How do you solve Count Numbers with Unique Digits on LeetCode?

Count Numbers with Unique Digits (LeetCode #357) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Unique digit numbers can be calculated based on the number of digits rather than generating each number.

What is the time complexity of Count Numbers with Unique Digits?

The optimal solution for Count Numbers with Unique Digits runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only loop through the number of digits, which is at most 8 (since n <= 8).

What is the brute force solution for Count Numbers with Unique Digits?

The brute force approach for Count Numbers with Unique Digits is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all numbers from 0 to 10^n and check each number to see if it has unique digits. This is straightforward but inefficient for larger n. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Numbers with Unique Digits?

Count Numbers with Unique Digits uses the following concepts: Math, Dynamic Programming, Backtracking. The recommended patterns to study are: Backtracking, Dynamic Programming.

What are the interview tips for Count Numbers with Unique Digits?

Always clarify the constraints and edge cases before diving into the solution. Think about the problem in terms of combinations and permutations, especially with unique items. Practice explaining your thought process clearly; it helps in interviews.

What are common mistakes when solving Count Numbers with Unique Digits?

Forgetting to handle the case when n = 0. Not considering the leading digit constraint (cannot be 0 for multi-digit numbers).

#357

Count Numbers with Unique Digits

Medium

MathDynamic ProgrammingBacktrackingBacktrackingDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of generating all numbers, we can calculate how many unique digit numbers exist based on the number of digits. This is much faster and avoids unnecessary checks.

⚙️

Algorithm

5 steps

1Step 1: If n is 0, return 1 (only the number 0).
2Step 2: Initialize count with 10 (for numbers 0-9).
3Step 3: For each digit from 1 to n-1, calculate the number of unique combinations.
4Step 4: Use a decreasing factor for available digits (9 for the first digit, then 9, 8, etc.).
5Step 5: Accumulate the count for each digit length.

solution.py13 lines

1# Full working Python code
2
3def countNumbersWithUniqueDigits(n):
4    if n == 0:
5        return 1
6    count = 10
7    uniqueDigits = 9
8    availableDigits = 9
9    for i in range(1, n):
10        count += uniqueDigits * availableDigits
11        uniqueDigits *= availableDigits
12        availableDigits -= 1
13    return count

ℹ

Complexity note: The time complexity is O(n) because we only loop through the number of digits, which is at most 8 (since n <= 8).

1Unique digit numbers can be calculated based on the number of digits rather than generating each number.
2The first digit has 9 options (1-9), and subsequent digits have decreasing options based on what has already been used.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.