Introduction

In this blog post, we will explore how to solve the LeetCode problem "1641" 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, return the number of strings of length n that consist only of vowels (a, e, i, o, u) and are lexicographically sorted. A string s is lexicographically sorted if for all valid i, s[i] is the same as or comes before s[i+1] in the alphabet. Example 1: Input: n = 1 Output: 5 Explanation: The 5 sorted strings that consist of vowels only are ["a","e","i","o","u"]. Example 2: Input: n = 2 Output: 15 Explanation: The 15 sorted strings that consist of vowels only are ["aa","ae","ai","ao","au","ee","ei","eo","eu","ii","io","iu","oo","ou","uu"]. Note that "ea" is not a valid string since 'e' comes after 'a' in the alphabet. Example 3: Input: n = 33 Output: 66045 Constraints: 1 <= n <= 50

Explanation

Here's the breakdown of the problem and the optimized solution:

• High-Level Approach:

  • Dynamic Programming: Build a table where dp[i][j] stores the number of sorted vowel strings of length i ending with the j-th vowel ('a'=0, 'e'=1, 'i'=2, 'o'=3, 'u'=4).
  • Base Case: For length 1, each vowel contributes 1 to the count.
  • Recursive Relation: dp[i][j] is the sum of dp[i-1][k] for all k from 0 to j (because the last vowel must be greater than or equal to the previous vowel to maintain lexicographical order).

• Complexity:

  • Runtime: O(n * 5) = O(n), as we iterate through the DP table, where n is the input length.
  • Storage: O(n * 5) = O(n), to store the DP table.

Code

    def countVowelStrings(n: int) -> int:
    """
    Given an integer n, return the number of strings of length n that consist only of vowels (a, e, i, o, u) and are lexicographically sorted.
    """
    dp = [[0] * 5 for _ in range(n + 1)]

    # Base case: n = 1
    for j in range(5):
        dp[1][j] = 1

    # Build the dp table
    for i in range(2, n + 1):
        for j in range(5):
            for k in range(j + 1):
                dp[i][j] += dp[i - 1][k]

    # Sum up the counts for all vowels to get the total
    total_count = sum(dp[n])
    return total_count