Find Duplicate Number

[Filomeno Robles]

Howto find Duplicate Elements in Array? I have array of phone numbers so in the phone numbers i should start searching from the right side to the left side and find similar 6 integers. then i should print them out.

There are many, many ways to implement this, so I would not focus on the implementation details above, but rather focus on the concept: Build cross reference original array by some key (e.g. phone number) and then filter results down to just those keys with duplicate values.

It sounds like you want to flatten this structure that reflects the duplicates, into a single array of contacts (I'm not sure why you'd want to do that, as you lose the structure identifying which are duplicates of each other), but if you want to do that, you can flatMap it:

In spite of that, taking into account of the cost of deep-copy or shallow-copy, I agree that @Patrick Perini's brilliant way would be better than this in some situations even the number of that would be N * N.

My concern is efficiency, i.e. quicker UI response, longer battery life, smaller memory footprint, etc. Avoiding unnecessary memory allocations and/or memory copies which are automatically done by Swift in the behind scene would be crucial if we are providing competitive products. (-;

Here's an efficient, O(n) method to do it. Some of the other answers here use .filter on a duplicates array or even the return value array, which makes the operation work in O(n^2) (using .contains is the same). Using a Set to store the duplicates we can make it O(n).

The other method that's been shown here is using a dictionary to first store the array elements. The idea being that a dictionary can't have duplicate elements. However, this doesn't guarantee the original order of the array to be preserved, so we need a different method.

There's still quite a bit of useful reusable stuff missing from Swift to make this simple, but OrderedCollections, which have not been used by the other answers yet, make it easier to get the duplicates "in order".

I have a SpreadSheet with a column with large number represented as text, and when I apply the duplicate operation to check ( I do not use any formula, I am using excel 2010 in-built functionality of "Conditional Formatting" -->"Highlight Cells Rule" --> "Duplicate Values") even distinct values are shown as duplicate values.

It seems that DupUnique property is converting the value to a number. I also note similar behavior with COUNTIF. Accordingly, I would suggest, in this situation, that you use the conditional format option to use a formula. The formula I would suggest (assuming that the range to check for duplicates is A2:A10, would be:

If you already have that kind of data in Excel, it may appear in Exponential values and chances are that Excel chomped it up to 15 digits numeric precision. You may have to re-enter the large data with apostrophe in front of them.

When you're working with lots of data in Apple Numbers on your Mac, you may run into a situation where you have duplicates. These could be names, email addresses, products, colors, or something else.

If you don't have a lot of data in your spreadsheet, you can sort it and check for duplicates manually. This method is faster than going through your Numbers cells one by one, but only if you don't have thousands of rows in your sheet.

There are two built-in functions in Apple Numbers that you can use to find duplicates. These are the IF and COUNTIF functions. Just like the IF and COUNTIF functions in Excel, IF can display duplicates as True or False or a word you assign and COUNTIF will display how many times an item appears to indicate duplicates.

To illustrate how the function works, our example data will show product names in column A, and our table has column headers in row 1. You need to have sorted your data per the above methods for this to work. Just follow these steps for your own spreadsheet:

If you prefer not to use a word of your own and just display True for duplicates and False for non-duplicates, you can simply enter (A2=A1) in the editor. This works without adding IF before it.

You should now see numbers in that new column showing how many times the item in your duplicate column appears. For our example in the screenshot above, you can see that Cap appears three times, Coat once, and Gloves twice.

You can manually remove duplicates using the IF or COUNTIF function, finding each cell that says Duplicate, True, or any number over 1 and deleting them one by one. Sorting the formula column lets you remove the duplicates much faster though, but you need to be careful you don't delete the originals as well as the duplicates.

Maybe you do want to remove the duplicates, but you also don't want to lose any data. For instance, you might have inventory data for products, like in our example. So you want to total those amounts before you delete the duplicates. To do this, you'll need to merge the data first. For this task, you'll use both a formula and a function in Numbers for Mac.

