#357

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Count Numbers with Unique Digits

The problem asks to count numbers with unique digits from 0 to 10^n using dynamic programming, math, and backtracking techniques.

math dynamic programming backtracking

Problem Statement

Given a non-negative integer n, you are tasked with counting all numbers x, where 0 <= x < 10^n, such that each number contains unique digits. For example, for n = 2, the range of numbers is 0 to 99, but numbers like 11, 22, 33, and so on, are excluded.

The problem requires you to find a method that can compute this count efficiently. The constraint on n is 0 <= n <= 8, which means you need to consider both smaller and larger values of n, ensuring the solution works for the entire input range.

Examples

Example 1

Input: n = 2

Output: 91

The answer should be the total numbers in the range of 0 ≤ x < 100, excluding 11,22,33,44,55,66,77,88,99

Example 2

Input: n = 0

Output: 1

Example details omitted.

Constraints

0 <= n <= 8

Solution Approach

State Transition Dynamic Programming

Using dynamic programming, we track the count of unique-digit numbers by progressively calculating possibilities for each position in the number. By starting with the smallest number (0), we iteratively count how many valid digits can be placed in each position, ensuring each digit is unique. This approach avoids redundant calculations, making it efficient.

Backtracking Approach

A backtracking approach is another valid method where we recursively build numbers, adding digits one by one while ensuring they remain unique. This approach explores all valid combinations, but it can be slower compared to dynamic programming due to repetitive calculations.

Math-Based Optimized Calculation

To further optimize, a direct mathematical formula based on permutations can be used. This formula calculates how many valid numbers can be formed for each digit length, considering the constraint of unique digits. This approach avoids explicitly generating the numbers and can be computationally efficient.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the specific approach used. For the dynamic programming approach, the time complexity is O(n), where n is the length of the number, due to the iterative counting process. The backtracking method typically has higher time complexity due to repeated state exploration, making it less efficient for larger n values. The mathematical approach offers a direct solution with O(1) time complexity after precomputation but may require space for storing precomputed values.

What Interviewers Usually Probe

The candidate demonstrates a good understanding of dynamic programming concepts and its application to state transitions.
The candidate recognizes the trade-offs between backtracking and dynamic programming, opting for the most efficient approach based on the problem constraints.
The candidate is aware of optimization techniques and can suggest mathematical shortcuts when applicable.

Common Pitfalls or Variants

Common pitfalls

Failing to account for repeated digits when forming numbers, especially when the input size is large.
Not considering edge cases such as n = 0, where the only valid number is 0.
Using an inefficient brute-force approach, which may lead to timeouts or excessive computation for larger values of n.

Follow-up variants

How would the solution change if n were larger (e.g., n = 10)?
What is the effect of reducing the range from 0 to 10^n to a more limited range, such as 0 to 10^5?
Can you extend this approach to work with non-decimal bases or larger sets of digits?

FAQ

How do I approach the 'Count Numbers with Unique Digits' problem efficiently?

You can use a dynamic programming approach to count numbers with unique digits, focusing on state transitions for each digit position. Backtracking can also be applied but may be less efficient.

What is the time complexity of solving 'Count Numbers with Unique Digits' using dynamic programming?

The time complexity of the dynamic programming approach is O(n), where n is the number of digits in the number, making it efficient for large inputs.

What is the primary pattern in the 'Count Numbers with Unique Digits' problem?

The primary pattern in this problem is state transition dynamic programming, where each state represents a valid count of unique-digit numbers for a given number length.

How does backtracking work for this problem, and when should it be used?

Backtracking works by recursively constructing numbers, checking at each step that no digit is repeated. It should be used when you're exploring all possibilities but might not be as efficient for large values of n.

How can I optimize the solution for 'Count Numbers with Unique Digits'?

You can optimize the solution by using a mathematical formula to calculate the count of unique-digit numbers directly, avoiding the need to generate each number.

terminal

Solution

Solution 1: State Compression + Digit DP

This problem essentially asks for the number of numbers in the given range $[l, ..r]$ that satisfy certain conditions. The conditions are related to the composition of the numbers rather than their size, so we can use the concept of Digit DP to solve it. In Digit DP, the size of the number has little impact on the complexity.

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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)

Continue Practicing

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