How do you solve Count Complete Substrings on LeetCode?

Count Complete Substrings (LeetCode #2953) 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 a sliding window can significantly reduce the time complexity from O(n²) to O(n).

What is the brute force solution for Count Complete Substrings?

The brute force approach for Count Complete Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of the input string to see if it meets the criteria of being complete. This is straightforward but inefficient, especially for long strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Complete Substrings?

Count Complete Substrings uses the following concepts: Hash Table, String, Sliding Window. The recommended patterns to study are: Sliding Window, Hash Map.

What are the interview tips for Count Complete Substrings?

Always think about the constraints and how they affect your approach. Practice writing out your thought process before coding. Discuss your approach with the interviewer to clarify your understanding.

What are common mistakes when solving Count Complete Substrings?

Not checking the conditions for adjacent characters correctly. Failing to reset the frequency map or window pointers properly.

#2953

Count Complete Substrings

Hard

Hash TableStringSliding WindowSliding WindowHash Map

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 to efficiently count substrings that meet the criteria. By maintaining a frequency map and adjusting the window size dynamically, we can avoid the overhead of generating all substrings.

⚙️

Algorithm

5 steps

1Step 1: Initialize a frequency map to count characters in the current window.
2Step 2: Use two pointers to define the window's start and end positions.
3Step 3: Expand the window by moving the end pointer and updating the frequency map.
4Step 4: Check if the current window meets the conditions for being complete. If so, count it.
5Step 5: If the window exceeds the valid conditions, move the start pointer to shrink the window.

solution.py15 lines

1def count_complete_substrings(word, k):
2    count = 0
3    n = len(word)
4    freq = {}
5    start = 0
6    for end in range(n):
7        freq[word[end]] = freq.get(word[end], 0) + 1
8        while any(v > k for v in freq.values()):
9            freq[word[start]] -= 1
10            if freq[word[start]] == 0:
11                del freq[word[start]]
12            start += 1
13        if all(v == k for v in freq.values()) and all(abs(ord(word[i]) - ord(word[i + 1])) <= 2 for i in range(start, end)): 
14            count += 1
15    return count

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once with the sliding window technique, adjusting the window size dynamically. The space complexity is O(n) due to the frequency map storing character counts.

1Using a sliding window can significantly reduce the time complexity from O(n²) to O(n).
2Maintaining a frequency map allows for quick checks on character counts.

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