5 Best Ways to Count Sorted Vowel Strings in Python

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💡 Problem Formulation: We need to determine the number of strings of a certain length that can be formed using the vowels (a, e, i, o, u) such that each string is sorted in non-decreasing order. For example, given a length of 2, the sorted vowel strings are “aa”, “ae”, “ai”, “ao”, “au”, “ee”, “ei”, “eo”, “eu”, “ii”, “io”, “iu”, “oo”, “ou”, “uu”, thus yielding the output 15.

Method 1: Recursive Approach

This method involves writing a recursive function that, given a starting vowel and remaining length, counts the sorted vowel strings. Recursive calls are made by passing the current vowel and decremented length, facilitating the construction of all possible sorted strings.

Here’s an example:

def count_vowel_strings(n, start_vowel='a'):
    if n == 0:
        return 1
    count = 0
    for vowel in 'aeiou'[start_vowel:]:
        count += count_vowel_strings(n-1, vowel)
    return count

print(count_vowel_strings(2))

Output: 15

This recursive function starts from the smallest vowel ‘a’ and counts all possible sorted strings of length ‘n’. The function decreases the length and iterates through the vowels greater than or equal to the current vowel to ensure sorted order.

Method 2: Dynamic Programming

In this approach, we use dynamic programming to avoid redundant computations, done by storing previously computed results in a matrix. This is more time-efficient for large inputs compared to the recursive approach.

Here’s an example:

def count_vowel_strings_dp(n):
    dp = [[0]*5 for _ in range(n+1)]
    for i in range(5):
        dp[1][i] = 5-i
    for i in range(2, n+1):
        for j in range(5):
            dp[i][j] = sum(dp[i-1][j:])
    return dp[n][0]

print(count_vowel_strings_dp(2))

Output: 15

The method uses a bottom-up approach to iterate over string lengths and vowels, summing the counts of shorter strings to form the final count. For each vowel, it accumulates the counts of all possible continuations.

Method 3: Using Combinatorics

By recognizing the problem as a combinatoric one, we can determine the number of sorted vowel strings as a combination problem. In particular, we need to choose ‘n’ positions for the vowels from ‘n + 4’ slots (considering the 5 vowels as dividers).

Here’s an example:

from math import comb

def count_vowel_strings_comb(n):
    return comb(n+4, n)

print(count_vowel_strings_comb(2))

Output: 15

The combination function, comb, from the math module calculates the number of ways we can choose ‘n’ items from ‘n+4’ total items, directly giving us the count of sorted vowel strings.

Method 4: Iterative Count

This method iteratively generates counts for strings of increasing length by progressively adding counts of strings ending with each vowel.

Here’s an example:

def count_vowel_strings_iter(n):
    counts = [1]*5
    for _ in range(1, n):
        for j in range(4, -1, -1):
            counts[j] = sum(counts[j:])
    return sum(counts)

print(count_vowel_strings_iter(2))

Output: 15

This code initializes a count of 1 for each vowel and then iteratively updates each count to reflect the number of strings ending with that vowel for the current length.

Bonus One-Liner Method 5: Pythonic Combinatorial Approach

Leveraging Python’s powerful one-liners, the combinatorial approach can be condensed into a single, readable line using lambda and reduce functions.

Here’s an example:

from functools import reduce

count_vowel_strings_one_liner = lambda n: reduce(lambda x, _: x * (x + 1) // 2, range(5), n + 4)
print(count_vowel_strings_one_liner(2))

Output: 15

The one-liner creates a lambda function that applies a reduce operation to simulate the combinatorial formula, each time computing part of the combination count until reaching the final value.

Summary/Discussion

  • Method 1: Recursive Approach. Strengths: Conceptually simple and intuitive. Weaknesses: Less efficient for large ‘n’ due to many recursive calls.
  • Method 2: Dynamic Programming. Strengths: More efficient than recursion, especially for larger ‘n’. Weaknesses: Requires additional memory for the DP matrix.
  • Method 3: Using Combinatorics. Strengths: Very efficient as it reduces the problem to a single combinatorial formula. Weaknesses: Might not be immediately intuitive to someone unfamiliar with combinatorics.
  • Method 4: Iterative Count. Strengths: Simpler and more intuitive than dynamic programming. Weaknesses: Slightly less efficient than the combinatorial method.
  • Method 5: Pythonic Combinatorial Approach. Strengths: Compact and elegant solution. Weaknesses: May be challenging to understand for beginners.