Count Beautiful Substrings I

Count Beautiful Substrings I - Problem

Hash Table Math String Enumeration Number Theory Prefix Sum Medium

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.

A string is beautiful if:

vowels == consonants
(vowels * consonants) % k == 0, in other terms the multiplication of vowels and consonants is divisible by k

Return the number of non-empty beautiful substrings in the given string s.

A substring is a contiguous sequence of characters in a string.

Vowel letters in English are 'a', 'e', 'i', 'o', and 'u'. Consonant letters in English are every letter except vowels.

Input & Output

Example 1 — Basic Case

$ Input: s = "baae", k = 2

› Output: 0

💡 Note: For s="baae" and k=2, we check all substrings: "b"(0v,1c), "a"(1v,0c), "a"(1v,0c), "e"(1v,0c) - none have equal vowels/consonants. "ba"(1v,1c) has equal counts but (1*1)%2=1≠0. "aa"(2v,0c), "ae"(2v,0c), "baa"(2v,1c), "aae"(2v,1c), "baae"(3v,1c) - none have equal vowels/consonants. Total beautiful substrings = 0.

Example 2 — Simple Beautiful

$ Input: s = "abab", k = 1

› Output: 4

💡 Note: For s="abab" and k=1, beautiful substrings are: "ab" at positions (0,1) with 1 vowel, 1 consonant, (1*1)%1=0 ✓; "ba" at positions (1,2) with 1 vowel, 1 consonant, (1*1)%1=0 ✓; "ab" at positions (2,3) with 1 vowel, 1 consonant, (1*1)%1=0 ✓; "abab" at positions (0,3) with 2 vowels, 2 consonants, (2*2)%1=0 ✓. Total = 4.

Example 3 — No Beautiful Substrings

$ Input: s = "abc", k = 3

› Output: 1

💡 Note: For s="abc" and k=3, checking all substrings: "a"(1v,0c), "b"(0v,1c), "c"(0v,1c) - none equal. "ab"(1v,1c) has equal counts but (1*1)%3=1≠0. "bc"(0v,2c), "abc"(1v,2c) - not equal. Total beautiful substrings = 0.

Constraints

1 ≤ s.length ≤ 1000
1 ≤ k ≤ 1000
s consists of only lowercase English letters

Visualization

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Asked in

G Google 25 M Microsoft 18 a Amazon 15 F Facebook 12

The key insight is that a beautiful substring needs both equal vowels/consonants AND their product divisible by k. The optimized approach uses running counts to avoid recalculating character frequencies. Time: O(n²), Space: O(1).

Common Approaches

✓ Sort First

⏱️ Time: N/A Space: N/A

Brute Force

⏱️ Time: O(n³) Space: O(1)

Generate all possible substrings and for each one, count vowels and consonants, then check if they are equal and their product is divisible by k.

Optimized with Running Counts

⏱️ Time: O(n²) Space: O(1)

Instead of recounting characters for each substring, maintain running counts of vowels and consonants as we extend the substring, reducing one level of nested loops.

Prefix Sum with Hash Map

⏱️ Time: O(n²) Space: O(n)

Convert vowels to +1 and consonants to -1, then use prefix sums to find subarrays with sum 0 (equal vowels/consonants). Use hash map to store prefix sums and their positions.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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