forgot to add that you can only use five comparisons. — carenekl
– carenekl, Commented Oct 7, 2014 at 21:21
given his constraints for only using five comparisons, this is probably a lesson of some kind — AlexanderBrevig
– AlexanderBrevig, Commented Oct 7, 2014 at 21:26
Is the output different than what you expect or why do you think that there is something wrong? We have a close reason for this kind of question, which reads "Questions seeking debugging help ("why isn't this code working?") must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself." — Felix Kling
– Felix Kling, Commented Oct 7, 2014 at 21:27
In your example, you're manually checking for every possible permutation, instead of letting the computer work for you. Why not just use Array.sort()? — thykka
– thykka, Commented Oct 7, 2014 at 21:32

georg · Accepted Answer · 2014-10-08 10:58:14Z

2

Some great jQuery answers here, I'll cover the comparisons part.

You don't need five comparisons, just three (or two if you're lucky). Compare a and b, and swap them if a > b. Then compare c and b. If c > b, you're done, otherwise compare c and a:

if (a > b)
  x = a, a = b, b = x;

if (c < b)
  if (c < a)
    result = [c, a, b];
  else
    result = [a, c, b];
else
  result = [a, b, c];

If all numbers are 32-bit positive ints, you can sort them without any comparisons at all:

min = function(a,b) { return b + ((a-b) & ((a-b)>>31)) }
max = function(a,b) { return a - ((a-b) & ((a-b)>>31)) }

x = min(a, min(b, c));
z = max(a, max(b, c));
y = (a + b + c) - (x + z);

result = [x, y, z];

edited Oct 8, 2014 at 10:58

answered Oct 7, 2014 at 23:18

georg

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Comments

Ningappa · Accepted Answer · 2014-10-07 21:40:53Z

0

This is also right, But Better you put all values into an array.

Get All element value and push value into an array, Use array.sort();;

to Sort numbers (numerically and ascending):

arrayname.sort(function(a, b){return a-b});

to Sort numbers (numerically and descending):

arrayname.sort(function(a, b){return b-a});

answered Oct 7, 2014 at 21:40

Ningappa

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Comments

suavidas · Accepted Answer · 2014-10-07 21:51:07Z

0

The biggest problem with the code is that it isn't well organized. Instead of using nested if statements to manually check every combination of values, try using the tools & methods that are already available to you.

To sort the values, you can put them in an array and call sort() on it.

//Function to compare numbers
function compareNumbers(a, b)
{
    return a - b;
}

var a = Number($("#a").val());
var b = Number($("#b").val());
var c = Number($("#c").val());

//let's say a = 2, b = 3, c = 1

var arr = [a,b,c];
//The array arr is now [2,3,1]

arr.sort(compareNumbers);
//The array arr is now [1,2,3]

Now you can set the message by grabbing elements from arr in order.

edited Oct 7, 2014 at 21:51

answered Oct 7, 2014 at 21:35

suavidas

1154 bronze badges

2 Comments

Jonathan Over a year ago

Major spoiler alert: Array.sort()

suavidas Over a year ago

Bergi: Whoops! Forgot about that. Updated.

Jonathan · Accepted Answer · 2014-10-07 22:02:53Z

0

Here's another (short) solution:

$(document).ready(function () {
    $("#sort").click(function () {
        var msg = [$("#a").val(), $("#b").val(), $("#c").val()].sort(
        function (a, b) {
            return a - b;
        });
        alert(msg);
    });
});

edited Oct 7, 2014 at 22:02

answered Oct 7, 2014 at 21:52

Jonathan

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Comments

guest271314 · Accepted Answer · 2015-04-12 13:44:25Z

-2

Try

$("#sort").on("click", function (e) {
    var vals = $.map($("#a, #b, #c"), function (v, k) {
        return Number(v.value)
    })
    , min = null
    , msg = "";
    do {
        min = Math.min.apply(Math, vals);
        msg += min;
        vals.splice($.inArray(min, vals), 1)
    } while (vals.length > 0);
    $("output").html(msg)
});

    $("#sort").on("click", function (e) {
        var vals = $.map($("#a, #b, #c"), function (v, k) {
            return Number(v.value)
        })
        , min = null
        , msg = "";
        do {
            min = Math.min.apply(Math, vals);
            msg += min;
            vals.splice($.inArray(min, vals), 1)
        } while (vals.length > 0);
        $("output").html(msg)
    });

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<input id="a" type="text" value="2" />
<input id="b" type="text" value="7" />
<input id="c" type="text" value="1" />
<button id="sort">click</button>
<br />
<output></output>

edited Apr 12, 2015 at 13:44

answered Oct 7, 2014 at 22:02

guest271314

1

17 Comments

Bergi Over a year ago

Lacks explanation as well as correctness (not even taking the horrible code into account)

guest271314 Over a year ago

Will update answer with description at comments . If possible , can detail "correctness" ? Thanks

Bergi Over a year ago

"All JS engines that someone tested with a specific example" is by no means a generic reason. In general, you cannot rely on this, and you should not - especially when answering a newbie question.

Bergi Over a year ago

Apart from that, there's no reason to do msg = Object.keys(n).join(""); inside of the loop; and delete n is total rubbish.

Bergi Over a year ago

The issue with delete is a) that you are using n in the next iteration of the loop b) that it doesn't work like that.

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