Introduction

In this blog post, we will explore how to solve the LeetCode problem "1189" 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

Given a string text, you want to use the characters of text to form as many instances of the word "balloon" as possible. You can use each character in text at most once. Return the maximum number of instances that can be formed. Example 1: Input: text = "nlaebolko" Output: 1 Example 2: Input: text = "loonbalxballpoon" Output: 2 Example 3: Input: text = "leetcode" Output: 0 Constraints: 1 <= text.length <= 104 text consists of lower case English letters only. Note: This question is the same as 2287: Rearrange Characters to Make Target String.

Explanation

Here's an efficient solution to determine the maximum number of "balloon" instances that can be formed from a given text.

• Count Character Frequencies: Count the occurrences of each character in the input string text.
• Determine Limiting Factors: The number of "balloon" instances is limited by the minimum count of the characters 'b', 'a', 'l', 'o', and 'n', considering that 'l' and 'o' appear twice in "balloon".

• Calculate Maximum Instances: Divide the counts of 'l' and 'o' by 2 and take the minimum of all the relevant character counts.

• Runtime Complexity: O(n), where n is the length of the input string. Storage Complexity: O(1), as we use a fixed-size dictionary to store character counts.

Code

    def max_number_of_balloons(text: str) -> int:
    """
    Given a string text, return the maximum number of instances that can be formed.
    """
    counts = {}
    for char in text:
        counts[char] = counts.get(char, 0) + 1

    balloon_counts = {'b': 0, 'a': 0, 'l': 0, 'o': 0, 'n': 0}
    for char in balloon_counts:
        if char in counts:
            balloon_counts[char] = counts[char]
        else:
            return 0

    l_count = balloon_counts['l'] // 2
    o_count = balloon_counts['o'] // 2

    return min(balloon_counts['b'], balloon_counts['a'], l_count, o_count, balloon_counts['n'])