How do you solve Count Beautiful Substrings I on LeetCode?

Count Beautiful Substrings I (LeetCode #2947) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the conditions for a beautiful substring is crucial to solving the problem efficiently.

What is the brute force solution for Count Beautiful Substrings I?

The brute force approach for Count Beautiful Substrings I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings of the input string and checking each one to see if it meets the criteria for being beautiful. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Beautiful Substrings I?

Always clarify the problem statement and constraints before jumping into coding. Think aloud while coding to show your thought process to the interviewer. Test your solution with different inputs, including edge cases, to ensure correctness.

What are common mistakes when solving Count Beautiful Substrings I?

Not considering edge cases, such as strings with no vowels or consonants. Failing to correctly update the hash map or misunderstanding how to use it.

#2947

Count Beautiful Substrings I

Medium

Hash TableMathStringEnumerationNumber TheoryPrefix SumHash MapPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses prefix sums and a hash map to efficiently count beautiful substrings without generating all possible substrings. This reduces the time complexity significantly.

⚙️

Algorithm

6 steps

1Step 1: Initialize a hash map to store the frequency of (vowel_count - consonant_count, product_mod_k) pairs.
2Step 2: Initialize variables for vowel and consonant counts and a result counter.
3Step 3: Loop through the string while updating vowel and consonant counts.
4Step 4: For each character, calculate the difference between vowel and consonant counts and the product modulo k.
5Step 5: Check if this pair exists in the hash map. If it does, add its frequency to the result.
6Step 6: Update the hash map with the current pair.

solution.py17 lines

1def count_beautiful_substrings(s, k):
2    vowels = 'aeiou'
3    count = 0
4    n = len(s)
5    freq_map = {}
6    v_count, c_count = 0, 0
7    freq_map[(0, 1)] = 1  # Base case for empty substring
8    for char in s:
9        if char in vowels:
10            v_count += 1
11        else:
12            c_count += 1
13        product_mod_k = (v_count * c_count) % k
14        key = (v_count - c_count, product_mod_k)
15        count += freq_map.get(key, 0)
16        freq_map[key] = freq_map.get(key, 0) + 1
17    return count

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Complexity note: The time complexity is O(n) because we make a single pass through the string, and the space complexity is O(n) due to the hash map storing the frequency of pairs.

1Understanding the conditions for a beautiful substring is crucial to solving the problem efficiently.
2Using a hash map to store intermediate results can significantly reduce time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.