What is the time complexity of Count Substrings That Can Be Rearranged to Contain a String I?

The optimal solution for Count Substrings That Can Be Rearranged to Contain a String I runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse `word1` once with a sliding window. The space complexity is O(n) due to the storage of character counts.

What is the brute force solution for Count Substrings That Can Be Rearranged to Contain a String I?

The brute force approach for Count Substrings That Can Be Rearranged to Contain a String I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings of `word1` and checking if each can be rearranged to start with `word2`. This is straightforward but inefficient due to the number of substrings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Substrings That Can Be Rearranged to Contain a String I?

Count Substrings That Can Be Rearranged to Contain a String 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 That Can Be Rearranged to Contain a String I?

Always clarify the problem constraints and edge cases. Think about how to optimize brute force solutions. Practice explaining your thought process clearly during the interview.

What are common mistakes when solving Count Substrings That Can Be Rearranged to Contain a String I?

Not considering all characters in `word2` when checking validity. Failing to reset counts correctly in the sliding window.

#3297

Count Substrings That Can Be Rearranged to Contain a String 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 approach combined with character frequency counting. This allows us to efficiently check if any substring can be rearranged to contain `word2` as a prefix without generating all substrings explicitly.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of characters in `word2`.
2Step 2: Use a sliding window of size equal to `word2` to traverse `word1`.
3Step 3: For each window, maintain a frequency count of characters and check if it can cover the frequency of `word2`.

solution.py16 lines

1def countValidSubstrings(word1, word2):
2    from collections import Counter
3    count2 = Counter(word2)
4    valid_count = 0
5    n, m = len(word1), len(word2)
6    count1 = Counter()
7    for i in range(n):
8        count1[word1[i]] += 1
9        if i >= m:
10            count1[word1[i - m]] -= 1
11            if count1[word1[i - m]] == 0:
12                del count1[word1[i - m]]
13        if i >= m - 1:
14            if all(count1[char] >= count2[char] for char in count2):
15                valid_count += 1
16    return valid_count

ℹ

Complexity note: The time complexity is O(n) because we traverse `word1` once with a sliding window. The space complexity is O(n) due to the storage of character counts.

1Character frequency is key to determining valid substrings.
2Sliding window helps efficiently manage substring checks.

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