Imagine you’re tasked with creating good strings 🧵 from two components (zero and one), while keeping their lengths within a range (low and high). The goal is to calculate how many such strings you can generate. This problem combines recursion, memoization, and creativity in dynamic programming. Let’s dive deep into this fascinating challenge! 🚀

📜 Problem Statement

Given integers low, high, zero, and one, the task is to count how many strings of lengths between low and high (inclusive) can be formed. A string length is defined as a sum of zero or one components.

Each component can be added repeatedly to form valid lengths. The result should be calculated modulo \(10^9 + 7\).

🛠 Example

Input:

low = 3, high = 10, zero = 2, one = 3  

Output:

5  

Explanation:
The valid lengths are 3, 5, 6, 8, and 10. Here’s how they can be formed:

• Length 3: 3 * one
• Length 5: 2 * zero + one
• Length 6: 3 * two
• Length 8: 4 * zero
• Length 10: 5 dot ..

Let’s continue expanding on this problem with all the details you asked for, including edge cases, example walkthroughs, and thorough explanations.

To summarize:

1. We are given integers low, high, zero, and one.
2. We need to count the number of valid string lengths \(L\) such that:
   • \[low \leq L \leq high\]
   • \(L\) can be formed by repeatedly adding zero and one.
3. The result should be returned modulo \(10^9 + 7\).

📝 Example Walkthrough

Input:

low = 3, high = 10, zero = 2, one = 3

Output:

5

Valid Lengths:

We iterate through possible lengths from low to high:

1. Length 3: This is possible because \(1 \times \text{one} = 3\).
2. Length 5: Achievable with \(1 \times \text{zero} + 1 \times \text{one} = 5\).
3. Length 6: Possible using \(2 \times \text{three}\).

Let’s refine the walkthrough and detail all aspects of this problem!

Got it! Let’s dive into the optimized solution and focus on explaining the intuition, approach, and solution in-depth while skipping the brute force method.

🌟 Count Good Strings 🌟

Challenge Yourself with Dynamic Programming Magic

🎯 Intuition Behind the Problem

At its core, this problem is about constructing lengths using two building blocks:

1. A smaller increment (zero).
2. A larger increment (one).

The challenge lies in efficiently counting valid lengths between low and high that can be constructed by repeatedly adding these two building blocks.

Here’s the thought process:

1. Recursive Exploration:
   Start with an initial length (e.g., 0) and explore all possible lengths by adding zero and one. This forms a tree-like structure where each node represents a possible length, and branches extend the length further.

2. Validating the Range:
   Only lengths between low and high count as valid. If a length exceeds high, we prune further exploration since it will lead to invalid results.

3. Avoid Redundant Work:
   Many intermediate lengths will be revisited multiple times through different paths. For example, a length of 6 can be reached by:
   • Adding zero three times.
   • Adding one twice.

   To avoid recalculating the count for the same length repeatedly, we use memoization.

4. Modulo Operation:
   Since the results can grow large, every intermediate calculation is taken modulo \(10^9 + 7\).

💡 Approach: Optimized Solution Using Dynamic Programming

We solve this problem with a recursive + memoization approach:

1. Recursive Function (DFS):
   • Start at length 0.
   • At each step, explore the two options:
     • Add zero to the current length.
     • Add one to the current length.
   • If the new length falls in the range [low, high], it contributes to the count.
2. Memoization:
   • Store results of previously computed lengths in a dictionary (dp).
   • This avoids recomputing the count for the same length and makes the solution efficient.
3. Base Cases:
   • If the current length exceeds high, return 0.
   • If the length is in dp, return the stored result.
4. Modulo Handling:
   • Use \(10^9 + 7\) for all additions to handle large numbers.

📝 Optimized Code

class Solution:
    def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
        dp = {}  # Memoization dictionary
        MOD = 10**9 + 7  # Modulo value

        def dfs(length):
            # Base case: length exceeds the valid range
            if length > high:
                return 0
            
            # Return memoized result
            if length in dp:
                return dp[length]

            # Start with current length's validity
            dp[length] = 1 if low <= length <= high else 0

            # Recursively explore the next lengths
            dp[length] += dfs(length + zero) + dfs(length + one)
            dp[length] %= MOD  # Apply modulo

            return dp[length]

        # Start recursion from length 0
        return dfs(0)

Let’s break these important steps down to understand their purpose and how they contribute to solving the problem.

Step 1: dp[length] = 1 if low <= length <= high else 0

Purpose:
This step checks whether the current length (length) falls within the valid range \([ \text{low}, \text{high} ]\):

1. Valid Length:
   If the length satisfies \(\text{low} \leq \text{length} \leq \text{high}\), then it is a valid “good string,” and we initialize the count for this length as 1.

2. Invalid Length:
   If the length does not fall in this range, it is not a “good string,” and its count is initialized as 0.

Example:

• Input: low = 3, high = 10
• Current length:
  • Case 1: length = 4
    • \(3 \leq 4 \leq 10\) is true, so dp[4] = 1.
  • Case 2: length = 2
    • \(3 \leq 2 \leq 10\) is false, so dp[2] = 0.

