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, 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

    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