Question

I’ve written a similar question which was closed I would like to ask not the code but an efficiency tip. I haven’t coded but if I can’t find any good hint in here I’ll go and code straightforward. My question:

Suppose you have a function listNums that take a as lower bound and b as upper bound.

For example a=120 and b=400

I want to print numbers between these numbers with one rule. 120’s permutations are 102,210,201 etc. Since I’ve got 120 I would like to skip printing 201 or 210.

Reason: The upper limit can go up to 10²⁰ and reducing the permutations would help the running time.

Again just asking for efficiency tips.

Answer 1

I am not sure how you are handling 0s (eg: after outputting 1 do you skip 10, 100 etc since technically 1=01=001..).

The trick is to select a number such that all its digits are in increasing order (from left to right).

You can do it recursively. AT every recursion add a digit and make sure it is equal to or higher than the one you recently added.

EDIT: If the generated number is less than the lower limit then permute it in such a way that it is greater than or equal to the lower limit. If A1A2A3..Ak is your number and it is lower than limit), then incrementally check if any of A2A1A3…Ak, A3A1A2…Ak, … , AkA1A2…Ak-1 are within limit. If need arises, repeat this step to with keeping Ak as first digit and finding a combination of A1A2..Ak-1.

Eg: Assume we are selecting 3 digits and lower limit is 99. If the combination is 012, then the lowest permutation that is higher than 99 is 102.