Maximum Number Of Balloons Leetcode Problem 1189 [Python Solution]

In this blog post, we will explore the LeetCode problem titled Maximum Number Of Balloons (Problem 1189) under the category of Arrays & Hashing.

We’ll provide a Python solution for this problem, delve into its problem overview, constraints, time and space complexity, and explore both the brute force and efficient approaches to solve it.

So, let’s get started with understanding the problem.

Problem Overview

The problem statement is as follows: Given a string text, you want to use the characters of text to form as many instances of the word “balloon” as possible.

However, you can use each character in text at most once.

Your task is to return the maximum number of instances of the word “balloon” that can be formed.

For instance, if we have the input string:

Input: text = "nlaebolko"

The maximum number of “balloon” instances that can be formed is 1.

Example 1:

Input: text = “nlaebolko”
Output: 1

Example 2:

Input: text = “loonbalxballpoon”
Output: 2

Example 3:

Input: text = “leetcode”
Output: 0

Understanding Constraints

Before we dive into the solution, let’s understand the constraints provided in the problem statement:

The length of the input string text ranges from 1 to 10,000.
The string text consists of lowercase English letters only.

Maximum Number of Balloons LeetCode Problem Solution

Now, let’s explore how to efficiently solve this problem using Python.

1. Bruteforce Approach

A straightforward approach is to iterate through the characters in the input string text and attempt to form the word “balloon”.

While iterating, we’ll maintain counts of each character and ensure that we can use each character at most once.

We’ll check how many times we can form “balloon” and return that count.

Here’s the Python code for the bruteforce approach:

def maxNumberOfBalloons(text: str) -> int:
    # Initialize character counts for 'balloon'
    balloon = {'b': 1, 'a': 1, 'l': 2, 'o': 2, 'n': 1}

    # Initialize character counts for the text
    countText = {}
    for char in text:
        countText[char] = countText.get(char, 0) + 1

    # Initialize the result to the maximum number of instances
    result = len(text)

    # Iterate through characters in 'balloon' and calculate the minimum instances
    for char in balloon:
        result = min(result, countText.get(char, 0) // balloon[char])

    return result

This code efficiently counts the instances of each character in both the input string and the word “balloon” and calculates the maximum number of times “balloon” can be formed.

2. Efficient Approach

In the efficient approach, we can make use of Python’s collections.Counter to count the occurrences of characters in the input string and “balloon.” This approach is more concise and relies on the ratios of character counts.

Here’s the Python code for the efficient approach:

View on GitHub ➹

def maxNumberOfBalloons(text: str) -> int:
    # Initialize character counts for 'balloon'
    balloon = Counter("balloon")

    # Initialize character counts for the text
    countText = Counter(text)

    # Initialize the result to the maximum number of instances
    result = len(text)

    # Iterate through characters in 'balloon' and calculate the minimum instances
    for char in balloon:
        result = min(result, countText[char] // balloon[char])

    return result

This approach is concise and leverages Python’s built-in Counter to efficiently count character occurrences.

Time and Space Complexity

Let’s analyze the time and space complexity of our efficient solution.

Time Complexity:

The time complexity of our solution is O(n), where n is the length of the input string text.

This is because we iterate through the input string once to count the character occurrences and perform a constant number of operations.

Space Complexity:

The space complexity is also O(n) because we create two Counter objects, one for the word “balloon” and one for the input string.

The space used is directly proportional to the size of the input string.

Reasoning Behind Our Approach

Our approach efficiently solves the problem by counting character occurrences and determining the maximum number of “balloon” instances that can be be formed.

We use a Counter to count characters, making it easier to perform the required calculations.

Conclusion

In this blog post, we explored the LeetCode problem Maximum Number Of Balloons (Problem 1189).

We provided both a brute force and an efficient Python solution to find the maximum number of times the word “balloon” can be formed using characters from the input string.

The efficient solution, which leverages Python’s Counter, is recommended for its simplicity and efficiency.

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If you have any questions, suggestions, or would like to discuss this problem further, feel free to comment below.

Happy coding!

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