Sponsored
Sponsored
We can use dynamic programming to solve this problem whereby we maintain a table dp[i][j] where i is the length of the string and j is the vowel index from the sorted vowels list [0: 'a', 1: 'e', 2: 'i', 3: 'o', 4: 'u']. Each cell dp[i][j] represents the number of valid strings of length i using the vowels from index j to the end. We initialize the base case for strings of length 1 and fill the table iteratively.
Time Complexity: O(n * 5 * 5) = O(n), as we have nested loops running across vowels and length.
Space Complexity: O(n * 5), for storing the table.
1def countVowelStrings(n: int) -> int:
2 dp = [[0] * 5 for _ in range(n + 1)]
3 for j in range(5):
4 dp[1][j] = 5 - j
5 for i in range(2, n + 1):
6 for j in range(4, -1, -1):
7 dp[i][j] = 0
8 for k in range(j, 5):
9 dp[i][j] += dp[i-1][k]
10 return dp[n][0]The Python implementation uses list comprehensions to set up dynamic programming tables with iterative replacement based on valid string extension counts.
Instead of counting strings iteratively, we can view the problem as a mathematical combination problem. Since the string is lexicographically sorted, the placement of 'n' vowels among 'a', 'e', 'i', 'o', 'u' is equivalent to distributing 'n' identical items into 5 distinct groups (vowel buckets). We use the combinatorial formula: C(n + m - 1, m - 1) where 'm' is the number of distinct vowels.
Time Complexity: O(n), due to the recursive calculation of combination.
Space Complexity: O(n), for the recursive call stack.
This solution provides a direct combinatory calculation using the formula for combinations. The helper function computes combinations recursively.