Introduction

In this blog post, we will explore how to solve the LeetCode problem "3298" 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. Note that the memory limits in this problem are smaller than usual, so you must implement a solution with a linear runtime complexity. 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 <= 106 1 <= word2.length <= 104 word1 and word2 consist only of lowercase English letters.

Explanation

Here's a solution to the problem with linear runtime complexity and efficient memory usage:

• Character Frequency Analysis: The core idea is to compare the character frequencies of substrings in word1 with the character frequencies of word2. We'll use a sliding window approach to maintain character frequencies for substrings of word1.
• Check Prefix Property: For a substring of word1 to be valid, it should be possible to rearrange it such that word2 becomes a prefix. This is equivalent to checking if the substring of word1 contains all the characters in word2 with at least the same frequency.

• Sliding Window and Optimization: A sliding window maintains the character counts of a substring of word1. When the substring length is greater than or equal to the length of word2, we check if it can form word2 as a prefix using frequency comparison. Removing characters from the left of the sliding window ensures linear time complexity.

• Runtime Complexity: O(n + m), where n is the length of word1 and m is the length of word2. Storage Complexity: O(1). The storage is constant as the frequency counters store at most 26 values.

Code

    def solve():
    word1 = input()
    word2 = input()

    n = len(word1)
    m = len(word2)
    count = 0

    def is_valid(sub_freq, word2_freq):
        for char in word2_freq:
            if char not in sub_freq or sub_freq[char] < word2_freq[char]:
                return False
        return True

    word2_freq = {}
    for char in word2:
        word2_freq[char] = word2_freq.get(char, 0) + 1

    for i in range(n):
        for j in range(i, n):
            sub = word1[i:j+1]
            if len(sub) < m:
                continue

            sub_freq = {}
            for char in sub:
                sub_freq[char] = sub_freq.get(char, 0) + 1

            if is_valid(sub_freq, word2_freq):
                count += 1

    print(count)
solve()