Before anything I want to preemptively thank anyone who drops by for their patience, I don't have any formal CS background so I'm probably going to use some of these terms wrong.

I have a puzzle: Given two numbers which define a set of continuous counting numbers of the same number of digits, between roughly 5 and 12 digits long (IE 50000 and 60000, 32325600000 and 32399999999), what's the fastest and most efficient way to condense this down to a set of prefixes which "contain" all permutations of subsequent digits?

The approach we've been using is a hybrid of treating these as numbers and character strings. First remove any pairs of matched 0's and 9's at the end of the start/end. Second create the full sequence copied to two columns, where the second column is always a substring with the rightmost digit removed relative to the first column. From there I can recursively count how many times any given one-digit-shorter substring occurs, keep the items where N-count<10, and where N-count>=10 remove another digit from both columns and repeat.

What I'm wondering is if there's a quicker and more efficient way to do this. String operations instead of math was an obvious quick fix, but the general approach still relies on recursively grouping and chopping off characters. I've considered making a full series of Prefix and N-count columns going back to the highest digit but at least instinctively that feels like it would be less efficient than operating recursively on a decreasing pool of numbers.


which becomes

Cycle one creates two sequence columns like this:

String   Prefix     N-Count
500        50          10
501        50          10
510        51          10
550        55          6
551        55          6
552        55          6
553        55          6   

Cycle two keeps everything where N-count<10 the same, but reduces the rest by 1
digit each and recalculates N-count (while getting rid of duplicates).

String   Prefix     N-Count
50        5          5
51        5          5
52        5          5         
53        5          5
54        5          5       
550       55         4
551       55         4
552       55         4
553       55         4  

Output:  50,51,52,53,54,55,550,551,552,553 
  • 1
    $\begingroup$ I don't understand your problem or your examples. What is the input (apparently, two non-negative integers $a \leq b$), and what is the desired output? $\endgroup$ Commented Apr 21, 2020 at 9:16
  • $\begingroup$ I don't know how to describe the problem any more clearly, there's probably some technical terminology for it that I don't know. The input was explicitly described with several potential examples given, as well as being noted right at the top of the example. I edited that to be even more explicit. I've also explicitly repeated exactly what the output will look like in addition to describing it. $\endgroup$
    – D3SL
    Commented Apr 21, 2020 at 10:55
  • 1
    $\begingroup$ Is the following correct? Let $S$ consist of the decimal encodings of all integers between $a$ and $b$, where $a,b$ are both $n$-digit long (without leading zeroes). You are looking for the smallest set $T$ such that an $n$-digit integer is in $S$ iff one of the strings in $T$ is its prefix. $\endgroup$ Commented Apr 21, 2020 at 12:27
  • $\begingroup$ I think that sounds right. Let me see if I get it: let's say Set S is every counting number from 500 to 599. Set T would consist solely of the string "5" because every number in Set S begins with the number "5". On the other hand if it were 500 to 600 then Set T would be "5" and "600", because every possible 3 digit permutation of 0-9 beginning with "5" is included in S however of those beginning with 6 only "600" is included in S. $\endgroup$
    – D3SL
    Commented Apr 21, 2020 at 12:39
  • $\begingroup$ Actually, in your example you shouldn't include 55. $\endgroup$ Commented Apr 21, 2020 at 12:52

1 Answer 1


Suppose the input is $a,b$, two $n$-digit long numbers. We allow leading zeroes (we will see in a moment why). Let $c$ be the longest common prefix of $a,b$, and let $a=cA$, $b=cB$.

If $A = 0^{n-|c|}$ and $B = 9^{n-|c|}$ then we simply output $c$ (this includes the case $|c|=n$).

Otherwise, let $d_A$ be the first digit of $A$, and let $d_B$ be the first digit of $B$.

Recursively find a solution for the ranges $[A,d_A 9^{|A|-1}]$ and $[d_B 0^{|B|-1},B]$, and prefix $c$ to everything. Also, add $c(d_A+1),\ldots,c(d_B-1)$.

Here is an unoptimized python implementation:

def prefixes(a,b,C=''):
     n, m = len(a), max(i for i in range(len(a)+1) if a[:i] == b[:i])
     c, A, B = C+a[:m], a[m:], b[m:]
     if A == '0'*len(A) and B == '9'*len(B):
         yield c
         yield from prefixes(A[1:],'9'*(len(A)-1),c+A[0])
         for i in range(int(A[0])+1,int(B[0])):
             yield f'{c}{i}'
         yield from prefixes('0'*(len(B)-1),B[1:],c+B[0])

For example, if you run list(prefixes('50000000','55399999')) then you get the following output: ['50', '51', '52', '53', '54', '550', '551', '552', '553']

  • $\begingroup$ If I understand the python right it's pretty much the same approach I'm already using. Clean up trailing 0's and 9's then count out groups of 10 and if there's a full group of 10 shorten by one digit and repeat until you've got less than 10. Those are the prefixes to keep. $\endgroup$
    – D3SL
    Commented Apr 21, 2020 at 13:14
  • $\begingroup$ You can in principle unroll the recursion into an explicit iteration. This would look more different from your approach. $\endgroup$ Commented Apr 21, 2020 at 13:24
  • $\begingroup$ In other words we've already found probably the most efficient procedure, any further gains will have to be in distribution of load between cores/machines and memory management. Thanks. $\endgroup$
    – D3SL
    Commented Apr 21, 2020 at 13:55

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