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Home/ Questions/Q 3451098
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T09:06:43+00:00 2026-05-18T09:06:43+00:00

I have got a sequence of strings – 0000001, 0000002, 0000003…. upto 2 million.

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I have got a sequence of strings – 0000001, 0000002, 0000003.... upto 2 million. They are not contiguous. Meaning there are gaps. Say after 0000003 the next string might be 0000006. I need to find out all these gaps. In the above case (0000004, 0000005).

This is what I have done so far –

gaps  = list()
total = len(curr_ids)

for i in range(total):
    tmp_id = '%s' %(str(i).zfill(7))
    if tmp_id in curr_ids:
        continue
    else:
        gaps.append(tmp_id)
return gaps

But as you would have guessed, this is slow since I am using list. If I use a dict, to pre-populate curr_ids it’ll be faster. But what’s the complexity to populating a hash-table? What’s the fastest way to do this.

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  1. Editorial Team
    Editorial Team
    2026-05-18T09:06:43+00:00Added an answer on May 18, 2026 at 9:06 am

    You could sort the list of ids and then step through it once only:

    def find_gaps(ids):
        """Generate the gaps in the list of ids."""
        j = 1
        for id_i in sorted(ids):
            while True:
                id_j = '%07d' % j
                j += 1
                if id_j >= id_i:
                    break
                yield id_j
    
    >>> list(find_gaps(["0000001", "0000003", "0000006"]))
    ['0000002', '0000004', '0000005']
    

    If the input list is already in order, then you can avoid the sorted (though it does little harm: Python’s adaptive mergesort is O(n) if the list is already sorted).

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