Count Numbers with Unique Digits — LeetCode 357 Python Solution
MediumMathDynamic ProgrammingBacktracking
- Problem
- #357
- Pattern
- Backtracking
- Reading time
- 3 min
- Source
- leetcode.com
The problem
Given an integer n, return the count of all numbers with unique digits, x, where 0 <= x < 10n.
Example
- Input
- n = 2
- Output
- 91
- Explanation
- The answer should be the total numbers in the range of 0 ≤ x < 100, excluding 11,22,33,44,55,66,77,88,99
Python solution
Python
class Solution:
def countNumbersWithUniqueDigits(self, n: int) -> int:
@cache
def dfs(i: int, mask: int, lead: bool) -> int:
if i < 0:
return 1
ans = 0
for j in range(10):
if mask >> j & 1:
continue
if lead and j == 0:
ans += dfs(i - 1, mask, True)
else:
ans += dfs(i - 1, mask | 1 << j, False)
return ans
return dfs(n - 1, 0, True)Complexity
| Measure | Complexity |
|---|---|
| Time | O(n \times 2^D \times D) |
| Space | O(n \times 2^D) auxiliary |
Pattern: Backtracking
Build candidates one choice at a time and abandon a branch the moment it cannot work. LeetCode 357. Count Numbers with Unique Digits is filed here because LeetCode tags it Backtracking, which is the vocabulary this hub collects.
The backtracking guide has the Python template for the pattern and the 105 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 357. Count Numbers with Unique Digits?
- LeetCode 357. Count Numbers with Unique Digits is rated Medium on LeetCode.
- What is the time complexity of LeetCode 357. Count Numbers with Unique Digits?
- The Python solution on this page runs in O(n \times 2^D \times D).
- What is the space complexity of LeetCode 357. Count Numbers with Unique Digits?
- The Python solution on this page uses O(n \times 2^D) auxiliary space.
- What topics does LeetCode 357. Count Numbers with Unique Digits cover?
- LeetCode 357. Count Numbers with Unique Digits is tagged Math, Dynamic Programming and Backtracking on LeetCode.