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

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

10

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

Efficient Approach: As we see that we only need current and previous state of the dp array, We can definitely space optimize the above solution. We can have 5 variables of previous state and 5 variables of current state that we need to compute with the given mapping relations.

Below is the implementation of 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)
{
      const int MOD = 1e9 + 7;
      // initialize current and previous state
      // variables.
    long a = 1, e = 1, i = 1, o = 1, u = 1, a_new, e_new, i_new, o_new, u_new;

    for(int j = 2; j <= n; j++) {
      // respective conditions
      a_new =  e;
      e_new = (a + i) % MOD;
      i_new = (a + e + o + u) % MOD;
      o_new = (i + u) % MOD;
      u_new =  a;
      // make current computed states
      // to make them previous for next computing
      // future states
      a = a_new, e = e_new, i = i_new, o = o_new, u = u_new;
    }
      // return the answer
    return (a + e + i + o + u) % MOD;

}

// Driver code
int main()
{
    int N = 2;

    cout << countVowelPermutation(N);
}

Java

public class VowelPermutation {

    static int countVowelPermutation(int n)
    {
        // Define the modulo value
        final int MOD = 1000000007;

        // Initialize current and previous state variables
        int a = 1, e = 1, i = 1, o = 1, u = 1;

        for (int j = 2; j <= n; j++) {
            // Calculate new states based on conditions
            int aNew = e;
            int eNew = (a + i) % MOD;
            int iNew = (a + e + o + u) % MOD;
            int oNew = (i + u) % MOD;
            int uNew = a;

            // Update current states for the next
            // computation
            a = aNew;
            e = eNew;
            i = iNew;
            o = oNew;
            u = uNew;
        }

        // Return the result
        return (a + e + i + o + u) % MOD;
    }

    public static void main(String[] args)
    {
        // Driver code
        int N = 2;
        System.out.println(countVowelPermutation(N));
    }
}

Python3

def countVowelPermutation(n):
    MOD = 10 ** 9 + 7
    # Initialize current and previous state variables.
    a = e = i = o = u = 1
    for j in range(2, n + 1):
        # Respective conditions
        a_new = e
        e_new = (a + i) % MOD
        i_new = (a + e + o + u) % MOD
        o_new = (i + u) % MOD
        u_new = a
        # Update current states for next computation
        a, e, i, o, u = a_new, e_new, i_new, o_new, u_new
    # Return the answer
    return (a + e + i + o + u) % MOD

# Driver code
N = 2
print(countVowelPermutation(N))

C#

using System;

class Program {
    // Function to find the number of vowel permutations
    // possible
    static int CountVowelPermutation(int n)
    {
        const int MOD = 1000000007;
        // Initialize current and previous state variables.
        long a = 1, e = 1, i = 1, o = 1, u = 1, a_new,
             e_new, i_new, o_new, u_new;

        for (int j = 2; j <= n; j++) {
            // Respective conditions
            a_new = e;
            e_new = (a + i) % MOD;
            i_new = (a + e + o + u) % MOD;
            o_new = (i + u) % MOD;
            u_new = a;

            // Make current computed states
            // to make them previous for next computing
            // future states
            a = a_new;
            e = e_new;
            i = i_new;
            o = o_new;
            u = u_new;
        }
        // Return the answer
        return (int)((a + e + i + o + u) % MOD);
    }

    // Driver code
    static void Main()
    {
        int N = 2;
        Console.WriteLine(CountVowelPermutation(N));
    }
}

Javascript

function countVowelPermutation(n) {
    const MOD = 10 ** 9 + 7;
    // Initialize current and previous state variables.
    let a = e = i = o = u = 1;
    for (let j = 2; j <= n; j++) {
        // Respective conditions
        const a_new = e;
        const e_new = (a + i) % MOD;
        const i_new = (a + e + o + u) % MOD;
        const o_new = (i + u) % MOD;
        const u_new = a;
        // Update current states for next computation
        a = a_new;
        e = e_new;
        i = i_new;
        o = o_new;
        u = u_new;
    }
    // Return the answer
    return (a + e + i + o + u) % MOD;
}

// Driver code
const N = 2;
console.log(countVowelPermutation(N));

Output

10

Time Complexity: O(N)

Space Complexity: O(1)