Leetcode #2063: Vowels of All Substrings
In this guide, we solve Leetcode #2063 Vowels of All Substrings in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a string word, return the sum of the number of vowels ('a', 'e', 'i', 'o', and 'u') in every substring of word. A substring is a contiguous (non-empty) sequence of characters within a string.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Math, String, Dynamic Programming, Combinatorics
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: word = "aba"
Output: 6
Explanation:
All possible substrings are: "a", "ab", "aba", "b", "ba", and "a".
- "b" has 0 vowels in it
- "a", "ab", "ba", and "a" have 1 vowel each
- "aba" has 2 vowels in it
Hence, the total sum of vowels = 0 + 1 + 1 + 1 + 1 + 2 = 6.
Python Solution
class Solution:
def countVowels(self, word: str) -> int:
n = len(word)
return sum((i + 1) * (n - i) for i, c in enumerate(word) if c in 'aeiou')
Complexity
The time complexity is , where is the length of the string . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.