# Solving Leetcode Interviews in Seconds with AI: Count Ways To Build Good Strings


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2466" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the solution to the problem:

*   **Dynamic Programming:** The core idea is to use dynamic programming to count the number of ways to form strings of different lengths. We build up a `dp` array where `dp[i]` represents the number of good strings of length `i`.
*   **Base Case and Transitions:** The base case is `dp[0] = 1` (empty string). The transition is `dp[i] = dp[i - zero] + dp[i - one]` if `i >= zero` and `i >= one` respectively. This represents appending '0' `zero` times or appending '1' `one` times.
*   **Counting Good Strings:** Iterate through the `dp` array from `low` to `high` and sum up the values.

*   **Time and Space Complexity:** O(high), O(high)

	
	# Code
	```python
	def countGoodStrings(low: int, high: int, zero: int, one: int) -> int:
    """
    Counts the number of good strings that can be constructed.

    Args:
        low: The minimum length of a good string.
        high: The maximum length of a good string.
        zero: The number of times to append '0'.
        one: The number of times to append '1'.

    Returns:
        The number of different good strings that can be constructed, modulo 10^9 + 7.
    """

    MOD = 10**9 + 7
    dp = [0] * (high + 1)
    dp[0] = 1

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

    count = 0
    for i in range(low, high + 1):
        count = (count + dp[i]) % MOD

    return count
	```