How do you solve Count Vowels Permutation on LeetCode?

Count Vowels Permutation (LeetCode #1220) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Dynamic programming allows us to build solutions incrementally, leveraging previously computed results.

What is the brute force solution for Count Vowels Permutation?

The brute force approach for Count Vowels Permutation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible strings of length n using the vowels and then filters them based on the given rules. While this method is straightforward, it becomes inefficient as n increases due to the exponential growth of combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Vowels Permutation?

Count Vowels Permutation uses the following concepts: Dynamic Programming. The recommended patterns to study are: Dynamic Programming, State Transition.

What are the interview tips for Count Vowels Permutation?

Always think about how to break down the problem into smaller subproblems — this is key for dynamic programming. Explain your thought process clearly to the interviewer; they are interested in your approach, not just the final answer. Practice writing out the state transitions for dynamic programming problems to solidify your understanding.

What are common mistakes when solving Count Vowels Permutation?

Students often try to brute-force the problem without considering more efficient approaches. Failing to properly initialize the base cases in dynamic programming can lead to incorrect results.

#1220

Count Vowels Permutation

Hard

Dynamic ProgrammingDynamic ProgrammingState Transition

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Using dynamic programming, we can build the solution iteratively by counting how many strings of length i can end with each vowel. This approach avoids generating all combinations and directly computes the count based on previous results.

⚙️

Algorithm

4 steps

1Step 1: Initialize a dp array where dp[i][j] represents the number of valid strings of length i ending with the j-th vowel.
2Step 2: Set base cases for strings of length 1 for each vowel.
3Step 3: For each length from 2 to n, calculate the number of valid strings based on the rules.
4Step 4: Sum the counts from the last row of the dp array to get the final result.

solution.py14 lines

1# Full working Python code
2def count_vowel_permutations(n):
3    MOD = 10**9 + 7
4    dp = [[0] * 5 for _ in range(n + 1)]
5    for j in range(5):
6        dp[1][j] = 1
7    for i in range(2, n + 1):
8        dp[i][0] = dp[i - 1][1] % MOD  # 'a' can only follow 'e'
9        dp[i][1] = (dp[i - 1][0] + dp[i - 1][2]) % MOD  # 'e' can follow 'a' or 'i'
10        dp[i][2] = (dp[i - 1][0] + dp[i - 1][1] + dp[i - 1][3] + dp[i - 1][4]) % MOD  # 'i' can follow any except 'i'
11        dp[i][3] = (dp[i - 1][2]) % MOD  # 'o' can only follow 'i'
12        dp[i][4] = (dp[i - 1][0]) % MOD  # 'u' can only follow 'a'
13    return sum(dp[n]) % MOD
14

ℹ

Complexity note: This complexity is efficient because we only need to calculate the counts iteratively based on the previous counts, leading to a linear growth in time with respect to n.

1Dynamic programming allows us to build solutions incrementally, leveraging previously computed results.
2Understanding the constraints of the problem helps in defining the state transitions effectively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.