Introduction

In this blog post, we will explore how to solve the LeetCode problem "3297" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two strings word1 and word2. A string x is called valid if x can be rearranged to have word2 as a prefix. Return the total number of valid substrings of word1. Example 1: Input: word1 = "bcca", word2 = "abc" Output: 1 Explanation: 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 Explanation: All the substrings except substrings of size 1 and size 2 are valid. Example 3: Input: word1 = "abcabc", word2 = "aaabc" Output: 0 Constraints: 1 <= word1.length <= 105 1 <= word2.length <= 104 word1 and word2 consist only of lowercase English letters.

Explanation

Here's a solution to the problem, incorporating efficiency and optimality considerations.

• Core Idea: Iterate through all possible substrings of word1. For each substring, check if it can be rearranged to have word2 as a prefix. This check involves comparing the character frequencies of word2 with the character frequencies of the substring.

• Frequency Comparison: Use dictionaries (or arrays for optimal constant-time access since we're dealing with lowercase letters) to store character frequencies. If the frequency of each character in word2 is less than or equal to its frequency in the substring, we can potentially rearrange the substring to have word2 as a prefix.

• Optimization: Avoid unnecessary computations. If the substring's length is less than word2's length, it cannot possibly have word2 as a prefix. Early exit conditions improve performance.

• Runtime and Space Complexity:

  • Runtime Complexity: O(n*m), where n is the length of word1 and m is the length of word2.
  • Space Complexity: O(1), since the frequency arrays/dictionaries have a fixed size (26 for lowercase English letters).

Code

    def count_valid_substrings(word1: str, word2: str) -> int:
    """
    Counts the number of valid substrings of word1 that can be rearranged to have word2 as a prefix.
    """
    n1 = len(word1)
    n2 = len(word2)
    count = 0

    for i in range(n1):
        for j in range(i, n1):
            substring = word1[i:j + 1]
            if len(substring) >= n2:
                if can_be_arranged_with_prefix(substring, word2):
                    count += 1

    return count


def can_be_arranged_with_prefix(substring: str, prefix: str) -> bool:
    """
    Checks if a substring can be rearranged to have a given prefix.
    """
    freq_substring = [0] * 26
    freq_prefix = [0] * 26

    for char in substring:
        freq_substring[ord(char) - ord('a')] += 1
    for char in prefix:
        freq_prefix[ord(char) - ord('a')] += 1

    for i in range(26):
        if freq_prefix[i] > freq_substring[i]:
            return False

    return True