feat: solve No.1220

Author: BaffinLee

1201-1300/1220. Count Vowels Permutation.md

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+# 1220. Count Vowels Permutation
+
+- Difficulty: Hard.
+- Related Topics: Dynamic Programming.
+- Similar Questions: .
+
+## Problem
+
+Given an integer `n`, your task is to count how many strings of length `n` can be formed under the following rules:
+
+
+	
+- Each character is a lower case vowel (`'a'`, `'e'`, `'i'`, `'o'`, `'u'`)
+	
+- Each vowel `'a'` may only be followed by an `'e'`.
+	
+- Each vowel `'e'` may only be followed by an `'a'` or an `'i'`.
+	
+- Each vowel `'i'` **may not** be followed by another `'i'`.
+	
+- Each vowel `'o'` may only be followed by an `'i'` or a `'u'`.
+	
+- Each vowel `'u'` may only be followed by an `'a'.`
+
+
+Since the answer may be too large, return it modulo `10^9 + 7.`
+
+ 
+Example 1:
+
+```
+Input: n = 1
+Output: 5
+Explanation: All possible strings are: "a", "e", "i" , "o" and "u".
+```
+
+Example 2:
+
+```
+Input: n = 2
+Output: 10
+Explanation: All possible strings are: "ae", "ea", "ei", "ia", "ie", "io", "iu", "oi", "ou" and "ua".
+```
+
+Example 3: 
+
+```
+Input: n = 5
+Output: 68
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= n <= 2 * 10^4`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var countVowelPermutation = function(n) {
+    var dp = Array(n).fill(0).map(() => ({}));
+    return helper('', n, dp);
+};
+
+var helper = function(lastChar, n, dp) {
+    if (n === 0) return 1;
+    if (dp[n - 1][lastChar] !== undefined) return dp[n - 1][lastChar];
+    var mod = Math.pow(10, 9) + 7;
+    var res = 0;
+    if (!lastChar || lastChar === 'e') res = (res + helper('a', n - 1, dp)) % mod;
+    if (!lastChar || lastChar === 'a' || lastChar === 'i') res = (res + helper('e', n - 1, dp)) % mod;
+    if (!lastChar || lastChar !== 'i') res = (res + helper('i', n - 1, dp)) % mod;
+    if (!lastChar || lastChar === 'i' || lastChar === 'u') res = (res + helper('o', n - 1, dp)) % mod;
+    if (!lastChar || lastChar === 'a') res = (res + helper('u', n - 1, dp)) % mod;
+    dp[n - 1][lastChar] = res;
+    return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).