0

I have following code for checking whether there is duplicate in an array. The code works fine. But it uses a new array named newUniqueArray. Is there a better code for this purpose without using a new array? Is there any optimization possible on this code?

Note: I have used inArray and in keywords from jQuery

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title></title>
<script type="text/javascript" src="http://ajax.aspnetcdn.com/ajax/jquery/jquery-1.4.1.js"></script>
<script type="text/javascript">
    $(document).ready(function () {
        $('#btnSave').click(function (e) {
            var reportRecipients = "A, a , b,";
            reportRecipients = reportRecipients.toLowerCase();
            checkDuplicate(reportRecipients);
        });

        function checkDuplicate(reportRecipients) {
            if (reportRecipients.length > 1) {
                var recipientsArray = reportRecipients.split(',');
                var newUniqueArray = [];

                for (a in recipientsArray) {
                    var email = $.trim(recipientsArray[a]);

                    if ($.inArray(email, newUniqueArray) == -1) {
                        newUniqueArray.push(email);
                    }
                }

                if (newUniqueArray.length < recipientsArray.length) {
                    alert('Duplicate Exists');
                }

                return false;
            }
        }
    });
</script>
</head>
<body>
<input name="txtName" type="text" id="txtName" />
<input type="submit" name="btnSave" value="Save" id="btnSave" />
</body>
</html>
4
  • Please read on how to use jsfiddle properly. Also, check the syntax highlighting. Commented Dec 12, 2012 at 8:17
  • You need to put your js code in the js frame and choose a proper framework(JQuery) on the left pane. See jsfiddle.net/jmDEZ/5, this is a updated version of your code and it works Commented Dec 12, 2012 at 8:20
  • What does your array consist of? Commented Dec 12, 2012 at 8:39
  • @badZoke Array consists of strings Commented Dec 12, 2012 at 9:39

2 Answers 2

3

If you just want to test on string arrays, you can use a JavaScript object's property to test. It used a hash-table to look up properties, which is faster than array iteration.

example: http://jsfiddle.net/jmDEZ/8/

function checkDuplicate(reportRecipients) {
    var recipientsArray = reportRecipients.split(','),
        textHash = {};
    for(var i=0; i<recipientsArray.length;i++){
        var key = $.trim(recipientsArray[i].toLowerCase());
        console.log("lower:" + key);
        if(textHash[key]){
            alert("duplicated:" + key);
            return true;
        }else{
            textHash[key] = true;
        }
    }
    alert("no duplicate");
    return false;
}​
Sign up to request clarification or add additional context in comments.

2 Comments

Thanks. What is the advantage of using textHash over newUniqueArray
@Lijo, object property lookup is often implemented by hash-table(O(1)), while in array you query over iteration(O(n)). See this question: stackoverflow.com/questions/7700987/…. Although, an object is not exactly a hash table, see this blog: devthought.com/2012/01/18/an-object-is-not-a-hash. But at most time when you know what you are testing, it works.
2

I can't see any reason to use jQuery for this purpose:

checkDuplicate = function (reportRecipients) {
    if (reportRecipients.length > 1) {
        var recipientsArray = reportRecipients.split(',');
        for (a in recipientsArray) {
            if(reportRecipients.indexOf(a) != reportRecipients.lastIndexOf(a)){
                return true;
            }
        }
    }
    return false;
}

$('#btnSave').click(function (e) {
            var reportRecipients = "A, a , b,";
            reportRecipients = reportRecipients.toLowerCase();
            if(checkDuplicate(reportRecipients)) alert('Duplicate Exists');
        });

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.