Numbers At Most N Given Digit Set - LeetCode
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
- code TLE
class Solution:
def atMostNGivenDigitSet(self, digits: List[str], n: int) -> int:
self.res = -1
def dfs(path):
if path and int("".join(path)) > n:
return
self.res += 1
for d in digits:
dfs(path + [d])
dfs([])
return self.res
- code
class Solution:
def atMostNGivenDigitSet(self, digits: List[str], n: int) -> int:
n_str = str(n)
n_len = len(n_str)
d_len = len(digits)
res = sum(d_len ** i for i in range(1, n_len))
for i, c in enumerate(n_str):
res += d_len ** (n_len - i - 1) * sum(d < c for d in digits)
if c not in digits: return res
return res + 1