How do you solve Count Substrings With K-Frequency Characters I on LeetCode?

Count Substrings With K-Frequency Characters I (LeetCode #3325) 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 frequency counts is crucial for solving substring problems.

What is the brute force solution for Count Substrings With K-Frequency Characters I?

The brute force approach for Count Substrings With K-Frequency Characters 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 given string and checking each one to see if it contains any character that appears at least k times. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Substrings With K-Frequency Characters I?

Count Substrings With K-Frequency Characters I uses the following concepts: Hash Table, String, Sliding Window. The recommended patterns to study are: Hash Map, Sliding Window.

What are the interview tips for Count Substrings With K-Frequency Characters I?

Always clarify the problem requirements and constraints before jumping into coding. Think aloud during the interview to show your thought process. Practice writing clean and efficient code, as readability is important.

What are common mistakes when solving Count Substrings With K-Frequency Characters I?

Failing to reset frequency counts when moving the left pointer. Not considering edge cases, such as when k is 1.

#3325

Count Substrings With K-Frequency Characters I

Medium

Hash TableStringSliding WindowHash MapSliding Window

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 sliding window technique combined with a frequency map to efficiently count valid substrings. By fixing the left index and expanding the right index, we can quickly determine when a substring meets the k-frequency condition without generating all possible substrings.

⚙️

Algorithm

3 steps

1Step 1: Initialize a frequency map and two pointers (left and right).
2Step 2: For each left index, expand the right pointer until the substring has a character with at least k occurrences.
3Step 3: Count all valid substrings starting from the current left index to the end of the string.

solution.py14 lines

1def countSubstrings(s, k):
2    count = 0
3    n = len(s)
4    for left in range(n):
5        freq = {}
6        valid_count = 0
7        for right in range(left, n):
8            freq[s[right]] = freq.get(s[right], 0) + 1
9            if freq[s[right]] == k:
10                valid_count += 1
11            if valid_count > 0:
12                count += n - right
13                break
14    return count

ℹ

Complexity note: This complexity is due to the linear traversal of the string with a sliding window approach, where we maintain a frequency map that can grow with the number of unique characters.

1Understanding frequency counts is crucial for solving substring problems.
2Sliding window techniques can significantly optimize the search for valid substrings.

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