<p>Given an integer <code>n</code>, your task is to count how many strings of length <code>n</code> can be formed under the following rules:</p>

<ul>
	<li>Each character is a lower case vowel&nbsp;(<code>&#39;a&#39;</code>, <code>&#39;e&#39;</code>, <code>&#39;i&#39;</code>, <code>&#39;o&#39;</code>, <code>&#39;u&#39;</code>)</li>
	<li>Each vowel&nbsp;<code>&#39;a&#39;</code> may only be followed by an <code>&#39;e&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;e&#39;</code> may only be followed by an <code>&#39;a&#39;</code>&nbsp;or an <code>&#39;i&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;i&#39;</code> <strong>may not</strong> be followed by another <code>&#39;i&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;o&#39;</code> may only be followed by an <code>&#39;i&#39;</code> or a&nbsp;<code>&#39;u&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;u&#39;</code> may only be followed by an <code>&#39;a&#39;</code>.</li>
</ul>

<p>Since the answer&nbsp;may be too large,&nbsp;return it modulo&nbsp;<code>10^9 + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 5
<strong>Explanation:</strong> All possible strings are: &quot;a&quot;, &quot;e&quot;, &quot;i&quot; , &quot;o&quot; and &quot;u&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 10
<strong>Explanation:</strong> All possible strings are: &quot;ae&quot;, &quot;ea&quot;, &quot;ei&quot;, &quot;ia&quot;, &quot;ie&quot;, &quot;io&quot;, &quot;iu&quot;, &quot;oi&quot;, &quot;ou&quot; and &quot;ua&quot;.
</pre>

<p><strong class="example">Example 3:&nbsp;</strong></p>

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 68</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 2 * 10^4</code></li>
</ul>

// #2 Class to implement the Bottom-up approach:
class Solution {
    vector<char> vowels = {'a','e','i','o','u'};
    const int MOD = 1e9+7;

public:
    // #1 Method to count how many such strings of length n that can be formed, using 2D tabulation - O(N*23*5) & O(N*26)
    int countVowelPermutation(int n) {
        // 2D DP table
        vector<vector<int>> dp(n + 1, vector<int>(26, 0));

        // Intialize the edge case: If you've created such string of length n then return 1
        for(int ch = 0; ch < 26; ++ch)
            dp[0][ch] = 1;

        // Fill the rest of the table
        for(int length = 1; length <= n; ++length) {
            for(char prevLetter = 'a'; prevLetter <= 'w'; ++prevLetter) {
                int count = 0;
                for(char v : vowels) {
                    if(prevLetter == 'w' || (prevLetter == 'a' && v == 'e') || (prevLetter == 'e' && (v == 'a' || v == 'i')) || (prevLetter == 'i' && v != 'i') || (prevLetter == 'o' && (v == 'i' || v == 'u')) || (prevLetter == 'u' && v == 'a')) {
                        count = count % MOD + dp[length - 1][v - 'a'] % MOD;
                    }
                }
                dp[length][prevLetter - 'a'] = count % MOD;
            }
        }

        // Return the result value
        return dp[n]['w' - 'a'];
    }

    // #2 Method to count how many such strings of length n that can be formed, using 1D tabulation - O(N*23*5) & O(2*26)
    int countVowelPermutation_V2(int n) {
        // 1D DP tables
        vector<int> prevRow(26, 0), currRow(26, 0);

        // Intialize the edge case: If you've created such string of length n then return 1
        for(int ch = 0; ch < 26; ++ch)
            prevRow[ch] = 1;

        // Fill the rest of the table
        for(int n = 1; n <= n; ++n) {
            for(char prevLetter = 'a'; prevLetter <= 'w'; ++prevLetter) {
                int count = 0;
                for(char v : vowels) {
                    if(prevLetter == 'w' || (prevLetter == 'a' && v == 'e') || (prevLetter == 'e' && (v == 'a' || v == 'i')) || (prevLetter == 'i' && v != 'i') || (prevLetter == 'o' && (v == 'i' || v == 'u')) || (prevLetter == 'u' && v == 'a')) {
                        count = count % MOD + prevRow[v - 'a'] % MOD;
                    }
                }
                currRow[prevLetter - 'a'] = count % MOD;
            }
            prevRow = currRow;
        }

        // Return the result value
        return prevRow['w' - 'a'];
    }
};
69588