How do you solve Number of Wonderful Substrings on LeetCode?

Number of Wonderful Substrings (LeetCode #1915) 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: Using bit manipulation allows us to efficiently track character frequencies.

What is the time complexity of Number of Wonderful Substrings?

The optimal solution for Number of Wonderful Substrings runs in O(n) time and uses O(n) space. The complexity is linear because we only traverse the string once and use a HashMap to store the counts, which allows for quick lookups.

What is the brute force solution for Number of Wonderful Substrings?

The brute force approach for Number of Wonderful Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible substring to see if it meets the criteria of being wonderful. This is straightforward but inefficient as it examines all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Wonderful Substrings?

Number of Wonderful Substrings uses the following concepts: Hash Table, String, Bit Manipulation, Prefix Sum. The recommended patterns to study are: Hash Map, Bit Manipulation.

What are the interview tips for Number of Wonderful Substrings?

Explain your thought process clearly while coding. Discuss edge cases, like empty strings or single-character strings. Practice recognizing patterns in problems to apply similar solutions.

What are common mistakes when solving Number of Wonderful Substrings?

Not considering all possible bitmask configurations. Overlooking the need to count substrings that differ by only one character.

#1915

Number of Wonderful Substrings

Medium

Hash TableStringBit ManipulationPrefix SumHash MapBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a bitmask to represent the frequency of characters, allowing us to efficiently check for wonderful substrings by leveraging previously computed results.

⚙️

Algorithm

3 steps

1Step 1: Initialize a HashMap to store the count of each bitmask configuration.
2Step 2: Iterate through the string, updating the bitmask based on character frequencies.
3Step 3: For each bitmask, check how many times it has occurred and how many times its neighbors (bitmask with one bit flipped) have occurred.

solution.py11 lines

1def wonderfulSubstrings(word):
2    count = 0
3    mask_count = {0: 1}
4    mask = 0
5    for char in word:
6        mask ^= 1 << (ord(char) - ord('a'))
7        count += mask_count.get(mask, 0)
8        for i in range(10):
9            count += mask_count.get(mask ^ (1 << i), 0)
10        mask_count[mask] = mask_count.get(mask, 0) + 1
11    return count

ℹ

Complexity note: The complexity is linear because we only traverse the string once and use a HashMap to store the counts, which allows for quick lookups.

1Using bit manipulation allows us to efficiently track character frequencies.
2HashMaps can be used to store previously seen states for quick lookups.

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