What is the time complexity of Count Vowels Permutation?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count Vowels Permutation?

This problem uses Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count Vowels Permutation?

Insight 1: Dynamic programming is the most efficient approach due to the overlapping subproblems. We can build the solution iteratively from smaller string lengths. Insight 2: Instead of trying to generate the strings, we can focus on counting the number of valid strings ending with each vowel for a given length.

Which companies ask LeetCode 1220. Count Vowels Permutation in interviews?

This problem is frequently asked at Atlassian.

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Count Vowels Permutation - LeetCode 1220 Solution

Count Vowels Permutation - Complete Solution Guide

Count Vowels Permutation is LeetCode problem 1220, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer n , your task is to count how many strings of length n can be formed under the following rules: Each character is a lower case vowel ( 'a' , 'e' , 'i' , 'o' , 'u' ) Each vowel 'a' may only be followed by an 'e' . Each vowel 'e' may only be followed by an 'a' or an 'i' . Each vowel 'i' may not be followed by another 'i' . Each vowel 'o' may only be followed by an 'i' or a 'u' . Each vowel 'u' may only be followed by an 'a' . Since the answer may be too large, return it modulo 10^

Detailed Explanation

The problem asks us to count the number of strings of length 'n' that can be formed using lowercase vowels ('a', 'e', 'i', 'o', 'u') under specific rules regarding which vowels can follow each other. For instance, 'a' can only be followed by 'e', 'e' can only be followed by 'a' or 'i', and so on. The output should be the count of valid strings modulo 10^9 + 7 to prevent integer overflow. The input is a single integer 'n' representing the length of the string.

Solution Approach

The provided solutions utilize dynamic programming to efficiently solve this problem. The core idea is to maintain counts of how many valid strings of a given length end with each of the five vowels. We initialize these counts to 1 for strings of length 1. Then, we iterate from length 2 up to 'n', updating the counts based on the allowed transitions between vowels. For example, the number of strings of length 'i' ending in 'a' is the sum of the number of strings of length 'i-1' ending in 'e', 'i', or 'u' (since only 'e', 'i', or 'u' can precede 'a'). After iterating through all lengths, the final answer is the sum of the counts for strings of length 'n' ending in each vowel, modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Initialize five variables (a_count, e_count, i_count, o_count, u_count) to 1. These represent the number of strings of length 1 ending in each vowel.
Step 2: Iterate from 2 to 'n' (inclusive).
Step 3: Inside the loop, calculate the next counts (a_count_next, e_count_next, i_count_next, o_count_next, u_count_next) based on the rules. For example, a_count_next = (e_count + i_count + u_count) % MOD. The '% MOD' operation prevents integer overflow.
Step 4: Update the current counts (a_count, e_count, i_count, o_count, u_count) with the next counts calculated in the previous step.
Step 5: After the loop finishes, calculate the total number of strings by summing the counts of strings ending in each vowel: total = (a_count + e_count + i_count + o_count + u_count) % MOD.
Step 6: Return the 'total'.

Key Insights

Insight 1: Dynamic programming is the most efficient approach due to the overlapping subproblems. We can build the solution iteratively from smaller string lengths.
Insight 2: Instead of trying to generate the strings, we can focus on counting the number of valid strings ending with each vowel for a given length.
Insight 3: The constraints defining valid vowel sequences are crucial and directly translate into the transition rules in the dynamic programming approach.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Dynamic Programming.

Companies

Asked at: Atlassian.

