902. Numbers At Most N Given Digit Set
Description
Given an array of digits
which is sorted in non-decreasing order. You can write numbers using each digits[i]
as many times as we want. For example, if digits = ['1','3','5']
, we may write numbers such as '13'
, '551'
, and '1351315'
.
Return the number of positive integers that can be generated that are less than or equal to a given integer n
.
Example 1:
Input: digits = ["1","3","5","7"], n = 100 Output: 20 Explanation: The 20 numbers that can be written are: 1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.
Example 2:
Input: digits = ["1","4","9"], n = 1000000000 Output: 29523 Explanation: We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers, 81 four digit numbers, 243 five digit numbers, 729 six digit numbers, 2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers. In total, this is 29523 integers that can be written using the digits array.
Example 3:
Input: digits = ["7"], n = 8 Output: 1
Constraints:
1 <= digits.length <= 9
digits[i].length == 1
digits[i]
is a digit from'1'
to'9'
.- All the values in
digits
are unique. digits
is sorted in non-decreasing order.1 <= n <= 109
Solutions
Solution 1: Digit DP
This problem is essentially asking for the count of positive integers generated by the numbers in digits
within the given range $[l,..r]$. The count is related to the number of digits and the digit at each position. We can use the idea of Digit DP to solve this problem. In Digit DP, the size of the number has a small impact on the complexity.
For the range $[l,..r]$, we generally convert it into $[1,..r]$ and then subtract $[1,..l - 1]$, i.e.,
$$ ans = \sum_{i=1}^{r} ans_i - \sum_{i=1}^{l-1} ans_i $$
However, for this problem, we only need to calculate the value of the range $[1,..r]$.
Here we use memoization search to implement Digit DP. We search from the starting point to the bottom layer to get the number of schemes, return the answer layer by layer and accumulate it, and finally get the final answer from the starting point of the search.
The basic steps are as follows:
- Convert the number $n$ into an int array $a$, where $a[1]$ is the lowest digit and $a[len]$ is the highest digit;
- Design the function $dfs()$ based on the problem information. For this problem, we define $dfs(pos, lead, limit)$, and the answer is $dfs(len, 1, true)$.
Where:
pos
represents the number of digits in the number, starting from the last digit or the first digit, usually chosen based on the digit construction property of the problem. For this problem, we choose to start from the high digit, so the initial value ofpos
islen
;lead
indicates whether the current number contains leading zeros. If it does, it is1
, otherwise it is0
. It is initialized to1
;limit
represents the restriction on the digits that can be filled. If there is no restriction, you can choose $[0,1,..9]$, otherwise, you can only choose $[0,..a[pos]]$. Iflimit
istrue
and the maximum value that can be taken has been taken, then the nextlimit
is alsotrue
. Iflimit
istrue
but the maximum value has not been taken, or iflimit
isfalse
, then the nextlimit
isfalse
.
For the implementation details of the function, you can refer to the code below.
The time complexity is $O(\log n)$.
Similar problems:
- 233. Number of Digit One
- 357. Count Numbers with Unique Digits
- 600. Non-negative Integers without Consecutive Ones
- 788. Rotated Digits
- 1012. Numbers With Repeated Digits
- 2376. Count Special Integers
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