Given an integer N, the task is to count the number of N-length strings consisting of lowercase vowels that can be generated based the following conditions:
- Each ‘a’ may only be followed by an ‘e’.
- Each ‘e’ may only be followed by an ‘a’ or an ‘i’.
- Each ‘i’ may not be followed by another ‘i’.
- Each ‘o’ may only be followed by an ‘i’ or a ‘u’.
- Each ‘u’ may only be followed by an ‘a’.
Examples:
Input: N = 1
Output: 5
Explanation: All strings that can be formed are: “a”, “e”, “i”, “o” and “u”.Input: N = 2
Output: 10
Explanation: All strings that can be formed are: “ae”, “ea”, “ei”, “ia”, “ie”, “io”, “iu”, “oi”, “ou” and “ua”.
Approach: The idea to solve this problem is to visualize this as a Graph Problem. From the given rules a directed graph can be constructed, where an edge from u to v means that v can be immediately written after u in the resultant strings. The problem reduces to finding the number of N-length paths in the constructed directed graph. Follow the steps below to solve the problem:
- Let the vowels a, e, i, o, u be numbered as 0, 1, 2, 3, 4 respectively, and using the dependencies shown in the given graph, convert the graph into an adjacency list relation where the index signifies the vowel and the list at that index signifies an edge from that index to the characters given in the list.
- Initialize a 2D array dp[N + 1][5] where dp[N][char] denotes the number of directed paths of length N which end at a particular vertex char.
- Initialize dp[i][char] for all the characters as 1, since a string of length 1 will only consist of one vowel in the string.
- For all possible lengths, say i, traverse over the directed edges using variable u and perform the following steps:
- Update the value of dp[i + 1][u] as 0.
- Traverse the adjacency list of the node u and increment the value of dp[i][u] by dp[i][v], that stores the sum of all the values such that there is a directed edge from node u to node v.
- After completing the above steps, the sum of all the values dp[N][i], where i belongs to the range [0, 5), will give the total number of vowel permutations.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h>using namespace std;// Function to find the number of // vowel permutations possible int countVowelPermutation(int n){ // To avoid the large output value int MOD = (int)(1e9 + 7); // Initialize 2D dp array long dp[n + 1][5]; // Initialize dp[1][i] as 1 since // string of length 1 will consist // of only one vowel in the string for(int i = 0; i < 5; i++) { dp[1][i] = 1; } // Directed graph using the // adjacency matrix vector<vector<int>> relation = { { 1 }, { 0, 2 }, { 0, 1, 3, 4 }, { 2, 4 }, { 0 } }; // Iterate over the range [1, N] for(int i = 1; i < n; i++) { // Traverse the directed graph for(int u = 0; u < 5; u++) { dp[i + 1][u] = 0; // Traversing the list for(int v : relation[u]) { // Update dp[i + 1][u] dp[i + 1][u] += dp[i][v] % MOD; } } } // Stores total count of permutations long ans = 0; for(int i = 0; i < 5; i++) { ans = (ans + dp[n][i]) % MOD; } // Return count of permutations return (int)ans;}// Driver codeint main(){ int N = 2; cout << countVowelPermutation(N);}// This code is contributed by Mohit kumar 29 |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG { // Function to find the number of // vowel permutations possible public static int countVowelPermutation(int n) { // To avoid the large output value int MOD = (int)(1e9 + 7); // Initialize 2D dp array long[][] dp = new long[n + 1][5]; // Initialize dp[1][i] as 1 since // string of length 1 will consist // of only one vowel in the string for (int i = 0; i < 5; i++) { dp[1][i] = 1; } // Directed graph using the // adjacency matrix int[][] relation = new int[][] { { 1 }, { 0, 2 }, { 0, 1, 3, 4 }, { 2, 4 }, { 0 } }; // Iterate over the range [1, N] for (int i = 1; i < n; i++) { // Traverse the directed graph for (int u = 0; u < 5; u++) { dp[i + 1][u] = 0; // Traversing the list for (int v : relation[u]) { // Update dp[i + 1][u] dp[i + 1][u] += dp[i][v] % MOD; } } } // Stores total count of permutations long ans = 0; for (int i = 0; i < 5; i++) { ans = (ans + dp[n][i]) % MOD; } // Return count of permutations return (int)ans; } // Driver Code public static void main(String[] args) { int N = 2; System.out.println( countVowelPermutation(N)); }} |