Count Vowels Permutation | LeetCode 1220

Hard

Dynamic Programming

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int countVowelPermutation(int n) {
        int MOD = (int)1e9 + 7;
        
        // Number of strings of length i ending with vowel...
        long a_count = 1;  // 'a'
        long e_count = 1;  // 'e'
        long i_count = 1;  // 'i'
        long o_count = 1;  // 'o'
        long u_count = 1;  // 'u'

        for (int j = 1; j < n; j++) {
            // Calculate the counts for the next length
            
            // A string ending in 'a' must be preceded by 'e', 'i', or 'u'.
            long a_count_next = (e_count + i_count + u_count) % MOD;
            
            // A string ending in 'e' must be preceded by 'a' or 'i'.
            long e_count_next = (a_count + i_count) % MOD;
            
            // A string ending in 'i' must be preceded by 'e' or 'o'.
            long i_count_next = (e_count + o_count) % MOD;
            
            // A string ending in 'o' must be preceded by 'i'.
            long o_count_next = i_count;
            
            // A string ending in 'u' must be preceded by 'i' or 'o'.
            long u_count_next = (i_count + o_count) % MOD;
            
            // Update counts for the new length
            a_count = a_count_next;
            e_count = e_count_next;
            i_count = i_count_next;
            o_count = o_count_next;
            u_count = u_count_next;
        }

        // Total number of strings of length n is the sum of strings ending in each vowel
        long total = (a_count + e_count + i_count + o_count + u_count) % MOD;
        
        return (int)total;
    }
}

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Problem Info

Description

Problem 1220: Count Vowels Permutation - Detailed explanation and solution approaches.

Topics

Dynamic Programming

Companies

Atlassian

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1220

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize five variables (a_count, e_count, i_count, o_count, u_count) to 1. These represent the number of strings of length 1 ending in each vowel.
2
Step 2: Iterate from 2 to 'n' (inclusive).
3
Step 3: Inside the loop, calculate the next counts (a_count_next, e_count_next, i_count_next, o_count_next, u_count_next) based on the rules. For example, a_count_next = (e_count + i_count + u_count) % MOD. The '% MOD' operation prevents integer overflow.
4
Step 4: Update the current counts (a_count, e_count, i_count, o_count, u_count) with the next counts calculated in the previous step.
5
Step 5: After the loop finishes, calculate the total number of strings by summing the counts of strings ending in each vowel: total = (a_count + e_count + i_count + o_count + u_count) % MOD.
6
Step 6: Return the 'total'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the dynamic programming solution involves a single loop that iterates from 1 to n. The space complexity is constant because the solution uses a fixed number of variables to store the counts of vowels, irrespective of the input 'n'.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Matrix Exponentiation

This approach formulates the transitions between vowels as a matrix and uses matrix exponentiation to calculate the counts for larger values of 'n' more efficiently than the basic dynamic programming approach for very large n.

Time: O(log n)Space: O(1)

Trade-offs: Matrix exponentiation has a slightly better time complexity for VERY large 'n' due to logarithmic time, but involves more complex implementation. The constant factor might make the simple O(n) DP solution faster for smaller n. The constant space usage remains an advantage. Requires knowledge of linear algebra.

✨

Key Insights

1Insight 1: Dynamic programming is the most efficient approach due to the overlapping subproblems. We can build the solution iteratively from smaller string lengths.
2Insight 2: Instead of trying to generate the strings, we can focus on counting the number of valid strings ending with each vowel for a given length.
3Insight 3: The constraints defining valid vowel sequences are crucial and directly translate into the transition rules in the dynamic programming approach.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The initialization ensures that the correct answer (5) is returned directly.
Edge case 2: Large n. The modulo operation is crucial to prevent integer overflow, which is a common issue with dynamic programming problems.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. Forgetting to take the modulo at each step can lead to incorrect results for larger values of 'n'.

✗Common mistake 2: Incorrectly implementing the transition rules. A small mistake in mapping which vowels can follow each other can lead to a significantly different result.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: DNA sequencing. This type of counting problem, with restrictions on sequences, can be adapted to model constraints in DNA or protein sequences.

🏦Real-world scenario 2: Speech recognition. Similar Markov models are used to predict sequences of phonemes (basic units of sound) in speech recognition.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 19, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Vowels Permutation"?

For this hard-level Dynamic Programming problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?