Step 2: dp[length] += dfs(length + zero) + dfs(length + one)

Purpose:
This step recursively explores further possible lengths that can be constructed by adding either zero or one to the current length. The results of these recursive calls are added to the current length’s count.

Breaking it Down:

1. Explore Length length + zero:
   Call dfs(length + zero) to compute the count of valid strings starting from length + zero.

2. Explore Length length + one:
   Similarly, call dfs(length + one) to compute the count of valid strings starting from length + one.

3. Accumulate Results:
   Add the counts obtained from these recursive calls to the current length’s count. This way, the dp[length] value accumulates:

   • The count of valid strings for length.
   • The count of valid strings that can be formed by extending length further using zero or one.

Why Recursion Works:

• Each recursive call explores deeper levels of possible lengths.
• If a length exceeds high, the recursion terminates early (base case).
• If a length falls within the valid range, it contributes to the count and triggers further exploration.

Example Walkthrough:

Input: low = 3, high = 10, zero = 2, one = 3
Let’s assume length = 3.

1. Initialize dp[3]:
   \(3 \geq \text{low}\) and \(3 \leq \text{high}\), so:
   \(dp[3] = 1\)

2. Explore length + zero (3 + 2 = 5):
   Call dfs(5). If dfs(5) returns 2 (valid paths from 5), we add this to dp[3].

3. Explore length + one (3 + 3 = 6):
   Call dfs(6). If dfs(6) returns 3 (valid paths from 6), we add this to dp[3].

4. Accumulate Results:
   After adding both contributions, dp[3] becomes:
   \(dp[3] = 1 + 2 + 3 = 6\)

This means starting from length = 3, there are 6 valid paths to construct lengths within the range [low, high].

Key Points

• Initialization: The line dp[length] = 1 if low <= length <= high else 0 ensures each valid length contributes at least 1 to the count.
• Recursive Accumulation: The line dp[length] += dfs(length + zero) + dfs(length + one) adds contributions from all valid extensions of the current length.

This combination ensures we efficiently calculate the total count of valid strings while avoiding redundant work with memoization. 🚀

🔍 Intuition Simplified

Imagine you’re building “good strings” as paths on a tree:

1. Start from Root: Begin at length 0.
2. Branch Out: Each step extends the current length by either zero or one.
3. Check Validity: At every node (length), check if it’s between low and high. If yes, it’s a valid string!
4. Prune the Tree: If a length exceeds high, stop further branching—it won’t contribute to the count.
5. Memoize: Once a length is calculated, store the result to reuse it for future branches.

This process ensures we only explore valid paths and reuse computations wherever possible.

📚 Example Walkthrough

Let’s break down the recursive process with an example:

Input:

low = 3, high = 10, zero = 2, one = 3

Execution Steps

1. Start at Length 0:
   \(\text{dfs}(0)\)
   \(\text{Add zero: dfs}(2)\)
   \(\text{Add one: dfs}(3)\)

2. Process Length 2:
   \(\text{dfs}(2)\)
   \(\text{Add zero: dfs}(4)\)
   \(\text{Add one: dfs}(5)\)

3. Process Length 3:
   \(\text{dfs}(3)\)
   Valid length (\(3 \geq \text{low}\)) contributes 1.
   \(\text{Add zero: dfs}(5)\)
   \(\text{Add one: dfs}(6)\)

4. Continue Recursion:

   • Explore lengths \(5, 6, 8, 10\) similarly.
   • Stop when a length exceeds 10.

Final Count:

The valid lengths are \(3, 5, 6, 8, 10\), resulting in a total count of \(5\).

Edge Cases

1. Small Ranges
   • Input: low = 1, high = 1, zero = 1, one = 1
   • Output: \(1\), as only length 1 is valid.
2. Zero-Length Range
   • Input: low = 5, high = 4, zero = 1, one = 2
   • Output: \(0\), as low > high.
3. Large Inputs
   • Input: low = 1, high = 10^6, zero = 2, one = 3
   • Ensure the solution handles large ranges efficiently.

🕒 Time Complexity

• Each unique length is computed once, resulting in \(O(\text{high})\).
• Space complexity: \(O(\text{high})\) for the memoization table.

📚 Example Walkthrough with Optimized Solution

Input

low = 3, high = 10, zero = 2, one = 3

Execution Steps

1. Start with length = 0.
2. Recursive calls:
   • length + zero = 2
   • length + one = 3
3. Continue recursion, storing results in dp:
   • Valid lengths: \(3, 5, 6, 8, 10\).
4. Result: \(dp[0] = 5\).

✨ Conclusion

• Brute Force: Simple but inefficient for large inputs.
• Dynamic Programming: Efficient and scalable, especially for large ranges.

This problem showcases how recursion and memoization can optimize solutions for combinatorial problems. 🧵 Dynamic programming reduces unnecessary computations, allowing us to handle even the largest inputs effectively. 🚀

Happy coding! 😊