Introduction

In this blog post, we will explore how to solve the LeetCode problem "3365" 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 s and t, both of which are anagrams of each other, and an integer k. Your task is to determine whether it is possible to split the string s into k equal-sized substrings, rearrange the substrings, and concatenate them in any order to create a new string that matches the given string t. Return true if this is possible, otherwise, return false. An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, using all the original letters exactly once. A substring is a contiguous non-empty sequence of characters within a string. Example 1: Input: s = "abcd", t = "cdab", k = 2 Output: true Explanation: Split s into 2 substrings of length 2: ["ab", "cd"]. Rearranging these substrings as ["cd", "ab"], and then concatenating them results in "cdab", which matches t. Example 2: Input: s = "aabbcc", t = "bbaacc", k = 3 Output: true Explanation: Split s into 3 substrings of length 2: ["aa", "bb", "cc"]. Rearranging these substrings as ["bb", "aa", "cc"], and then concatenating them results in "bbaacc", which matches t. Example 3: Input: s = "aabbcc", t = "bbaacc", k = 2 Output: false Explanation: Split s into 2 substrings of length 3: ["aab", "bcc"]. These substrings cannot be rearranged to form t = "bbaacc", so the output is false. Constraints: 1 <= s.length == t.length <= 2 * 105 1 <= k <= s.length s.length is divisible by k. s and t consist only of lowercase English letters. The input is generated such that s and t are anagrams of each other.

Explanation

Here's the breakdown of the solution:

• Anagram Check is Implicit: Since s and t are guaranteed to be anagrams, and the substrings must form t after rearrangement, we don't need to explicitly check if s and t are anagrams. This is inherent in the problem conditions.

• Substrings' Anagram Condition: The core idea is that if s can be split into k equal-sized substrings and rearranged to form t, then each substring of s (of length len(s) // k) must be an anagram of the corresponding substring that could be formed within t. More specifically, we can form k potential substrings by dividing t into equal parts. The problem boils down to verifying that the substrings of s can rearranged to form each of the equal sized substrings of t. Therefore, the substrings of s, when counted, should match the counts of substrings formed by t.

• Counting Substrings and Comparing: We iterate through all possible splits of s and compare the frequency of anagram-equivalent substrings in s and t. We check if s and t are composed of the same count of each substring anagram.

• Runtime Complexity: O(n), Storage Complexity: O(n) where n is the length of the strings s and t.

Code

    from collections import Counter

def solve():
    s = input()
    t = input()
    k = int(input())

    n = len(s)
    sub_len = n // k

    if n % k != 0:
        return False

    s_substrings = []
    for i in range(0, n, sub_len):
        s_substrings.append("".join(sorted(s[i:i + sub_len])))

    t_substrings = []
    for i in range(0, n, sub_len):
        t_substrings.append("".join(sorted(t[i:i + sub_len])))

    s_counts = Counter(s_substrings)
    t_counts = Counter(t_substrings)

    return s_counts == t_counts

print(solve())