Problem Description

You need to construct strings using a specific process and count how many "good" strings can be formed.

String Construction Rules:

• Start with an empty string
• At each step, you can do one of two operations:
  • Append zero number of '0' characters to the string
  • Append one number of '1' characters to the string
• You can repeat these operations any number of times

What is a Good String: A good string is any string constructed using the above process that has a length between low and high (inclusive).

Task: Count the total number of different good strings that can be constructed. Since this number can be very large, return the result modulo 10^9 + 7.

Example Understanding: If zero = 2, one = 3, low = 3, high = 5:

• You can append 2 zeros at a time or 3 ones at a time
• Valid good strings would have lengths 3, 4, or 5
• For length 3: You can create "111" (append 3 ones once)
• For length 4: You can create "0011" or "1100" (append 2 zeros and 2 ones in different orders)
• For length 5: You can create "00111", "11100", "01110", etc.
• Count all such distinct possibilities

The solution uses dynamic programming with memoization, where dfs(i) represents the number of good strings that can be constructed starting from length i. Starting from length 0 (empty string), we explore all possible ways to add zero zeros or one ones, counting valid strings when their length falls within the [low, high] range.

Intuition

Think of this problem as building strings step by step, where at each point you're deciding whether to add zero zeros or one ones to your current string length.

The key insight is that we don't need to track the actual strings themselves - we only care about their lengths. Why? Because the problem asks for the count of different strings, and two strings are different if they have different sequences of operations, even if they end up with the same length.

Starting from an empty string (length 0), we can visualize this as a decision tree:

• From length 0, we can go to length zero (by adding zeros) or length one (by adding ones)
• From length zero, we can go to length zero + zero or zero + one
• And so on...

This naturally leads to a recursive approach: from any current length i, we can transition to length i + zero or i + one. Whenever we reach a length within [low, high], we've found a good string, so we count it.

However, multiple paths can lead to the same length. For example, if zero = 2 and one = 3, we can reach length 5 by:

• Adding 2 zeros then 3 ones (0→2→5)
• Adding 3 ones then 2 zeros (0→3→5)

These are different strings even though they have the same length! But here's the crucial observation: once we've calculated how many good strings can be formed starting from a particular length i, that count remains the same regardless of how we reached length i. This is why memoization works perfectly here.

The function dfs(i) represents: "Starting from length i, how many good strings can I create?" We stop exploring when i > high because any string longer than high cannot contribute to our answer. For lengths within [low, high], we add 1 to our count (representing the current valid string) and continue exploring further possibilities.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses memoization search (top-down dynamic programming) to efficiently count the number of good strings.

Implementation Details:

1. Main Function Structure:

   • Define a recursive function dfs(i) that calculates the number of good strings that can be constructed starting from length i
   • The base case: if i > high, return 0 since we've exceeded the maximum allowed length
   • Use the @cache decorator to memoize results and avoid redundant calculations

2. Counting Logic:

   if low <= i <= high:
       ans += 1

   • When the current length i falls within the valid range [low, high], we've found a good string, so increment the count by 1

3. Recursive Transitions:

   ans += dfs(i + zero) + dfs(i + one)

   • From current length i, we can either:
     • Add zero zeros to reach length i + zero
     • Add one ones to reach length i + one
   • The total count is the sum of good strings from both branches

4. Modulo Operation:

   return ans % mod

   • Apply modulo 10^9 + 7 to keep the numbers manageable and meet the problem requirements

5. Starting Point:

   return dfs(0)

   • Start from length 0 (empty string) and explore all possibilities

Why Memoization Works:

The key insight is that many different sequences of operations can lead to the same length. For instance, if we've already calculated dfs(5), we don't need to recalculate it whether we reached length 5 by adding 2+3 or 3+2. The @cache decorator automatically stores and retrieves these computed values.

Time Complexity: O(high) - We compute at most high + 1 different states (lengths from 0 to high)

Space Complexity: O(high) - For the memoization cache and recursion stack

Example Walkthrough

Let's walk through a concrete example with zero = 2, one = 3, low = 3, high = 5.

Starting from an empty string (length 0), we'll trace through the recursive calls:

Initial Call: dfs(0)

• Current length: 0
• Is 0 in [3, 5]? No, so don't count this
• Explore two branches:
  • dfs(0 + 2) = dfs(2) (add 2 zeros)
  • dfs(0 + 3) = dfs(3) (add 3 ones)

Branch 1: dfs(2)

• Current length: 2
• Is 2 in [3, 5]? No, so don't count this
• Explore two branches:
  • dfs(2 + 2) = dfs(4) (add 2 more zeros)
  • dfs(2 + 3) = dfs(5) (add 3 ones)

Branch 1.1: dfs(4)

• Current length: 4
• Is 4 in [3, 5]? Yes! Count = 1
• Explore two branches:
  • dfs(4 + 2) = dfs(6) (exceeds high=5, returns 0)
  • dfs(4 + 3) = dfs(7) (exceeds high=5, returns 0)
• Return: 1 + 0 + 0 = 1

