Problem Description

In this problem, we are given a non-negative integer n, and we are asked to find out how many integers there are with unique digits such that the integer x satisfies 0 <= x < 10^n. Unique digits mean that no digit in the number repeats. For example, the number 123 has unique digits, while the number 112 does not because the digit 1 is repeated.

Flowchart Walkthrough

First, let's use the algorithm Flowchart to determine the appropriate algorithm pattern for solving Leetcode 357. Count Numbers with Unique Digits. Here’s the reasoning flow:

1. Is it a graph?

   • No: This problem does not involve graph traversal or connectivity problems.

2. Need to solve for kth smallest/largest?

   • No: This problem is not about finding the kth smallest or largest element; it is about counting numbers.

3. Involves Linked Lists?

   • No: This problem does not involve linked list operations or manipulation.

4. Does the problem have small constraints?

   • Yes: The problem is limited to counting numbers with up to a certain number of digits, usually not very large (e.g., typically less than 10).

5. Brute force / Backtracking?

   • Yes: Backtracking or brute force is particularly suitable to build numbers and ensure their digits are unique.

Conclusion: The flowchart suggests using a backtracking approach to generate and count all numbers within the constraint that has all unique digits. This resolves by sequentially adding digits and checking the uniqueness requirement via backtracking.

Intuition

To solve this problem, we can approach it by counting the number of valid numbers rather than generating each possible number, which would be inefficient.

• For n == 0, the only number we can have is 0 itself, hence only one unique number.
• For n == 1, any digit from 0 to 9 is valid, which means there are 10 unique numbers (including 0).

As soon as n is greater than 1, we start with 10 possibilities (from 0 to 9) and choose the second digit. There are only 9 possible choices left for the second digit since it has to be different from the first (excluding the case where the first digit is 0, as we've counted that in n == 1). For the third digit, there's one less choice than for the second (since now two digits are taken), and so on.

The solution follows these steps for n > 1:

• Start the answer with 10 cases (all single-digit numbers plus the number 0).
• For each additional digit place, we multiply our current count of unique digits by the decreasing number of options available (starting from 9 for the second digit, 8 for the third, etc.).

The formula for the number of unique digit numbers that can be formed with i+1 digits is f(i+1) = f(i) * (10 - i) where f(i) is the number of unique digit numbers with i digits and i begins at 1 and increments until n-1.

The solution code uses a loop to count the number of unique digit numbers for each number of digits from 1 up to n and adds them up to accumulate the total count.

Learn more about Math, Dynamic Programming and Backtracking patterns.

Solution Approach

The implementation of the solution for counting unique digit numbers consists of the following steps:

• Start by checking for the base cases. If n is 0, return 1 because only the number 0 fits the criteria. If n is 1, return 10 because the numbers 0 through 9 are the only valid possibilities and they all have unique digits.

• For numbers with more than one digit (n > 1), we'll need to calculate the possibilities using a loop. Initialize the ans (answer) variable with 10, to cover the one-digit numbers. Also, initialize a variable cur to 9, representing the number of choices for the first digit, excluding 0.

• Loop from 0 to n - 1. In each iteration, we will calculate the number of unique numbers that can be created with an additional digit. Multiply cur by 9 - i, where i is the current iteration's index. This represents the decrease in available choices as we fix more digits in the number.

• Add the result of the multiplication to ans, updating it to include the count of unique numbers with the new number of digits.

• Continue this process until the loop ends.

• Finally, return ans, which now holds the total count of unique-digit numbers for all lengths up to n digits.

Python Solution Code

class Solution:
    def countNumbersWithUniqueDigits(self, n: int) -> int:
        if n == 0:
            return 1
        if n == 1:
            return 10
        ans, cur = 10, 9
        for i in range(n - 1):
            cur *= 9 - i
            ans += cur
        return ans

This solution employs a mathematical pattern without using any complex data structures. The loop efficiently calculates the count for each number of digits, and the use of multiplication (cur *= 9 - i) within the loop follows the pattern of the decreasing number of choices for each subsequent digit.

Example Walkthrough

Let's illustrate the solution approach with n = 3. The task is to count numbers with unique digits where 0 <= x < 1000 (since 10^3 = 1000).

For n = 0, there's only one number, 0, so the answer is 1.

For n = 1, any single digit number, 0 to 9, is valid and unique. That's 10 possibilities.

Now, for n > 2, we need to calculate the possibilities for numbers having 2 and 3 digits.

• Two-digit numbers (10 to 99):

  • Start with 10 total unique numbers from the n = 1 case.
  • For the first digit (tens place), we have 9 choices (1 to 9, as we're not including 0 here since that's accounted for in the n = 1 case).
  • For the second digit (ones place), we have 9 choices again because it can be any digit except the one chosen for the tens place. This includes 0.
  • So for two-digit numbers, we have 9 * 9 = 81 possibilities.
  • Now, our total is 10 + 81 = 91.

• Three-digit numbers (100 to 999):

  • Continuing from 91 unique numbers.
  • For the first digit (hundreds place), we still have 9 choices (1 to 9).
  • For the second digit (tens place), we have 9 choices.
  • Now, for the third digit (ones place), we have 8 choices because two digits are already used.
  • Multiplying these together, for three-digit numbers, we have 9 * 9 * 8 = 648.
  • Our total now is 91 + 648 = 739.

Adding all these up, for n = 3, we would have 739 unique digit numbers where 0 <= x < 1000. Using the pattern described in the Solution Approach, the loop calculates this same total. The pseudo-code for the loop would look like:

• Initialize ans with 10 (for n = 1).
• For each additional digit place (i from 0 to n - 1 = 2):
  • set cur to 9 for the first iteration.
  • multiply cur with 9 - i to account for the already chosen digits.
  • add the result to ans
  • if i = 0 (2 digits), cur = 9 * 9, add 81 to ans; ans becomes 91
  • if i = 1 (3 digits), cur = 9 * 9 * 8, add 648 to ans; ans becomes 739

Therefore, for n = 3, the countNumbersWithUniqueDigits function returns 739.

Solution Implementation

Time and Space Complexity

The provided Python code defines a function countNumbersWithUniqueDigits which calculates the number of n-digit integers that have unique digits.

• Time Complexity: The time complexity of the function is primarily determined by the for loop that iterates n - 1 times. Within the for loop, there are only constant-time operations. Therefore, the overall time complexity is O(n).

• Space Complexity: The space complexity of the function is O(1) because the space used does not grow with the input size n. The function only uses a constant amount of additional space for variables ans and cur.

Learn more about how to find time and space complexity quickly using problem constraints.