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 essentially asks for the number of positive integers that can be generated from the digits in digits within the given range $[l, .., r]$. The count depends on the number of digits and the value of each digit. We can solve this problem using the Digit DP approach. In Digit DP, the size of the number has little impact on the complexity.
For the range $[l, .., r]$ problem, we generally convert it to the problem of $[1, .., r]$ and then subtract the result of $[1, .., l - 1]$, i.e.,
$$ ans = \sum_{i=1}^{r} ans_i - \sum_{i=1}^{l-1} ans_i $$
However, for this problem, we only need to find the value for the range $[1, .., r]$.
Here, we use memoization to implement Digit DP. We start searching from the top, get the number of solutions at the bottom, and return the answers layer by layer until we get the final answer from the starting point.
The basic steps are as follows:
We convert the number $n$ into a string $s$ and denote the length of the string $s$ as $m$.
Next, we design a function $\textit{dfs}(i, \textit{lead}, \textit{limit})$ to represent the number of solutions from the current $i$-th digit to the last digit of the string. Here:
- The integer $i$ represents the current position in the string $s$.
- The boolean $\textit{lead}$ indicates whether the current number contains only leading zeros.
- The boolean $\textit{limit}$ indicates whether the current position is restricted by the upper bound.
The function executes as follows:
If $i$ is greater than or equal to $m$, it means we have processed all digits. If $\textit{lead}$ is true, it means the current number is a leading zero, and we should return $0$. Otherwise, we should return $1$.
Otherwise, we calculate the upper bound $\textit{up}$. If $\textit{limit}$ is true, then $\textit{up}$ is the digit corresponding to $s[i]$. Otherwise, $\textit{up}$ is $9$.
Then, we enumerate the current digit $j$ in the range $[0, \textit{up}]$. If $j$ is $0$ and $\textit{lead}$ is true, we recursively calculate $\textit{dfs}(i + 1, \text{true}, \textit{limit} \wedge j = \textit{up})$. Otherwise, if $j$ is in $\textit{digits}$, we recursively calculate $\textit{dfs}(i + 1, \text{false}, \textit{limit} \wedge j = \textit{up})$. We accumulate all the results as the answer.
Finally, we return $\textit{dfs}(0, \text{true}, \text{true})$.
The time complexity is $O(\log n \times D)$, and the space complexity is $O(\log n)$. Here, $D = 10$.
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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