How do you solve Count Residue Prefixes on LeetCode?

Count Residue Prefixes (LeetCode #3803) 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: Distinct character count must match length modulo 3.

What is the time complexity of Count Residue Prefixes?

The optimal solution for Count Residue Prefixes runs in O(n) time and uses O(n) space. We traverse the string once, maintaining a set of distinct characters, leading to O(n) complexity.

What is the brute force solution for Count Residue Prefixes?

The brute force approach for Count Residue Prefixes is Brute Force, with O(n²) time complexity and O(1) space complexity. We check each prefix of the string to count distinct characters and compare it with the prefix length modulo 3. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Residue Prefixes?

Count Residue Prefixes uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Residue Prefixes?

Explain your thought process clearly. Consider edge cases like empty strings. Optimize your solution step by step.

What are common mistakes when solving Count Residue Prefixes?

Not using a set to track distinct characters. Confusing 0-based and 1-based indexing for modulo calculations.

#3803

Count Residue Prefixes

Easy

Hash TableStringHash MapArray

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)

We maintain a set to track distinct characters as we build prefixes, allowing us to check the condition in constant time.

⚙️

Algorithm

3 steps

1Step 1: Initialize a set to keep track of distinct characters and a count for residue prefixes.
2Step 2: Iterate through the string, adding each character to the set.
3Step 3: Check if the size of the set equals the current index + 1 modulo 3, and update the count accordingly.

solution.py8 lines

1def countResiduePrefixes(s):
2    count = 0
3    distinct_chars = set()
4    for i, char in enumerate(s):
5        distinct_chars.add(char)
6        if len(distinct_chars) == (i + 1) % 3:
7            count += 1
8    return count

ℹ

Complexity note: We traverse the string once, maintaining a set of distinct characters, leading to O(n) complexity.

1Distinct character count must match length modulo 3.
2Set operations allow efficient tracking of distinct characters.

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