Create: 1220-count-vowels-permutation.cpp

Author: ashutosh2706

cpp/1220-count-vowels-permutation.cpp

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+/*
+Given an integer n, our 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, we have to return it modulo 10^9 + 7.
+
+Example. For n = 2, Output = 10
+	 
+	Explanation: All possible strings of length 2 that can be formed as per the given rules are: "ae", "ea", "ei", "ia", "ie", "io", "iu",
+	"oi", "ou" and "ua".
+	So we return 10 as our answer. 
+
+
+Time: O(n)
+Space: O(1)
+
+*/
+
+
+class Solution {
+const unsigned int mod = 1e9+7;
+public:
+    int countVowelPermutation(int n) {
+        vector<int> prev(5,1), curr(5, 0);
+        for(int i=1; i<n; i++) {
+            curr[0] = prev[1] % mod;
+            curr[1] = (prev[0] + prev[2]) % mod;
+            curr[2] = ((prev[0] % mod) + (prev[1] % mod) + (prev[3] % mod) + (prev[4] % mod)) % mod;
+            curr[3] = (prev[4] + prev[2]) % mod;
+            curr[4] = prev[0] % mod;
+            prev = curr;
+        }
+        int ans = 0;
+        for(auto &a:prev) ans = (ans + a) % mod;
+        return ans;
+    }
+};