Problem Description

You need to count how many "good" binary strings exist within certain constraints.

A binary string consists only of 0s and 1s. When we look at these strings, consecutive identical digits form "blocks". For example, in the string 0011101111100, we have blocks of zeros [2, 1, 2] (at the beginning, middle, and end) and blocks of ones [3, 5].

A binary string is considered good if it meets three conditions:

1. Length constraint: The string's length must be between minLength and maxLength (inclusive).

2. Ones constraint: Every block of consecutive 1s must have a size that's a multiple of oneGroup. For instance, if oneGroup = 3, then blocks of 1s can only be of size 3, 6, 9, etc.

3. Zeros constraint: Every block of consecutive 0s must have a size that's a multiple of zeroGroup. For instance, if zeroGroup = 2, then blocks of 0s can only be of size 2, 4, 6, etc.

Important note: The value 0 is considered a multiple of every number. This means an empty block (no consecutive 0s or 1s at all) is valid.

Your task is to find the total number of good binary strings. Since this number can be very large, return the result modulo 10^9 + 7.

For example, if minLength = 2, maxLength = 3, oneGroup = 1, and zeroGroup = 2:

• The string 00 is good (one block of 2 zeros, which is a multiple of 2)
• The string 11 is good (one block of 2 ones, which is a multiple of 1)
• The string 001 is good (one block of 2 zeros and one block of 1 one)
• And so on...

You need to count all such valid strings and return the count.

Intuition

Think about how we build a good binary string step by step. At any point, we can either add a complete block of consecutive 1s or a complete block of consecutive 0s. Since blocks must be multiples of oneGroup or zeroGroup, we can only add blocks of size oneGroup, 2*oneGroup, 3*oneGroup... for 1s, and similarly for 0s.

But here's the key insight: instead of thinking about adding variable-sized blocks, we can simplify by only considering adding blocks of the minimum valid size - exactly oneGroup 1s or exactly zeroGroup 0s at a time. Why? Because any larger valid block (like 2*oneGroup) can be formed by adding the minimum block multiple times consecutively.

This transforms our problem into something similar to the "climbing stairs" problem. If we want to form a string of length i, we can either:

• Take a valid string of length i - oneGroup and add a block of oneGroup ones
• Take a valid string of length i - zeroGroup and add a block of zeroGroup zeros

This leads us to dynamic programming. Let f[i] represent the number of ways to form a valid string of exactly length i. Then:

• f[0] = 1 (empty string is valid)
• For any i > 0: f[i] = f[i - oneGroup] + f[i - zeroGroup] (if the indices are valid)

The beauty of this approach is that it automatically handles all valid combinations. When we add a block to an existing string, we don't need to worry about whether we're extending an existing block or starting a new one - the math works out correctly either way because we're always adding complete valid blocks.

Finally, since we want strings of length between minLength and maxLength, we sum up all f[i] where minLength ≤ i ≤ maxLength.

Learn more about Dynamic Programming patterns.

Solution Approach

We implement the dynamic programming solution using a 1D array f where f[i] represents the number of valid binary strings of length i.

Initialization:

• Create an array f of size maxLength + 1
• Set f[0] = 1 (empty string is valid)
• All other positions start at 0

State Transition: For each position i from 1 to maxLength, we calculate f[i] based on two possible transitions:

1. Adding a block of ones: If i >= oneGroup, we can form a string of length i by taking any valid string of length i - oneGroup and appending oneGroup ones to it. This contributes f[i - oneGroup] ways.

2. Adding a block of zeros: If i >= zeroGroup, we can form a string of length i by taking any valid string of length i - zeroGroup and appending zeroGroup zeros to it. This contributes f[i - zeroGroup] ways.

The recurrence relation is:

f[i] = f[i - oneGroup] + f[i - zeroGroup]

(where we only add each term if the index is valid)

Implementation Details:

f = [1] + [0] * maxLength  # f[0] = 1, rest are 0
for i in range(1, len(f)):
    if i - oneGroup >= 0:
        f[i] += f[i - oneGroup]
    if i - zeroGroup >= 0:
        f[i] += f[i - zeroGroup]
    f[i] %= mod  # Apply modulo to prevent overflow

Final Answer: Sum all values from f[minLength] to f[maxLength] (inclusive). This gives us the total count of valid strings with lengths in the required range.

return sum(f[minLength:]) % mod

The time complexity is O(maxLength) as we iterate through each position once. The space complexity is also O(maxLength) for storing the DP array.

Example Walkthrough

Let's walk through a small example with minLength = 2, maxLength = 4, oneGroup = 2, and zeroGroup = 1.

We need to find all valid binary strings of length 2, 3, or 4 where:

• Every block of 1s has size that's a multiple of 2 (so blocks of size 2, 4, 6...)
• Every block of 0s has size that's a multiple of 1 (so blocks of size 1, 2, 3...)

