LeetCode #1220 — HARD

Count Vowels Permutation

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

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 10^9 + 7.

Example 1:

Input: n = 1
Output: 5
Explanation: All possible strings are: "a", "e", "i" , "o" and "u".

Example 2:

Input: n = 2
Output: 10
Explanation: All possible strings are: "ae", "ea", "ei", "ia", "ie", "io", "iu", "oi", "ou" and "ua".

Example 3:

Input: n = 5
Output: 68

Constraints:

1 <= n <= 2 * 10^4

Step 01

Brute Force Baseline

Problem summary: 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 10^9 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming

Example 1

Example 2

Core Insight

What unlocks the optimal approach

Use dynamic programming.
Let dp[i][j] be the number of strings of length i that ends with the j-th vowel.
Deduce the recurrence from the given relations between vowels.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1220: Count Vowels Permutation
class Solution {
    public int countVowelPermutation(int n) {
        long[] f = new long[5];
        Arrays.fill(f, 1);
        final int mod = (int) 1e9 + 7;
        for (int i = 1; i < n; ++i) {
            long[] g = new long[5];
            g[0] = (f[1] + f[2] + f[4]) % mod;
            g[1] = (f[0] + f[2]) % mod;
            g[2] = (f[1] + f[3]) % mod;
            g[3] = f[2];
            g[4] = (f[2] + f[3]) % mod;
            f = g;
        }
        long ans = 0;
        for (long x : f) {
            ans = (ans + x) % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #1220: Count Vowels Permutation
func countVowelPermutation(n int) (ans int) {
	const mod int = 1e9 + 7
	f := make([]int, 5)
	for i := range f {
		f[i] = 1
	}
	for i := 1; i < n; i++ {
		g := make([]int, 5)
		g[0] = (f[1] + f[2] + f[4]) % mod
		g[1] = (f[0] + f[2]) % mod
		g[2] = (f[1] + f[3]) % mod
		g[3] = f[2] % mod
		g[4] = (f[2] + f[3]) % mod
		f = g
	}
	for _, x := range f {
		ans = (ans + x) % mod
	}
	return
}

# Accepted solution for LeetCode #1220: Count Vowels Permutation
class Solution:
    def countVowelPermutation(self, n: int) -> int:
        f = [1] * 5
        mod = 10**9 + 7
        for _ in range(n - 1):
            g = [0] * 5
            g[0] = (f[1] + f[2] + f[4]) % mod
            g[1] = (f[0] + f[2]) % mod
            g[2] = (f[1] + f[3]) % mod
            g[3] = f[2]
            g[4] = (f[2] + f[3]) % mod
            f = g
        return sum(f) % mod

// Accepted solution for LeetCode #1220: Count Vowels Permutation
struct Solution;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn count_vowel_permutation(n: i32) -> i32 {
        let n = n as usize;
        let mut a = vec![0; n];
        let mut e = vec![0; n];
        let mut i = vec![0; n];
        let mut o = vec![0; n];
        let mut u = vec![0; n];
        a[0] = 1;
        e[0] = 1;
        i[0] = 1;
        o[0] = 1;
        u[0] = 1;
        for j in 1..n {
            e[j] += a[j - 1];

            a[j] += e[j - 1];
            i[j] += e[j - 1];

            a[j] += i[j - 1];
            e[j] += i[j - 1];
            o[j] += i[j - 1];
            u[j] += i[j - 1];

            i[j] += o[j - 1];
            u[j] += o[j - 1];

            a[j] += u[j - 1];

            a[j] %= MOD;
            e[j] %= MOD;
            i[j] %= MOD;
            o[j] %= MOD;
            u[j] %= MOD;
        }
        ((a[n - 1] + e[n - 1] + i[n - 1] + o[n - 1] + u[n - 1]) % MOD) as i32
    }
}

#[test]
fn test() {
    let n = 1;
    let res = 5;
    assert_eq!(Solution::count_vowel_permutation(n), res);
    let n = 2;
    let res = 10;
    assert_eq!(Solution::count_vowel_permutation(n), res);
    let n = 5;
    let res = 68;
    assert_eq!(Solution::count_vowel_permutation(n), res);
}

// Accepted solution for LeetCode #1220: Count Vowels Permutation
function countVowelPermutation(n: number): number {
    const f: number[] = Array(5).fill(1);
    const mod = 1e9 + 7;
    for (let i = 1; i < n; ++i) {
        const g: number[] = Array(5).fill(0);
        g[0] = (f[1] + f[2] + f[4]) % mod;
        g[1] = (f[0] + f[2]) % mod;
        g[2] = (f[1] + f[3]) % mod;
        g[3] = f[2];
        g[4] = (f[2] + f[3]) % mod;
        f.splice(0, 5, ...g);
    }
    return f.reduce((a, b) => (a + b) % mod);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(C)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.