Leetcode #2947: Count Beautiful Substrings I
In this guide, we solve Leetcode #2947 Count Beautiful Substrings I 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
You are given a string s and a positive integer k. Let vowels and consonants be the number of vowels and consonants in a string.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Hash Table, Math, String, Enumeration, Number Theory, Prefix Sum
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: s = "baeyh", k = 2
Output: 2
Explanation: There are 2 beautiful substrings in the given string.
- Substring "baeyh", vowels = 2 (["a",e"]), consonants = 2 (["y","h"]).
You can see that string "aeyh" is beautiful as vowels == consonants and vowels * consonants % k == 0.
- Substring "baeyh", vowels = 2 (["a",e"]), consonants = 2 (["b","y"]).
You can see that string "baey" is beautiful as vowels == consonants and vowels * consonants % k == 0.
It can be shown that there are only 2 beautiful substrings in the given string.
Python Solution
class Solution:
def beautifulSubstrings(self, s: str, k: int) -> int:
n = len(s)
vs = set("aeiou")
ans = 0
for i in range(n):
vowels = 0
for j in range(i, n):
vowels += s[j] in vs
consonants = j - i + 1 - vowels
if vowels == consonants and vowels * consonants % k == 0:
ans += 1
return ans
Complexity
The time complexity is O(n). The space complexity is O(n).
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.