LeetCode 1641: Count Sorted Vowel Strings

Explanation

To solve this problem, we can use dynamic programming. We can maintain a 2D array dp where dp[i][j] represents the number of valid strings of length i+1 ending with the vowel at index j (0-indexed) in the vowels array ['a', 'e', 'i', 'o', 'u']. We can calculate the values in the dp array iteratively based on the previous values.

At each step, the number of valid strings ending with a vowel at index j is the sum of the number of valid strings ending with a vowel at index j or any vowel before it. This is because the strings are required to be lexicographically sorted.

Finally, we sum up the values in the last row of the dp array to get the total number of valid strings of length n.

Time complexity: O(n)
Space complexity: O(5n) = O(n)

class Solution {
    public int countVowelStrings(int n) {
        int[][] dp = new int[n][5];
        
        for (int i = 0; i < 5; i++) {
            dp[0][i] = 1;
        }
        
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < 5; j++) {
                for (int k = 0; k <= j; k++) {
                    dp[i][j] += dp[i-1][k];
                }
            }
        }
        
        int result = 0;
        for (int i = 0; i < 5; i++) {
            result += dp[n-1][i];
        }
        
        return result;
    }
}