Introduction

In this blog post, we will explore how to solve the LeetCode problem "1220" 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 an integer n, your task is to count how many strings of length n can be formed under the following rules: Each character is a lower case vowel ('a', 'e', 'i', 'o', 'u') Each vowel 'a' may only be followed by an 'e'. Each vowel 'e' may only be followed by an 'a' or an 'i'. Each vowel 'i' may not be followed by another 'i'. Each vowel 'o' may only be followed by an 'i' or a 'u'. Each vowel 'u' may only be followed by an 'a'. Since the answer may be too large, return it modulo 10^9 + 7. Example 1: Input: n = 1 Output: 5 Explanation: All possible strings are: "a", "e", "i" , "o" and "u". Example 2: Input: n = 2 Output: 10 Explanation: All possible strings are: "ae", "ea", "ei", "ia", "ie", "io", "iu", "oi", "ou" and "ua". Example 3: Input: n = 5 Output: 68 Constraints: 1 <= n <= 2 * 10^4

Explanation

Here's the solution to the problem, incorporating efficiency and clarity:

• Dynamic Programming: The core idea is to use dynamic programming to build up the counts of valid strings of increasing lengths. We maintain counts of strings ending in each vowel.
• State Transition: The transitions are based on the allowed vowel sequences. For instance, the number of strings of length k ending in 'e' is the sum of strings of length k-1 ending in 'a'.

• Space Optimization: Instead of storing counts for all lengths from 1 to n, we only need to keep track of the counts for the current length. This significantly reduces the space complexity.

• Runtime Complexity: O(n)

• Storage Complexity: O(1)

Code

    def countVowelPermutation(n: int) -> int:
    """
    Counts the number of strings of length n formed by vowels with given constraints.

    Args:
        n: The length of the strings.

    Returns:
        The number of valid strings modulo 10^9 + 7.
    """

    MOD = 10**9 + 7

    a, e, i, o, u = 1, 1, 1, 1, 1  # Initialize counts for length 1

    for _ in range(1, n):
        a_new = (e + i + u) % MOD
        e_new = (a + i) % MOD
        i_new = (e + o) % MOD
        o_new = i % MOD
        u_new = (i + o) % MOD

        a, e, i, o, u = a_new, e_new, i_new, o_new, u_new

    return (a + e + i + o + u) % MOD