Branch 1.2: dfs(5)

• Current length: 5
• Is 5 in [3, 5]? Yes! Count = 1
• Explore two branches:
  • dfs(5 + 2) = dfs(7) (exceeds high=5, returns 0)
  • dfs(5 + 3) = dfs(8) (exceeds high=5, returns 0)
• Return: 1 + 0 + 0 = 1

Back to dfs(2): Returns 1 + 1 = 2

Branch 2: dfs(3)

• Current length: 3
• Is 3 in [3, 5]? Yes! Count = 1
• Explore two branches:
  • dfs(3 + 2) = dfs(5) (already computed via memoization, returns 1)
  • dfs(3 + 3) = dfs(6) (exceeds high=5, returns 0)
• Return: 1 + 1 + 0 = 2

Final Result: dfs(0) = dfs(2) + dfs(3) = 2 + 2 = 4

The 4 good strings correspond to:

1. Length 3: "111" (path: 0→3)
2. Length 4: "0011" (path: 0→2→4)
3. Length 5: "00111" (path: 0→2→5)
4. Length 5: "11100" (path: 0→3→5)

Note how dfs(5) was reached through two different paths (0→2→5 and 0→3→5), but memoization ensured we only computed it once.

Solution Implementation

Time and Space Complexity

The time complexity is O(high). The recursive function dfs(i) explores all possible values from 0 to high. Due to memoization with @cache, each unique value of i from 0 to high is computed exactly once. Since we can reach at most high + 1 different states (values from 0 to high), and each state performs constant time operations (checking conditions and making at most two recursive calls that are either cached or will be cached), the total time complexity is O(high).

The space complexity is O(high). This comes from two sources:

1. The memoization cache stores results for each unique value of i that the function visits, which can be at most high + 1 values (from 0 to high).
2. The recursion stack depth in the worst case can also be O(high), when we have consecutive calls like dfs(0) → dfs(1) → dfs(2) → ... → dfs(high) if either zero or one equals 1.

Therefore, the overall space complexity is O(high).

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

Common Pitfalls

1. Forgetting to Apply Modulo at Each Step

Pitfall: A common mistake is only applying modulo at the final return, not during intermediate calculations:

# WRONG - Can cause integer overflow
@cache
def dfs(current_length: int) -> int:
    if current_length > high:
        return 0
  
    answer = 0
    if low <= current_length <= high:
        answer += 1
  
    # This addition can overflow before modulo is applied
    answer += dfs(current_length + zero) + dfs(current_length + one)
  
    return answer  # Missing modulo here!

# Then applying modulo only at the end
return dfs(0) % MOD

Solution: Apply modulo at each recursive step to prevent overflow:

# CORRECT
@cache
def dfs(current_length: int) -> int:
    if current_length > high:
        return 0
  
    answer = 0
    if low <= current_length <= high:
        answer += 1
  
    answer += dfs(current_length + zero) + dfs(current_length + one)
  
    return answer % MOD  # Apply modulo here

2. Using Bottom-Up DP with Incorrect Range

Pitfall: When converting to bottom-up DP, iterating only from low to high:

# WRONG - Misses valid combinations
dp = [0] * (high + 1)
for i in range(low, high + 1):  # Starting from low is incorrect
    if i - zero >= 0:
        dp[i] += dp[i - zero]
    if i - one >= 0:
        dp[i] += dp[i - one]

Solution: Start from 0 and build up all possible lengths:

# CORRECT
dp = [0] * (high + 1)
dp[0] = 1  # Empty string as base case

for i in range(1, high + 1):
    if i - zero >= 0:
        dp[i] = (dp[i] + dp[i - zero]) % MOD
    if i - one >= 0:
        dp[i] = (dp[i] + dp[i - one]) % MOD

result = sum(dp[low:high + 1]) % MOD

3. Incorrect Base Case Handling

Pitfall: Not properly handling the case when low = 0:

# WRONG - Doesn't count empty string when low = 0
@cache
def dfs(current_length: int) -> int:
    if current_length > high:
        return 0
  
    answer = 0
    if low < current_length <= high:  # Wrong condition
        answer += 1

Solution: Use inclusive comparison for both bounds:

# CORRECT
if low <= current_length <= high:  # Inclusive on both ends
    answer += 1

4. Cache Clearing Between Test Cases

Pitfall: In competitive programming platforms with multiple test cases, forgetting to clear the cache:

# WRONG - Cache persists across test cases
@cache
def dfs(current_length: int) -> int:
    # ... implementation
  
# Multiple test cases will use stale cached values

Solution: Clear cache or use local function definition:

# CORRECT - Define function inside the main method
def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
    @cache
    def dfs(current_length: int) -> int:
        # ... implementation
  
    return dfs(0)
    # Cache is local to this function call