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

The optimal solution for Count Substrings That Can Be Rearranged to Contain a String II runs in O(n) time and uses O(n) space. This complexity is achieved because we only traverse `word1` once with the sliding window, maintaining counts in constant time.

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

The brute force approach for Count Substrings That Can Be Rearranged to Contain a String II 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 any of them can be rearranged to contain `word2` as a prefix. 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 That Can Be Rearranged to Contain a String II?

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

Clarify the problem requirements before jumping to code. Discuss your thought process and any trade-offs in your approach. Practice explaining your code as you write it to simulate the interview environment.

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

Not considering all possible substrings or miscounting valid ones. Overcomplicating the character frequency checks.

#3298

Count Substrings That Can Be Rearranged to Contain a String II

Hard

Hash TableStringSliding WindowHash MapSliding Window

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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 approach uses a sliding window technique to maintain a count of characters in the current window of `word1` and checks if it can satisfy the character requirements of `word2`. This reduces the need to generate all substrings explicitly.

⚙️

Algorithm

5 steps

1Step 1: Create a frequency map of characters in `word2`.
2Step 2: Initialize a sliding window on `word1` with two pointers.
3Step 3: Expand the window by moving the right pointer and update the frequency count.
4Step 4: When the window size exceeds the length of `word2`, move the left pointer to shrink the window.
5Step 5: Count valid substrings when the current window can rearrange to contain `word2`.

solution.py19 lines

1# Full working Python code
2from collections import Counter
3
4def count_valid_substrings(word1, word2):
5    word2_count = Counter(word2)
6    window_count = Counter()
7    valid_count = 0
8    left = 0
9    for right in range(len(word1)):
10        window_count[word1[right]] += 1
11        while right - left + 1 > len(word2):
12            window_count[word1[left]] -= 1
13            if window_count[word1[left]] == 0:
14                del window_count[word1[left]]
15            left += 1
16        if all(window_count[c] >= word2_count[c] for c in word2_count):
17            valid_count += right - left + 1
18    return valid_count
19

ℹ

Complexity note: This complexity is achieved because we only traverse `word1` once with the sliding window, maintaining counts in constant time.

1Understanding character frequency is crucial for rearrangement problems.
2Sliding window technique is effective for substring problems with constraints.

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