#3298

Hard

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Count Substrings That Can Be Rearranged to Contain a String II

Count Substrings That Can Be Rearranged to Contain a String II involves identifying valid substrings with a sliding window approach.

hash table string sliding window

Problem Statement

You are given two strings, word1 and word2. A substring of word1 is valid if it can be rearranged to start with word2. Your task is to return the total number of valid substrings of word1.

To achieve this, you can use a sliding window technique where the window size is based on word2’s length. Adjust the window as you check different substrings of word1, ensuring that at each step the substring can be rearranged to have word2 as a prefix.

Examples

Example 1

Input: word1 = "bcca", word2 = "abc"

Output: 1

The only valid substring is "bcca" which can be rearranged to "abcc" having "abc" as a prefix.

Example 2

Input: word1 = "abcabc", word2 = "abc"

Output: 10

All the substrings except substrings of size 1 and size 2 are valid.

Example 3

Input: word1 = "abcabc", word2 = "aaabc"

Output: 0

Example details omitted.

Constraints

1 <= word1.length <= 106
1 <= word2.length <= 104
word1 and word2 consist only of lowercase English letters.

Solution Approach

Sliding Window with Hash Map

The sliding window approach will help efficiently check substrings of word1. Maintain a hash map to count character frequencies and update the window as you slide through the string. This allows for quick validation of whether a substring can be rearranged to contain word2.

Two-pointer Technique

Use two pointers to manage the window’s bounds. The first pointer keeps track of the start of the substring, and the second pointer expands the window until the substring size matches word2's length. Then, update the window by shifting the start pointer, ensuring the substring always maintains valid rearrangements.

Efficient Frequency Comparison

As the window slides, maintain a frequency count of characters within the window. Compare this count with the character count of word2 to check if the current substring can be rearranged to have word2 as a prefix. This reduces unnecessary recalculations and ensures the solution scales with larger inputs.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the sliding window approach with the two-pointer technique, leading to O(n) for processing word1. Space complexity is O(k), where k is the length of word2, due to the storage needed for the frequency hash map.

What Interviewers Usually Probe

Look for familiarity with sliding window problems and hash table usage.
Test if the candidate can handle large input sizes efficiently.
Check for understanding of how to maintain running state updates within the window.

Common Pitfalls or Variants

Common pitfalls

Forgetting to update the hash map efficiently as the window slides.
Not adjusting the window size correctly when shifting the pointers.
Overcomplicating the frequency comparison step, leading to inefficiencies.

Follow-up variants

Handling different string lengths for word1 and word2, particularly when word2 is much larger.
Optimizing the space complexity by avoiding unnecessary data structures.
Considering edge cases, such as when no valid substrings exist or when word2 is much shorter than word1.

FAQ

How do I solve the Count Substrings That Can Be Rearranged to Contain a String II problem?

Use a sliding window approach with a hash map to track character frequencies, updating the window as you iterate through word1.

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

The time complexity is O(n), where n is the length of word1, as the window slides across the string in a linear pass.

Can the sliding window technique be applied to other string problems?

Yes, the sliding window technique is widely used in substring search, anagram checks, and various other string manipulation problems.

What edge cases should I consider for this problem?

Consider edge cases where no valid substrings exist or when word2 is much longer or shorter than word1.

How can I optimize the space complexity of this solution?

You can optimize space by minimizing the hash map’s size, tracking only the necessary characters and updating it efficiently as the window slides.

terminal

Solution

Solution 1: Sliding Window

The problem is essentially to find how many substrings in $\textit{word1}$ contain all the characters in $\textit{word2}$. We can use a sliding window to handle this.

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class Solution:
    def validSubstringCount(self, word1: str, word2: str) -> int:
        if len(word1) < len(word2):
            return 0
        cnt = Counter(word2)
        need = len(cnt)
        ans = l = 0
        win = Counter()
        for c in word1:
            win[c] += 1
            if win[c] == cnt[c]:
                need -= 1
            while need == 0:
                if win[word1[l]] == cnt[word1[l]]:
                    need += 1
                win[word1[l]] -= 1
                l += 1
            ans += l
        return ans

Continue Practicing

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