onenoughtone

Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

Dynamic Programming (2D) - Complex approach
Dynamic Programming with Space Optimization - Complex approach
Matrix Exponentiation - Complex approach

Approach 1: Dynamic Programming (2D)

The key insight for solving this problem is to use dynamic programming to count the number of valid strings of each length ending with each vowel.

Let's define a 2D DP array dp[i][j] where:

i represents the length of the string (from 1 to n)
j represents the last vowel (0 for 'a', 1 for 'e', 2 for 'i', 3 for 'o', 4 for 'u')

The value dp[i][j] represents the number of valid strings of length i that end with vowel j.

The base cases are:

dp[1][j] = 1 for all j from 0 to 4 (there is one valid string of length 1 for each vowel)

For the recurrence relation, we need to consider the rules for each vowel:

For strings ending with 'a' (j = 0): they can only be formed by appending 'a' to strings ending with 'e', 'i', or 'u'.
For strings ending with 'e' (j = 1): they can only be formed by appending 'e' to strings ending with 'a' or 'i'.
For strings ending with 'i' (j = 2): they can only be formed by appending 'i' to strings ending with 'e' or 'o'.
For strings ending with 'o' (j = 3): they can only be formed by appending 'o' to strings ending with 'i'.
For strings ending with 'u' (j = 4): they can only be formed by appending 'u' to strings ending with 'i' or 'o'.

This gives us the following recurrence relations:

dp[i][0] = dp[i-1][1] + dp[i-1][2] + dp[i-1][4] (strings ending with 'a')
dp[i][1] = dp[i-1][0] + dp[i-1][2] (strings ending with 'e')
dp[i][2] = dp[i-1][1] + dp[i-1][3] (strings ending with 'i')
dp[i][3] = dp[i-1][2] (strings ending with 'o')
dp[i][4] = dp[i-1][2] + dp[i-1][3] (strings ending with 'u')

The final answer is the sum of dp[n][j] for all j from 0 to 4.

Algorithm:

Initialize a 2D DP array dp of size (n+1) × 5, where dp[i][j] represents the number of valid strings of length i that end with vowel j.
Set the base cases: dp[1][j] = 1 for all j from 0 to 4.
For each length i from 2 to n:
a. dp[i][0] = (dp[i-1][1] + dp[i-1][2] + dp[i-1][4]) % MOD (strings ending with 'a')
b. dp[i][1] = (dp[i-1][0] + dp[i-1][2]) % MOD (strings ending with 'e')
c. dp[i][2] = (dp[i-1][1] + dp[i-1][3]) % MOD (strings ending with 'i')
d. dp[i][3] = dp[i-1][2] % MOD (strings ending with 'o')
e. dp[i][4] = (dp[i-1][2] + dp[i-1][3]) % MOD (strings ending with 'u')
Calculate the final answer as the sum of dp[n][j] for all j from 0 to 4, modulo 10^9 + 7.

Time Complexity:

O(n)

We need to compute the DP values for each length from 1 to n and for each of the 5 vowels, which requires a constant amount of work for each length.

Space Complexity:

O(n)

We use a 2D DP array of size (n+1) × 5 to store the number of valid strings for each length and each vowel.

Approach 2: Dynamic Programming with Space Optimization

We can optimize the space complexity by observing that to compute the DP values for the current length, we only need the DP values for the previous length.

Instead of using a 2D DP array, we can use two 1D arrays: curr and prev, each of size 5, to store the DP values for the current length and the previous length, respectively.

After computing the DP values for the current length, we update prev to curr and reset curr for the next length.

The recurrence relations remain the same:

curr[0] = (prev[1] + prev[2] + prev[4]) % MOD (strings ending with 'a')
curr[1] = (prev[0] + prev[2]) % MOD (strings ending with 'e')
curr[2] = (prev[1] + prev[3]) % MOD (strings ending with 'i')
curr[3] = prev[2] % MOD (strings ending with 'o')
curr[4] = (prev[2] + prev[3]) % MOD (strings ending with 'u')

The final answer is the sum of prev[j] for all j from 0 to 4 after the last iteration.

Algorithm:

Initialize two 1D arrays prev and curr, each of size 5, where prev[j] represents the number of valid strings of length i-1 that end with vowel j.
Set the base cases: prev[j] = 1 for all j from 0 to 4.
For each length i from 2 to n:
a. curr[0] = (prev[1] + prev[2] + prev[4]) % MOD (strings ending with 'a')
b. curr[1] = (prev[0] + prev[2]) % MOD (strings ending with 'e')
c. curr[2] = (prev[1] + prev[3]) % MOD (strings ending with 'i')
d. curr[3] = prev[2] % MOD (strings ending with 'o')
e. curr[4] = (prev[2] + prev[3]) % MOD (strings ending with 'u')
f. Update prev = curr.
Calculate the final answer as the sum of prev[j] for all j from 0 to 4, modulo 10^9 + 7.

Time Complexity:

O(n)

The time complexity remains the same as the 2D DP approach, as we still need to compute the DP values for each length and each vowel.

Space Complexity:

O(1)

We only use two 1D arrays of size 5, which is a constant space regardless of the input size.

Approach 3: Matrix Exponentiation

For large values of n, we can use matrix exponentiation to compute the result more efficiently.

The recurrence relations can be represented using a 5×5 transition matrix M:

M = [ [0, 1, 1, 0, 1], // a can be followed by e, i, u [1, 0, 1, 0, 0], // e can be followed by a, i [0, 1, 0, 1, 0], // i can be followed by e, o [0, 0, 1, 0, 0], // o can be followed by i [0, 0, 1, 1, 0] // u can be followed by i, o ]

Let's define a column vector v of size 5, where v[j] = 1 for all j from 0 to 4, representing the base case.

Then, the number of valid strings of length n is given by the sum of all elements in the vector M^(n-1) * v.

We can compute M^(n-1) efficiently using binary exponentiation, which takes O(log n) time.

Algorithm:

Define the transition matrix M as described above.
Define a column vector v of size 5, where v[j] = 1 for all j from 0 to 4.
Compute M^(n-1) using binary exponentiation.
Compute the result vector res = M^(n-1) * v.
Calculate the final answer as the sum of all elements in res, modulo 10^9 + 7.

Time Complexity:

O(log n)

Using binary exponentiation, we can compute the matrix power in logarithmic time.

Space Complexity:

O(1)

We use a constant amount of space to store the matrices and vectors, regardless of the input size.

Pseudocode

solution.pseudo

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function countVowelPermutation(n):
    MOD = 10^9 + 7
    
    // Initialize DP array
    dp = new 2D array of size (n+1) × 5, initialized with 0
    
    // Base cases
    for j from 0 to 4:
        dp[1][j] = 1
    
    // Fill DP array
    for i from 2 to n:
        dp[i][0] = (dp[i-1][1] + dp[i-1][2] + dp[i-1][4]) % MOD  // a
        dp[i][1] = (dp[i-1][0] + dp[i-1][2]) % MOD               // e
        dp[i][2] = (dp[i-1][1] + dp[i-1][3]) % MOD               // i
        dp[i][3] = dp[i-1][2] % MOD                              // o
        dp[i][4] = (dp[i-1][2] + dp[i-1][3]) % MOD               // u
    
    // Calculate the final answer
    result = 0
    for j from 0 to 4:
        result = (result + dp[n][j]) % MOD
    
    return result

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Learning Objective

Understand different approaches to solve the count vowels permutation problem and analyze their efficiency.