Count all possible N-length vowel permutations that can be generated based on the given conditions

0 0 7 minutes read

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 code
int main()
{
    int N = 2;
     
    cout << countVowelPermutation(N);
}
 
// This code is contributed by Mohit kumar 29

Java

// Java program for the above approach
 
import 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 code
if __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:

Time Complexity: O(N)
Auxiliary Space: O(N)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!