Count Beautiful Substrings II - Practice Coding Problems

Count Beautiful Substrings II - Problem

Hash Table Math String Number Theory Prefix Sum Hard

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, which means 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 = "baeyh", k = 2

› Output: 2

💡 Note: Beautiful substrings are "baey" (V=2, C=2, 2*2=4, 4%2=0) and "aey" (V=2, C=1... wait, this doesn't work). Actually: "ae" (V=2, C=0) fails balance. Let me recalculate: "ba" (V=1, C=1, 1*1=1, 1%2≠0), "bae" (V=2, C=1) fails balance. The beautiful substrings are "baey" (V=2, C=2, 2*2%2=0) and "aeyh" (V=2, C=2, 2*2%2=0).

Example 2 — Single Character

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

› Output: 0

💡 Note: Single vowel "a" has V=1, C=0, so vowels ≠ consonants. No beautiful substrings possible.

Example 3 — Minimum Beautiful

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

› Output: 1

💡 Note: Substring "ab" has V=1, C=1 (balanced) and 1*1=1, 1%1=0 (divisible). This is beautiful.

Constraints

1 ≤ s.length ≤ 5 × 10⁴
1 ≤ k ≤ 1000
s consists of only lowercase English letters

Visualization

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Understanding the Visualization

Input Analysis

String s and divisor k

Balance Check

Find substrings where vowels = consonants

Divisibility Check

Verify (vowels × consonants) % k = 0

Key Takeaway

🎯 Key Insight: Use prefix differences to efficiently track vowel-consonant balance and apply number theory to optimize divisibility checks

Asked in

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

The key insight is to use prefix differences to track vowel-consonant balance and apply number theory for efficient divisibility checking. The optimal approach uses a hash map to store states (difference, position mod √k) achieving O(n√k) time complexity with O(n√k) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Substrings	O(n³)	O(1)	Check every possible substring for both balance and divisibility conditions
Advanced Prefix Sum with Hash Map	O(n × √k)	O(n × √k)	Use prefix differences and hash map to track balance states efficiently
Optimized with Number Theory	O(n²)	O(n)	Use prefix sums and number theory to reduce time complexity to O(n²)

Brute Force - Check All Substrings — Algorithm Steps

Step 1: Generate all possible substrings using nested loops
Step 2: For each substring, count vowels and consonants
Step 3: Check if vowels == consonants and (vowels * consonants) % k == 0
Step 4: Count all valid beautiful substrings

Visualization

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Step-by-Step Walkthrough

Generate Substrings

Check all possible start-end combinations

Count V & C

Count vowels and consonants in each substring

Check Conditions

Verify balance and divisibility

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool isVowel(char c) {
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}

int solution(char* s, int k) {
    int count = 0;
    int n = strlen(s);
    
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            int vowels = 0;
            int consonants = 0;
            
            for (int idx = i; idx <= j; idx++) {
                if (isVowel(s[idx])) {
                    vowels++;
                } else {
                    consonants++;
                }
            }
            
            if (vowels == consonants && vowels > 0 && (vowels * consonants) % k == 0) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char s[1001];
    int k;
    
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    scanf("%d", &k);
    
    int result = solution(s, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops: O(n²) substrings × O(n) to count each substring

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables for counting

✓ Linear Space

12.5K Views

Medium Frequency

~35 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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