Count Ways To Build Good Strings

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Dynamic Programming

Given the integers zero, one, low, and high, we can construct a string by starting with an empty string, and then at each step perform either of the following:

Append the character '0' zero times.
Append the character '1' one times.

This can be performed any number of times.

A good string is a string constructed by the above process having a length between low and high (inclusive).

Return the number of different good strings that can be constructed satisfying these properties. Since the answer can be large, return it modulo 10⁹ + 7.

Example 1:

Input: low = 3, high = 3, zero = 1, one = 1
Output: 8
Explanation: 
One possible valid good string is "011". 
It can be constructed as follows: "" -> "0" -> "01" -> "011". 
All binary strings from "000" to "111" are good strings in this example.

Example 2:

Input: low = 2, high = 3, zero = 1, one = 2
Output: 5
Explanation: The good strings are "00", "11", "000", "110", and "011".

Constraints:

1 <= low <= high <= 10⁵
1 <= zero, one <= low

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the input values `zero` and `one`? Are they guaranteed to be non-negative integers?
What are the upper bounds for `low` and `high`? Are `low` and `high` guaranteed to be non-negative and is `low` always less than or equal to `high`?
Should the result be modulo'd by a specific number if it gets very large? If so, what is the modulo value?
Is it possible for `zero` or `one` to be equal to 0?
Could you provide an example where multiple good strings can be built with the same length, and how the count would be affected?

Brute Force Solution

Approach

To solve this problem using brute force, we essentially try building every possible string combination. We explore every option to see which ones meet our specific criteria of being a 'good' string, and then we simply count how many of these good strings we find.

Here's how the algorithm would work step-by-step:

Start by considering the empty string.
From the empty string, try adding either a string of zeros (of a certain length) or a string of ones (of a different length).
Each time you add a string, check if the overall length is within the desired range.
If it's too short, keep adding more strings of zeros or ones.
If it's too long, discard that particular combination and try a different one.
If it's exactly the right length, you've found one 'good' string; keep track of it.
Repeat this process, exploring all possible combinations of zeros and ones, until you've exhausted every option.
Finally, count all the 'good' strings you found during this exhaustive search.

Code Implementation

def count_ways_to_build_good_strings_brute_force(
    low, high, zero_length, one_length
):
    number_of_good_strings = 0

    def build_string(current_string):
        nonlocal number_of_good_strings
        string_length = len(current_string)

        # If length is in the desired range, count the good string
        if low <= string_length <= high:
            number_of_good_strings += 1

        # If the length exceeds the maximum, stop building this string.
        if string_length > high:
            return

        # Recursively try adding zeros

        build_string(current_string + "0" * zero_length)

        # Recursively try adding ones
        build_string(current_string + "1" * one_length)

    build_string("")

    return number_of_good_strings

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible combinations of adding substrings of length zero or one until the total length exceeds 'high'. In the worst case, we explore a binary tree where each node represents a decision to append a string of zeros or a string of ones. The depth of the tree can be up to 'high' (let's denote this as 'n' for simplicity), leading to potentially 2^n different paths to explore. Thus, the time complexity is exponential, specifically O(2^n), as we are examining all possible combinations of building strings up to length 'n'.

Space Complexity

O(L) – The brute force approach described builds strings recursively. The maximum depth of the recursion is bounded by L, where L is the high value in the input range, because that's the maximum length the string can reach. Each recursive call stores a new string being constructed, which takes up space. Thus, the maximum space occupied by the recursion call stack and the strings being constructed is proportional to L, giving a space complexity of O(L).

Optimal Solution

Approach

The problem can be solved efficiently using a technique similar to dynamic programming. We build the solution from the ground up, remembering the number of ways to reach each possible length of the string. This avoids redundant calculations by reusing previously computed results.

Here's how the algorithm would work step-by-step:

Imagine building your string piece by piece. We will keep track of how many different ways we can reach a specific length.
Start with a string of length zero. There's one way to have nothing: do nothing.
Now, consider adding either a sequence of zeros or a sequence of ones to the current string. These sequences have fixed lengths.
For each possible length of the current string, add all possible combinations of sequences of zeros and ones to get to a new length.
Keep track of how many ways there are to reach each new length by adding the number of ways to reach the previous length that allows creating the new length.
Repeat this process until you've built all possible strings that fit within the allowed length range.
Finally, count up all the strings whose lengths fall within the given length range (between the lower and upper bound).

