2466. Count Ways To Build Good Strings

Problem Description

In this problem, we are given four integers: zero, one, low, and high. Our goal is to count how many distinct strings can be constructed with the following rules:

1. We begin with an empty string.
2. At each step, we can either add the character '0' exactly zero times or add the character '1' exactly one times to the string.
3. This process can be repeated any number of times.

A "good" string is one that is built by these rules and has a length within the low and high range, inclusive. We are asked to return the number of distinct good strings that can be created subject to these rules. Because the answer could be very large, the final result should be returned modulo 10^9 + 7.

Intuition

The intuition behind solving this problem lies in understanding that it is a question of combinatorial mathematics but modeled as a depth-first search (DFS) problem. Essentially, from any string of length i, it can extend to a string of length i + zero by appending zero number of '0's or to a string of length i + one by appending one number of '1's.

The key point is to explore all the possible paths of constructing strings from length 0 to up to high. However, if the length of a constructed string ever exceeds high, it can no longer be part of a valid solution, so that branch is terminated.

To solve this efficiently, we use a recursive approach with memoization (caching results), which is implemented using the dfs function in this case.

The dfs function works as follows:

• It takes the current length i as the input.
• It checks if the current length is greater than high, and if so, it ends the current branch (returns 0).
• If the length i is within the range [low, high], it is counted as a valid string and increments the ans by 1.
• Then it calls itself twice recursively: once with the new length i + zero, and once with i + one, adding the results to the ans.
• The answer is kept within the limit of modulo 10^9 + 7 and returned.

The final result starts with a call to dfs(0) since we start with an empty string of length 0. The function will then recursively compute the total number of distinct good strings. The use of the @cache decorator implies that Python will automatically remember the results of the dfs function calls with particular arguments, thus optimizing the number of computations by avoiding repeated calculation of the same function call.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution for this problem utilizes a recursive depth-first search (DFS) approach with memoization. Here's a step-by-step implementation breakdown:

1. Memoization Decorator (@cache): The function dfs is decorated with @cache, which is Python's built-in memoization decorator available from the functools module. This automatically remembers the results of the dfs function when it is called with specific arguments, so if the same length i is encountered again, it will not recompute the results. This reduces the complexity significantly as repeated function calls with the same arguments use the cached results.

2. Recursive dfs Function: This is the function that will be doing the heavy lifting. It receives an integer i which represents the current length of the string being constructed.

   • When the function is called, it first checks if i exceeds high, in which case it returns 0 as this branch cannot produce any good strings.

   • If i is within the low to high range, the function increments ans by 1 to account for the valid string of this particular length.

   • The function then makes a recursive call to itself for the next possible lengths, i + zero and i + one, and adds their results to ans. These recursive calls will branch out and cover all possible string lengths that can be created from this point.

3. Modulo Operation (% mod): The result of every increment and addition is taken modulo 10^9 + 7 to ensure that the result never exceeds the specified limit. This is important because the number of strings can be quite large, and taking the result modulo 10^9 + 7 keeps the numbers within integer limits. It's also a common requirement in programming problems to prevent integer overflow and to simplify large number arithmetic.

4. Initialization and Result: The function dfs is initially called with 0 as it starts with an empty string. The result of this call gives us the total count of valid "good" strings that can be constructed within the given parameters.

Here's the key part of the implementation that accomplishes this:

1def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
2    @cache
3    def dfs(i):
4        if i > high:
5            return 0
6        ans = 0
7        if low <= i <= high:
8            ans += 1
9        ans += dfs(i + zero) + dfs(i + one)
10        return ans % mod
11
12    mod = 10**9 + 7
13    return dfs(0)

Through these steps, the algorithm efficiently enumerates and counts all possible "good" strings that can be formed based on the input parameters. The use of memoization in this context optimizes the recursive exploration, making this a practical approach for potentially large input values.

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have the following inputs:

• low = 1
• high = 2
• zero = 1
• one = 2

We want to count distinct strings of '0's and '1's that can be constructed using zero number of '0's and one number of '1's, and the length of the string must be between low and high.

1. We begin with an empty string s = "" and the current length i = 0.

2. We call dfs(i) with i = 0. Since i is not greater than high, we do not return 0.

3. Now, i is not within low and high, so we do not increment ans.

4. We will call dfs(i + zero) which is dfs(1) and dfs(i + one) which is dfs(2).

5. For dfs(1):

   • i = 1 is within the range [1, 2], so we increment ans by 1.
   • We make recursive calls dfs(1 + 1) and dfs(1 + 2) which are dfs(2) and dfs(3).
   • dfs(2) we already will calculate separately, so let's look at dfs(3):
     • i = 3 is greater than high, so this call returns 0 and does not contribute to ans.

6. For dfs(2) (from the initial dfs(0) call):

   • i = 2 is within the range [1, 2], so we increment ans by 1.
   • We make recursive calls dfs(2 + 1) and dfs(2 + 2) which are dfs(3) and dfs(4).
   • Both dfs(3) and dfs(4) are greater than high and will return 0, contributing nothing to ans.

7. At this point, we have ans = 1 for dfs(1) and ans = 1 for dfs(2). Our dfs(0) call will thus return ans = 1 + 1.

8. The memoization ensures that during our recursive calls, the result of dfs(2) was calculated only once, saving computational time.

9. With the modulo operation, the final result would simply be 2 % (10^9 + 7) = 2.

10. Hence, the function countGoodStrings would return 2 as there are 2 distinct strings ("01", "11") that can be constructed within the given parameters and rules.

Note that in the actual implementation, the function will not recompute dfs(2) from both dfs(0) and dfs(1) due to the memoization decorator, as the result of dfs(2) would have already been cached and retrieved from the memo on the second call.

Solution Implementation

Time and Space Complexity

The time complexity and space complexity analysis of the given code are as follows:

Time Complexity

The time complexity of the dfs function primarily depends on the number of unique states it will visit during its execution. In this scenario, each state is represented by a value of i that is checked against low and high. Since we increment i by either zero or one, and we use memoization (@cache), each state is computed only once.

The maximum number of unique states that can be visited can be roughly estimated as (high - low) / min(zero, one), because we're incrementally increasing i starting from 0 to high in steps of zero or one. However, we do perform extra checks when i exceeds high, so some overhead is present.

Therefore, the time complexity can be considered as O((high - low) / min(zero, one)).

Space Complexity

The space complexity involves the stack space used by the recursive DFS calls and the space used by the memoization cache. Since we cache every unique state and each state calls dfs twice (once with i + zero and once with i + one), the depth of the recursion tree can go up to high / min(zero, one) in the worst case.

Therefore, the space complexity for the recursive stack is O(high / min(zero, one)). Additionally, the cache will store each unique state, contributing a space complexity that is also in the order of O(high / min(zero, one)).

Considering both the stack and the caching, the total space complexity can also be approximated as O(high / min(zero, one)).

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