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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T23:06:56+00:00 2026-05-15T23:06:56+00:00

I have an unsorted array, what is the best method to remove all the

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I have an unsorted array, what is the best method to remove all the duplicates of an element if present?

e.g:

a[1,5,2,6,8,9,1,1,10,3,2,4,1,3,11,3]

so after that operation the array should look like

 a[1,5,2,6,8,9,10,3,4,11]
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  1. Editorial Team
    Editorial Team
    2026-05-15T23:06:57+00:00Added an answer on May 15, 2026 at 11:06 pm

    Check every element against every other element

    The naive solution is to check every element against every other element. This is wasteful and yields an O(n2) solution, even if you only go “forward”.

    Sort then remove duplicates

    A better solution is sort the array and then check each element to the one next to it to find duplicates. Choose an efficient sort and this is O(n log n).

    The disadvantage with the sort-based solution is order is not maintained. An extra step can take care of this however. Put all entries (in the unique sorted array) into a hashtable, which has O(1) access. Then iterate over the original array. For each element, check if it is in the hash table. If it is, add it to the result and delete it from the hash table. You will end up with a resultant array that has the order of the original with each element being in the same position as its first occurrence.

    Linear sorts of integers

    If you’re dealing with integers of some fixed range you can do even better by using a radix sort. If you assume the numbers are all in the range of 0 to 1,000,000 for example, you can allocate a bit vector of some 1,000,001. For each element in the original array, you set the corresponding bit based on its value (eg a value of 13 results in setting the 14th bit). Then traverse the original array, check if it is in the bit vector. If it is, add it to the result array and clear that bit from the bit vector. This is O(n) and trades space for time.

    Hash table solution

    Which leads us to the best solution of all: the sort is actually a distraction, though useful. Create a hashtable with O(1) access. Traverse the original list. If it is not in the hashtable already, add it to the result array and add it to the hash table. If it is in the hash table, ignore it.

    This is by far the best solution. So why the rest? Because problems like this are about adapting knowledge you have (or should have) to problems and refining them based on the assumptions you make into a solution. Evolving a solution and understanding the thinking behind it is far more useful than regurgitating a solution.

    Also, hash tables are not always available. Take an embedded system or something where space is VERY limited. You can implement an quick sort in a handful of opcodes, far fewer than any hash table could be.

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