Code Implementation

def count_ways_to_build_good_strings(low_length, high_length, zero_length, one_length):    modulo_value = 10**9 + 7
    ways_to_reach_length = [0] * (high_length + 1)

    # Initialize base case: one way to have length zero
    ways_to_reach_length[0] = 1

    for current_length in range(high_length):
        if ways_to_reach_length[current_length] > 0:

            # Try adding a sequence of zeros
            new_length_zeros = current_length + zero_length
            if new_length_zeros <= high_length:
                ways_to_reach_length[new_length_zeros] = (ways_to_reach_length[new_length_zeros] + ways_to_reach_length[current_length]) % modulo_value

            # Try adding a sequence of ones
            new_length_ones = current_length + one_length
            if new_length_ones <= high_length:
                ways_to_reach_length[new_length_ones] = (ways_to_reach_length[new_length_ones] + ways_to_reach_length[current_length]) % modulo_value

    total_ways = 0

    # Sum up ways to reach lengths within the specified range.
    for length in range(low_length, high_length + 1):
        total_ways = (total_ways + ways_to_reach_length[length]) % modulo_value

    return total_ways

Big(O) Analysis

Time Complexity

O(maxLength) – The core of the algorithm involves iterating through all possible string lengths up to a maximum length (maxLength) determined by the upper bound (high). For each length, a fixed number of calculations (at most two: adding zeroGroup or oneGroup) are performed to update the number of ways to reach subsequent lengths. Because the number of computations is directly proportional to the maximum length, the time complexity is O(maxLength), or often noted as O(n) when discussing sequence length.

Space Complexity

O(high) – The algorithm uses a dynamic programming approach to store the number of ways to reach each length up to the 'high' bound, as described in step 5. This requires an array or similar data structure to hold these intermediate results. The size of this auxiliary data structure is directly proportional to the upper bound 'high', representing the maximum possible string length. Therefore, the space complexity is O(high), where 'high' is the upper bound of the string length.

Edge Cases

Case	How to Handle
low > high	Return 0 since no string length can satisfy the condition.
zero = one = 0	If both zero and one are zero, an infinite number of strings can be built; handle appropriately (e.g., throw exception or return a special value like -1).
low or high is extremely large	Use modular arithmetic to prevent integer overflow during the counting process.
zero or one is extremely large (close to high)	The number of possible strings might be very limited, but the algorithm should still work without overflow issues if modular arithmetic is used.
low = high	Calculate the number of strings with exactly that length.
Both zero and one are 1	Each step extends the string by one character, so the number of good strings is simply high - low + 1.
low = 0	Need to account for empty string if low is 0, potentially adding 1 to the initial count, depending on the problem definition.
Memory constraints for large high values when using dynamic programming	Consider using tabulation with optimized space complexity, potentially O(high - low + max(zero, one)).

low > high

How to Handle:

Return 0 since no string length can satisfy the condition.

zero = one = 0

How to Handle:

If both zero and one are zero, an infinite number of strings can be built; handle appropriately (e.g., throw exception or return a special value like -1).

low or high is extremely large

How to Handle:

Use modular arithmetic to prevent integer overflow during the counting process.

zero or one is extremely large (close to high)

How to Handle:

The number of possible strings might be very limited, but the algorithm should still work without overflow issues if modular arithmetic is used.

low = high

How to Handle:

Calculate the number of strings with exactly that length.

Both zero and one are 1

How to Handle:

Each step extends the string by one character, so the number of good strings is simply high - low + 1.

low = 0

How to Handle:

Need to account for empty string if low is 0, potentially adding 1 to the initial count, depending on the problem definition.

Memory constraints for large high values when using dynamic programming

How to Handle:

Consider using tabulation with optimized space complexity, potentially O(high - low + max(zero, one)).