Introduction

In this blog post, we will explore how to solve the LeetCode problem "2287" 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 0-indexed strings s and target. You can take some letters from s and rearrange them to form new strings. Return the maximum number of copies of target that can be formed by taking letters from s and rearranging them. Example 1: Input: s = "ilovecodingonleetcode", target = "code" Output: 2 Explanation: For the first copy of "code", take the letters at indices 4, 5, 6, and 7. For the second copy of "code", take the letters at indices 17, 18, 19, and 20. The strings that are formed are "ecod" and "code" which can both be rearranged into "code". We can make at most two copies of "code", so we return 2. Example 2: Input: s = "abcba", target = "abc" Output: 1 Explanation: We can make one copy of "abc" by taking the letters at indices 0, 1, and 2. We can make at most one copy of "abc", so we return 1. Note that while there is an extra 'a' and 'b' at indices 3 and 4, we cannot reuse the letter 'c' at index 2, so we cannot make a second copy of "abc". Example 3: Input: s = "abbaccaddaeea", target = "aaaaa" Output: 1 Explanation: We can make one copy of "aaaaa" by taking the letters at indices 0, 3, 6, 9, and 12. We can make at most one copy of "aaaaa", so we return 1. Constraints: 1 <= s.length <= 100 1 <= target.length <= 10 s and target consist of lowercase English letters. Note: This question is the same as 1189: Maximum Number of Balloons.

Explanation

Here's the breakdown of the solution:

• Count character frequencies: Determine the frequency of each character in both the source string s and the target string target.
• Calculate maximum copies: For each character in target, determine how many copies of target can be made based on the available count of that character in s. This is done by dividing the frequency of the character in s by its frequency in target.

• Find the minimum: The overall maximum number of copies is limited by the character with the fewest available copies. Therefore, take the minimum of all the calculated copy counts.

• Runtime Complexity: O(m + n), where n is the length of s and m is the length of target.

• Storage Complexity: O(1), since character counts use a fixed amount of space (26 lowercase English letters).

Code

    def max_number_of_string(s: str, target: str) -> int:
    """
    Given two 0-indexed strings s and target. You can take some letters from s and rearrange them to form new strings.
    Return the maximum number of copies of target that can be formed by taking letters from s and rearranging them.
    """
    s_counts = {}
    target_counts = {}

    for char in s:
        s_counts[char] = s_counts.get(char, 0) + 1

    for char in target:
        target_counts[char] = target_counts.get(char, 0) + 1

    max_copies = float('inf')
    for char, count in target_counts.items():
        if char not in s_counts:
            return 0
        max_copies = min(max_copies, s_counts[char] // count)

    return max_copies