Problem Description

This problem requires finding the number of integers between two positive integers a and b (inclusive) that have unique digits. An integer has unique digits if no digit is repeated within that integer. For example, '123' has unique digits, but '112' does not because the digit '1' is repeated.

The challenge lies in efficiently calculating this number over a potentially large range without having to check every integer individually. This problem can be approached algorithmically with combinatory and dynamic programming principles to avoid brute force enumeration, which would be computationally expensive.

Intuition

To intuitively approach this problem, we can think of it in terms of counting. If we need to count the number of unique digit numbers within a range, it's logical to break down the range into smaller parts that are easier to handle. If we count from '1' to the upper bound 'b' and from '1' to 'a-1' and find the difference, we can effectively count the unique digit numbers between 'a' and 'b'.

Now, we can take advantage of digit dynamic programming (digit DP), which often involves solving number-related problems by considering one digit at a time, generally moving from the most significant digit to the least significant digit. This is combined with state compression, which efficiently tracks the state of our computation, particularly which digits have been used so far.

Digit DP can help us on two levels:

1. It reduces the problem from considering ranges of numbers to considering actual digits, which is a significantly smaller sample space.
2. The use of memoization prevents us from recalculating states that we've previously solved, which significantly speeds up our computations.

The binary representation of the digit states (state compression) allows us to use bitwise operations to track which digits have been used, allowing us to easily check if adding a new digit would violate the uniqueness condition.

In essence, we use recursion with memoization to count unique digit numbers by building them digit by digit, ensuring that we do not reuse any digits. Our recursive function, dfs, keeps track of which digits are used and whether we are bounded by the current digit of our limit (either 'a - 1' or 'b') at every recursive call. The tricky part is to handle this current digit bounding correctly, which we do with the limit flag.

By comparing the count for the upper and lower bounds, we determine the count of unique digit numbers in the specified range. This approach eliminates the need to explicitly check every number in the range and allows us to solve the problem with greater efficiency.

Learn more about Math and Dynamic Programming patterns.

Solution Approach

The solution provided employs state compression and digit DP to efficiently solve the problem. Here's a step-by-step breakdown of the algorithms and patterns used in the implementation, which will help us understand the approach mentioned in the reference solution:

1. Converting to String for Easy Digit Access: We start by converting the integer representation of our bounds 'a-1' and 'b' into string format. This allows us to easily access each digit without the need for additional calculations or conversions.

2. Digit Dynamic Programming (DP): We create a recursive function dfs(pos, mask, limit) that explores each digit position pos of our current number. The mask argument is an integer with bits representing digits used in our current recursive path (state compression). The limit flag tells us if we need to be bound by the digit at the current position, ensuring we do not exceed our upper bound limits.

3. Memoization and Caching: By leveraging the cache decorator, we can store the results of calculations we've already performed. This significantly reduces the number of computations by eliminating redundant calculations, essential for the overall efficiency of recursive algorithms.

4. Implementing the Function for Bounds: We compute the count of unique digit numbers upto 'b' by calling dfs(0, 0, True) on the string representation of 'b'. Similarly, we calculate the count upto 'a - 1'. Subtracting these two values yields the answer.

5. Bitwise Operations for State Compression: In mask, we use bitwise operations to check whether a digit is used (mask >> i & 1) and to update the used digits (mask | 1 << i). This is the essence of state compression: representing the state of used digits with bits in an integer.

6. Limit Management: While exploring digits, we need to manage the limit to know if we are bound by the current digit of our upper bound. If limit is True, we can only explore digits up to num[pos]. If limit is False, there are no such restrictions. When we opt for a digit equal to num[pos], the subsequent recursive call will still have limit as True.

As a result of this approach, we have a powerful and efficient solution that doesn't directly inspect each number in the range but instead builds potential numbers digit by digit while abiding by the constraints. This greatly cuts down the search space and optimizes calculation times to solve even large input ranges in a reasonable timeframe.

Example Walkthrough

Let's walk through an example to illustrate the solution approach described. Suppose we are trying to find the number of unique digit integers between a=100 and b=130, inclusive.

1. Converting to String for Easy Digit Access: First, we convert our integer bounds a-1 and b to strings. Hence a-1 becomes '99' and b becomes '130'.

2. Digit Dynamic Programming (DP): We use a recursive function dfs(pos, mask, limit) where pos starts at 0 (most significant digit), mask starts as 0 (no digits used), and limit starts as True to respect the boundary.

3. Memoization and Caching: Assume we have a memory cache to store results from the dfs calls indexed by pos, mask, and limit. We do this to not repeat calculations.

4. Implementing the Function for Bounds: We run the dfs function twice: once up to '99' (representing a-1) and once up to '130' (representing b). We'll store the results as variables, say count_upto_a_minus_1 and count_upto_b.

5. Bitwise Operations for State Compression: As we build numbers in dfs, say we want to use digit '1', we check if it's already been used with (mask >> 1) & 1. If not, we use it and update the mask to mask | (1 << 1).

6. Limit Management: If at any pos the limit is True, we can only use digits up to the same digit in our bound ('1', '3', or '0' for '130'). If limit is False, we can use any digit from 0 to 9, provided we haven't used it before.

To calculate the result, we let the algorithm run and simulate using the smaller example below:

• For '99', the dfs might explore making the first digit '0' through '9' recursively, keeping track of used digits and limits.
• Once done with '99', we have the count of unique number integers up to 99.
• For '130', the dfs must start with '1' (since limit is True and '1' is the most significant digit of the upper bound) and can then explore '0' through '3' for the second digit and '0' for the last digit due to the limit flag.
• Any recursive call that leads to a digit already used in mask will return 0, disallowing repeated digits.

By subtracting count_upto_a_minus_1 from count_upto_b, we get the count of unique digit numbers between 100 and 130, which is our desired answer. In this case, these numbers would be 102, 103, 104, ..., up to 130, except for 110, 120, and 122 through 129 because these numbers contain repeated digits. The dfs function checks all possibilities in a reduced space instead of checking every number individually, which makes it efficient.

Solution Implementation

Time and Space Complexity

The given code defines a function numberCount that computes the count of numbers between two integers a and b such that no digit repeats within those numbers.

• Time Complexity:

  The time complexity is derived from the number of recursive calls made by the function dfs. Here m represents the number of digits in the larger number b. For each digit position, there are up to 10 possible digits to choose from (0 through 9). The mask parameter can represent any combination of the 10 digits, meaning there are 2^10 possible mask values. As the recursion can happen for each position of the digit with each possible combination of the masks, and for each digit choice, the total number of calls is in the worst case O(m * 2^10 * 10).

  The constraint limit ensures that we are only considering valid numbers less than or equal to b. Whenever limit is True, it means the current number we are forming has to be less than or equal to the corresponding digits in b. If limit becomes False, we can use any digit at the current position, up to 9. This doesn't affect the worst case scenario much, since the number of operations is still bounded by the number of digits in b and the possible states of the mask.

• Space Complexity:

  The space complexity is mostly influenced by the space needed to store the results of the recursive calls, which is the memoization cache that the @cache decorator creates. The maximum depth of the recursion is m, which is the number of digits in b. At each level, there can be 2^10 different states of the mask. Therefore, the space needed to store the intermediate results is O(m * 2^10).

In conclusion, the time complexity is O(m * 2^10 * 10) and the space complexity is O(m * 2^10).

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