No of numbers less than a given number with no repeating digits

Question

How can we find the number of numbers less than a given number with no repeating digits in it?

For example the number of such numbers less than 100 is 90. (11, 22, 3

Answer 1

You can consider two cases:

• numbers shorter than the limit
• numbers that that differ from the limit at some digit

The count of d-digit numbers is 9*9*8*... = 9*9!/(9-d)! (the first digit may not be zero). The count of all numbers shorter than d is the count of 0-digit numbers + .. count of d-1-digit numbers. These sums may be precomputed (or even hard-coded).

The count of d-digit numbers with f first digits given is (10-f)*...*(10-(d-1)) = (10-f)!/(10-d)!. You can precomupte the factorials as well.

Pseudocode :

To precompute fac:
  - fac = int[10];
  - fac[0] = 1;
  - for i in 1..10:
    - fac[i] = fac[i-1] * i;

To precompute count_shorter:
  - cs = int[10];
  - cs[0] = 0;
  - cs[1] = 1; // if zero is allowed
  - for i in 1..10:
    - cs[i+1] = cs[i] + 9 * fac[9] / fac[10-i]
  - count_shorter = cs;

To determine the count of numbers smaller than d:
  - sl = strlen(d)
  - if sl > 10
    - return count_shorter[11]
  - else
    - sum = 0
    account for shorter numbers:
    - sum += count_shorter[sl]
    account for same-length numbers; len=count of digits shared with the limit:
    - sum += 9* fac[9] / fac[10-sl];
    - for every len in 1..{sl-1}:
      count the unused digits less than d[len]; credits to @MvG for noting:
      - first_opts = d[len]-1;
      - for every i in 0..{len-1}:
        - if d[i] < d[len]
          - first_opts -= 1;
      - sum += first_opts * fac[9-len] / fac[10-sl] 
  - return sum