Step 1: Initialize DP array Create array f where f[i] = number of valid strings of length i:

• f[0] = 1 (empty string is valid)
• f[1] = 0, f[2] = 0, f[3] = 0, f[4] = 0 (initially)

Step 2: Build up the DP table

For i = 1:

• Can we add a block of 2 ones? No, because 1 - 2 = -1 (invalid index)
• Can we add a block of 1 zero? Yes, from f[0]: f[1] = f[0] = 1
• Valid strings of length 1: "0"

For i = 2:

• Can we add a block of 2 ones? Yes, from f[0]: contributes f[0] = 1
• Can we add a block of 1 zero? Yes, from f[1]: contributes f[1] = 1
• f[2] = 1 + 1 = 2
• Valid strings of length 2: "11" (from f[0] + two 1s), "00" (from f[1] + one 0)

For i = 3:

• Can we add a block of 2 ones? Yes, from f[1]: contributes f[1] = 1
• Can we add a block of 1 zero? Yes, from f[2]: contributes f[2] = 2
• f[3] = 1 + 2 = 3
• Valid strings of length 3: "011" (from "0" + two 1s), "110" (from "11" + one 0), "000" (from "00" + one 0)

For i = 4:

• Can we add a block of 2 ones? Yes, from f[2]: contributes f[2] = 2
• Can we add a block of 1 zero? Yes, from f[3]: contributes f[3] = 3
• f[4] = 2 + 3 = 5
• Valid strings of length 4: "1111" (from "11" + two 1s), "0011" (from "00" + two 1s), "1100" (from "110" + one 0), "0110" (from "011" + one 0), "0000" (from "000" + one 0)

Step 3: Calculate final answer Sum values for lengths in range [2, 4]:

• f[2] + f[3] + f[4] = 2 + 3 + 5 = 10

The answer is 10 valid binary strings.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n = maxLength. The algorithm uses dynamic programming with a single loop that iterates from 1 to maxLength (inclusive). Within each iteration, it performs constant-time operations: at most two additions, two array accesses, and one modulo operation. Therefore, the overall time complexity is linear with respect to maxLength.

The space complexity is O(n), where n = maxLength. The algorithm creates an array f of size maxLength + 1 to store the dynamic programming states. This is the dominant space usage in the algorithm. The additional variables (mod, i, and the sum computation) use constant space. Thus, the total space complexity is linear with respect to maxLength.

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

Common Pitfalls

1. Misunderstanding the DP State Transition

Pitfall: A common mistake is thinking that dp[i] represents strings ending with a specific digit (0 or 1), leading to incorrect state transitions like trying to track the last character or attempting to alternate between 0s and 1s.

Why it happens: The problem mentions "blocks" of consecutive digits, which might mislead you into thinking you need to track what digit the string ends with to ensure valid block formations.

Correct Understanding: The DP state dp[i] represents ALL valid binary strings of length i, regardless of what digit they end with. When we add oneGroup ones or zeroGroup zeros, we're adding a complete valid block to any valid string, and the block constraint is automatically satisfied because we only add blocks of the required sizes.

2. Incorrect Modulo Application

Pitfall: Forgetting to apply modulo during intermediate calculations, only applying it at the final return:

# WRONG - can cause integer overflow
for length in range(1, len(dp)):
    if length - oneGroup >= 0:
        dp[length] += dp[length - oneGroup]
    if length - zeroGroup >= 0:
        dp[length] += dp[length - zeroGroup]
    # Missing modulo here!

return sum(dp[minLength:]) % MOD  # Only applying modulo at the end

Solution: Apply modulo after each update to prevent overflow:

# CORRECT
for length in range(1, len(dp)):
    if length - oneGroup >= 0:
        dp[length] += dp[length - oneGroup]
    if length - zeroGroup >= 0:
        dp[length] += dp[length - zeroGroup]
    dp[length] %= MOD  # Apply modulo after each calculation

3. Edge Case: Same Group Sizes

Pitfall: Double-counting when oneGroup == zeroGroup. Some might think special handling is needed to avoid counting the same transitions twice.

Why it's not a problem: Even when oneGroup == zeroGroup, the transitions represent different actions (adding ones vs adding zeros), creating different binary strings. For example, if both groups are size 2, adding "11" vs "00" to an empty string creates different valid strings.

4. Overthinking Block Combinations

Pitfall: Trying to explicitly enumerate or track different combinations of blocks (like "first add 2 zeros, then 3 ones, then 2 zeros...").

Why it's unnecessary: The DP approach automatically handles all valid combinations. When we compute dp[i], it includes ALL possible ways to form valid strings of length i through any valid sequence of block additions. The beauty of this solution is that it abstracts away the complexity of block ordering and combinations.