LeetCode #2949 — HARD

Count Beautiful Substrings II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

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.

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.

Example 1:

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.

Example 2:

Input: s = "abba", k = 1
Output: 3
Explanation: There are 3 beautiful substrings in the given string.
- Substring "abba", vowels = 1 (["a"]), consonants = 1 (["b"]).
- Substring "abba", vowels = 1 (["a"]), consonants = 1 (["b"]).
- Substring "abba", vowels = 2 (["a","a"]), consonants = 2 (["b","b"]).
It can be shown that there are only 3 beautiful substrings in the given string.

Example 3:

Input: s = "bcdf", k = 1
Output: 0
Explanation: There are no beautiful substrings in the given string.

Constraints:

1 <= s.length <= 5 * 10⁴
1 <= k <= 1000
s consists of only English lowercase letters.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Math

Example 1

"baeyh"
2

Example 2

"abba"
1

Example 3

"bcdf"
1

Step 02

Core Insight

What unlocks the optimal approach

For the given <code>k</code> find all the <code>x</code> integers such that <code>x^2 % k == 0</code>. Notice, that there aren’t many such candidates.
We can iterate over all such <code>x</codes> values and count the number of substrings such that <code>vowels == consonants == x</code>.
This can be done with prefix sums and hash map.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2949: Count Beautiful Substrings II
class Solution {
  // Same as 2947. Count Beautiful Substrings I
  public int beautifulSubstrings(String s, int k) {
    final int root = getRoot(k);
    int ans = 0;
    int vowels = 0;
    int vowelsMinusConsonants = 0;
    // {(vowels, vowelsMinusConsonants): count}
    Map<Pair<Integer, Integer>, Integer> prefixCount = new HashMap<>();
    prefixCount.put(new Pair<>(0, 0), 1);

    for (final char c : s.toCharArray()) {
      if (isVowel(c)) {
        vowels = (vowels + 1) % root;
        ++vowelsMinusConsonants;
      } else {
        --vowelsMinusConsonants;
      }
      Pair<Integer, Integer> prefix = new Pair<>(vowels, vowelsMinusConsonants);
      ans += prefixCount.getOrDefault(prefix, 0);
      prefixCount.merge(prefix, 1, Integer::sum);
    }

    return ans;
  }

  private boolean isVowel(char c) {
    return "aeiou".indexOf(c) != -1;
  }

  private int getRoot(int k) {
    for (int i = 1; i <= k; ++i)
      if (i * i % k == 0)
        return i;
    throw new IllegalArgumentException();
  }
}

// Accepted solution for LeetCode #2949: Count Beautiful Substrings II
package main

// https://space.bilibili.com/206214
func pSqrt(n int) int {
	res := 1
	for i := 2; i*i <= n; i++ {
		i2 := i * i
		for n%i2 == 0 {
			res *= i
			n /= i2
		}
		if n%i == 0 {
			res *= i
			n /= i
		}
	}
	if n > 1 {
		res *= n
	}
	return res
}

func beautifulSubstrings(s string, k int) (ans int64) {
	k = pSqrt(k * 4)

	type pair struct{ i, sum int }
	cnt := map[pair]int{{k - 1, 0}: 1} // k-1 和 -1 同余
	sum := 0
	const aeiouMask = 1065233
	for i, c := range s {
		bit := aeiouMask >> (c - 'a') & 1
		sum += bit*2 - 1
		p := pair{i % k, sum}
		ans += int64(cnt[p])
		cnt[p]++
	}
	return
}

# Accepted solution for LeetCode #2949: Count Beautiful Substrings II
class Solution:
  # Same as 2947. Count Beautiful Substrings I
  def beautifulSubstrings(self, s: str, k: int) -> int:
    VOWELS = 'aeiou'
    root = self._getRoot(k)
    ans = 0
    vowels = 0
    vowelsMinusConsonants = 0
    # {(vowels, vowelsMinusConsonants): count}
    prefixCount = collections.Counter({(0, 0): 1})

    for c in s:
      if c in VOWELS:
        vowelsMinusConsonants += 1
        vowels = (vowels + 1) % root
      else:
        vowelsMinusConsonants -= 1
      ans += prefixCount[(vowels, vowelsMinusConsonants)]
      prefixCount[(vowels, vowelsMinusConsonants)] += 1

    return ans

  def _getRoot(self, k: int) -> int:
    for i in range(1, k + 1):
      if i * i % k == 0:
        return i

// Accepted solution for LeetCode #2949: Count Beautiful Substrings II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2949: Count Beautiful Substrings II
// class Solution {
//   // Same as 2947. Count Beautiful Substrings I
//   public int beautifulSubstrings(String s, int k) {
//     final int root = getRoot(k);
//     int ans = 0;
//     int vowels = 0;
//     int vowelsMinusConsonants = 0;
//     // {(vowels, vowelsMinusConsonants): count}
//     Map<Pair<Integer, Integer>, Integer> prefixCount = new HashMap<>();
//     prefixCount.put(new Pair<>(0, 0), 1);
// 
//     for (final char c : s.toCharArray()) {
//       if (isVowel(c)) {
//         vowels = (vowels + 1) % root;
//         ++vowelsMinusConsonants;
//       } else {
//         --vowelsMinusConsonants;
//       }
//       Pair<Integer, Integer> prefix = new Pair<>(vowels, vowelsMinusConsonants);
//       ans += prefixCount.getOrDefault(prefix, 0);
//       prefixCount.merge(prefix, 1, Integer::sum);
//     }
// 
//     return ans;
//   }
// 
//   private boolean isVowel(char c) {
//     return "aeiou".indexOf(c) != -1;
//   }
// 
//   private int getRoot(int k) {
//     for (int i = 1; i <= k; ++i)
//       if (i * i % k == 0)
//         return i;
//     throw new IllegalArgumentException();
//   }
// }

// Accepted solution for LeetCode #2949: Count Beautiful Substrings II
function beautifulSubstrings(s: string, k: number): number {
    const l = pSqrt(k * 4);
    const n = s.length;
    let sum = n;
    let ans = 0;
    const counter = new Map();
    counter.set(((l - 1) << 17) | sum, 1);
    for (let i = 0; i < n; i++) {
        const char = s[i];
        const bit = (AEIOU_MASK >> (char.charCodeAt(0) - 'a'.charCodeAt(0))) & 1;
        sum += bit * 2 - 1; // 1 -> 1    0 -> -1
        const key = ((i % l) << 17) | sum;
        ans += counter.get(key) || 0; // ans += cnt[(i%k,sum)]++
        counter.set(key, (counter.get(key) ?? 0) + 1);
    }
    return ans;
}
const AEIOU_MASK = 1065233;

function pSqrt(n: number) {
    let res = 1;
    for (let i = 2; i * i <= n; i++) {
        let i2 = i * i;
        while (n % i2 == 0) {
            res *= i;
            n /= i2;
        }
        if (n % i == 0) {
            res *= i;
            n /= i;
        }
    }
    if (n > 1) {
        res *= n;
    }
    return res;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.