What is the brute force solution for Number of Substrings Containing All Three Characters?

The brute force approach for Number of Substrings Containing All Three Characters 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 contains at least one occurrence of 'a', 'b', and 'c'. This is straightforward but inefficient for longer strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Substrings Containing All Three Characters?

Number of Substrings Containing All Three Characters uses the following concepts: Hash Table, String, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Number of Substrings Containing All Three Characters?

Always clarify the problem requirements and constraints before jumping into coding. Discuss your thought process and approach with the interviewer to demonstrate your understanding. Practice explaining your code as you write it, as this mimics the interview environment.

What are common mistakes when solving Number of Substrings Containing All Three Characters?

Not resetting the frequency counts correctly when moving the left pointer. Overlooking edge cases where the string might not contain all three characters.

#1358

Number of Substrings Containing All Three Characters

Medium

Hash TableStringSliding WindowSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a sliding window approach to efficiently count valid substrings. By maintaining a window that contains at least one 'a', 'b', and 'c', we can quickly calculate how many substrings can be formed from this window.

⚙️

Algorithm

4 steps

1Step 1: Initialize a frequency map to count occurrences of 'a', 'b', and 'c'.
2Step 2: Use two pointers (left and right) to represent the current window.
3Step 3: Expand the right pointer to include characters until the window contains at least one of each character.
4Step 4: Once valid, count all substrings that can be formed from the left pointer to the right pointer, then move the left pointer to find new valid windows.

solution.py16 lines

1# Full working Python code
2
3def count_substrings(s):
4    count = 0
5    left = 0
6    freq = {'a': 0, 'b': 0, 'c': 0}
7    for right in range(len(s)):
8        freq[s[right]] += 1
9        while freq['a'] > 0 and freq['b'] > 0 and freq['c'] > 0:
10            count += len(s) - right
11            freq[s[left]] -= 1
12            left += 1
13    return count
14
15# Example usage
16print(count_substrings('abcabc'))

ℹ

Complexity note: The time complexity is O(n) because we traverse the string with two pointers, and each character is processed a limited number of times. The space complexity is O(1) since we only use a fixed-size array for character counts.

1Using a sliding window allows for efficient substring counting without redundant checks.
2The frequency array helps track the presence of required characters in constant time.

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