Python3
# Python 3 program for the above approach # Function to find the number of # vowel permutations possible def countVowelPermutation(n): # To avoid the large output value MOD = 1e9 + 7 # Initialize 2D dp array dp = [[0 for i in range(5)] for j in range(n + 1)] # Initialize dp[1][i] as 1 since # string of length 1 will consist # of only one vowel in the string for i in range(5): dp[1][i] = 1 # Directed graph using the # adjacency matrix relation = [[1],[0, 2], [0, 1, 3, 4], [2, 4],[0]] # Iterate over the range [1, N] for i in range(1, n, 1): # Traverse the directed graph for u in range(5): dp[i + 1][u] = 0 # Traversing the list for v in relation[u]: # Update dp[i + 1][u] dp[i + 1][u] += dp[i][v] % MOD # Stores total count of permutations ans = 0 for i in range(5): ans = (ans + dp[n][i]) % MOD # Return count of permutations return int(ans)# Driver codeif __name__ == '__main__': N = 2 print(countVowelPermutation(N)) # This code is contributed by bgangwar59. |
C#
// C# program to find absolute difference // between the sum of all odd frequency and // even frequent elements in an array using System;using System.Collections.Generic; class GFG { // Function to find the number of // vowel permutations possible static int countVowelPermutation(int n) { // To avoid the large output value int MOD = (int)(1e9 + 7); // Initialize 2D dp array long[,] dp = new long[n + 1, 5]; // Initialize dp[1][i] as 1 since // string of length 1 will consist // of only one vowel in the string for (int i = 0; i < 5; i++) { dp[1, i] = 1; } // Directed graph using the // adjacency matrix List<List<int>> relation = new List<List<int>>(); relation.Add(new List<int> { 1 }); relation.Add(new List<int> { 0, 2 }); relation.Add(new List<int> { 0, 1, 3, 4 }); relation.Add(new List<int> { 2, 4 }); relation.Add(new List<int> { 0 }); // Iterate over the range [1, N] for (int i = 1; i < n; i++) { // Traverse the directed graph for (int u = 0; u < 5; u++) { dp[i + 1, u] = 0; // Traversing the list foreach(int v in relation[u]) { // Update dp[i + 1][u] dp[i + 1, u] += dp[i, v] % MOD; } } } // Stores total count of permutations long ans = 0; for (int i = 0; i < 5; i++) { ans = (ans + dp[n, i]) % MOD; } // Return count of permutations return (int)ans; } // Driver code static void Main() { int N = 2; Console.WriteLine(countVowelPermutation(N)); }}// This code is contributed by divyesh072019. |
Javascript
<script>// JavaScript program to implement// the above approach // Function to find the number of // vowel permutations possible function countVowelPermutation(n) { // To avoid the large output value let MOD = (1e9 + 7); // Initialize 2D dp array let dp = new Array(n + 1); // Loop to create 2D array using 1D array for (var i = 0; i < dp.length; i++) { dp[i] = new Array(2); } // Initialize dp[1][i] as 1 since // string of length 1 will consist // of only one vowel in the string for (let i = 0; i < 5; i++) { dp[1][i] = 1; } // Directed graph using the // adjacency matrix let relation = [ [ 1 ], [ 0, 2 ], [ 0, 1, 3, 4 ], [ 2, 4 ], [ 0 ] ]; // Iterate over the range [1, N] for (let i = 1; i < n; i++) { // Traverse the directed graph for (let u = 0; u < 5; u++) { dp[i + 1][u] = 0; // Traversing the list for (let v in relation[u]) { // Update dp[i + 1][u] dp[i + 1][u] += dp[i][v] % MOD; } } } // Stores total count of permutations let ans = 0; for (let i = 0; i < 5; i++) { ans = (ans + dp[n][i]) % MOD; } // Return count of permutations return ans; } // Driver code let N = 2; document.write( countVowelPermutation(N));</script> |
Output:
10
Time Complexity: O(N)
Auxiliary Space: O(N)
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