Count Vowels Permutation LeetCode solution in Java

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class Solution {
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  private long[][] dp;
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  private int mod = (int) 1e9 + 7;
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  public int countVowelPermutation(int n) {
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    dp = new long[6][n + 1];
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    if (n == 1) return 5;
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    for (int i = 0; i < 5; i++) dp[i][0] = 1;
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    helper(n, 'z');
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    return (int) dp[5][n];
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  }
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  private long helper(int n, char vowel) {
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    long ans = 0;
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    if (n == 0) return 1;
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    if (vowel == 'z') // we are using z for our convenience just to add Permutations of all Vowels
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    {
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      ans =
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          (ans
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                  + helper(n - 1, 'a')
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                  + helper(n - 1, 'e')
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                  + helper(n - 1, 'i')
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                  + helper(n - 1, 'o')
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                  + helper(n - 1, 'u'))
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              % mod;
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      dp[5][n] = ans;
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    }
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    // from here as per our assumptions of Vowels we will make calls & store results
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    else if (vowel == 'a') // for Nth number we would store Result for "a" in dp[0][n]
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    {
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      if (dp[0][n] != 0) return dp[0][n];
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      ans = (ans + helper(n - 1, 'e')) % mod;
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      dp[0][n] = ans;
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    } else if (vowel == 'e') // for Nth number we would store Result for "e" in dp[1][n]
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    {
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      if (dp[1][n] != 0) return dp[1][n];
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      ans = (ans + helper(n - 1, 'a') + helper(n - 1, 'i')) % mod;
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      dp[1][n] = ans;
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    } else if (vowel == 'i') // for Nth number we would store Result for "i" in dp[2][n]
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    {
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      if (dp[2][n] != 0) return dp[2][n];
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      ans =
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          (ans + helper(n - 1, 'a') + helper(n - 1, 'e') + helper(n - 1, 'o') + helper(n - 1, 'u'))
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              % mod;
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      dp[2][n] = ans;
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    } else if (vowel == 'o') // for Nth number we would store Result for "o" in dp[3][n]
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    {
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      if (dp[3][n] != 0) return dp[3][n];
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      ans = (ans + helper(n - 1, 'i') + helper(n - 1, 'u')) % mod;
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      dp[3][n] = ans;
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    } else // for Nth number we would store Result for "u" in dp[4][n]
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    {
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      if (dp[4][n] != 0) return dp[4][n];
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      ans = (ans + helper(n - 1, 'a')) % mod;
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      dp[4][n] = ans;
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    }
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    return ans;
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  }
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}