To remove the duplicates from your Mac Numbers sheet after you merge data, you'll use the sort action again. But first, you need to create new columns to copy and paste the data results as values so that they are no longer formulas.

Now, you can sort by the remaining Duplicate indicator column using the sorting instructions at the beginning of this tutorial. You should see all your duplicates grouped together so you can delete those rows.

It's easy to have duplicates in your Numbers spreadsheets, whatever you're calculating. Hopefully, the steps we've outlined above help you quickly identify any duplicates, merge relevant data about them, and remove them for a clean sheet.

I did see binary search being listed as a solution, probably to reduce time complexity by half(so O(log n)) with constant space complexity. But I was more interested in another solution and that was a bit challenging. Somehow, it took me about a day to understand this.

The idea is to have two pointers - slow and fast. These would move through the list by using the current number as the index to the next element to iterate through. And as always, fast moves two steps ahead of slow. The iterations would create a linked list that contains a cycle due to the duplicate element.

Constructing our linked list

We would now find the sequence of numbers by iterating through nums(given above) and creating a linked list. The duplicate would have the same value and link back to the original value thus creating the cycle. The explanation of how each next number is identified is mentioned in () next to the number.

Find if there's a loop - while iterating through the list, if at some point, both the slow and fast pointers are pointing to the same node, then yes, a loop is present. Since fast is moving two nodes ahead, fast and slow would point to the same node only if there is a loop.

Why?

The intersection node is not necessarily the duplicate element/node as this intersection node is a node in the cycle where the fast caught up with the slow or where the fast and slow pointers randomly met - this just proves that there's a cycle in the linked list.

To find this origin, we need to push the slow pointer to the start of the list and move both fast and slow pointers at the same rate, one at a time, and when they now meet at the same node, that is the start of the cycle/loop.

Some math-ish stuff

Assuming D = the number of steps/distance from the start of the linked list to the start of the cycle and K = the number of steps/distance from the start of the cycle to the intersection point we determined in Step 1. Since slow moves to the start(0), its position after D steps is D. The fast pointer continues from the intersection point, so its position after D steps would be, nC(some number of cycles) - K, which is equal to D => nC - K = D.

I agree that is probably the easiest way for Manager to Manage the possibility of duplicate reference numbers. As it is covers their generation from any source and could be applied where ever reference numbers are used.

Yes, finding out you have a duplicate after you have created it is not has good as identifying it right at the beginning. The same way the codes come up when you type in a sales quote, the items could come up when you type in the code for a new inventory item?

When you click the notice, the list is filtered so the first duplicates are shown. Duplicates are identified either by reference numbers (in case of transactions) or codes (in case of customers, suppliers, employees, items etc.)

In the attached VI, I am attempting to find the number of duplicates in a string array, then return a new array with the number and the corresponding number of times that string appears. In the VI, for the given array:

I have been trying to build a formula that will find duplicate numbers. Most solutions that I found in the community will only work for text. Can anyone help with finding what I'm doing wrong. The formula shown on my screenshot is the same the entire column. Cells highlighted yellow should show as a duplicate.

I'm trying include a date range with counting the number of applicants within various depts, in certain date ranges, but it's saying incorrect argument set. =COUNTIFS(DISTINCT([Name of Requestor]:[Name of Requestor], [Submission Date]:[Submission Date], AND(@cell > DATE (2023, 9, 30), @cell

Actually there shouldn't be such a thing as duplicates if you're re-imaging the Macs when someone leaves. The Casper Suite uses the computer's MAC address (currently) to identify it, so if it gets re-imaged, the old computer record just gets replaced by the newly imaged Mac's record in the database.

Going forward, in Casper Suite 9, they'll be using the UUID of the Mac as its unique identifier but the process should be the same, since its still a hardware identifier.

Sounds like you may have another issue at work then. Do you happen to be using USB to Ethernet dongles with these when you image them or some other type of connector that could be causing some problems? You really should not see duplicates.

Also, what version of theJSS are you using? Not that it should matter. I'm nearly certain its worked this way for years now.

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