The “Maximum Number of Balloons” problem is an intriguing text manipulation and counting problem that appears on Leetcode. This problem is an excellent exercise for those interested in honing their string manipulation and data structure skills. It requires one to not just perform simple string operations but also to think about the data structures best suited for the task.

In this article, we will delve deep into multiple approaches to solve this problem, covering their time and space complexities in detail. By the end, you will have a holistic understanding of how to tackle problems involving string manipulation and counting.

## Problem Statement

### Description

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.

### Constraints

`1 <= text.length <= 10^4`

`text`

consists of lowercase English letters only.

## Understanding the Problem

The problem asks for the maximum number of times we can form the word “balloon” using the characters from the given string `text`

. This means we need to keep track of the occurrence of each letter in `text`

and in “balloon”, and then see how many times “balloon” can be constructed using these occurrences.

## Approaches

### Approach 1: Counting Frequencies Using Dictionary

One straightforward way to solve this problem is to count the frequencies of each character in `text`

and then check how many times we can form the word “balloon” using these frequencies.

#### Algorithm

- Count the frequency of each letter in
`text`

. - Count the frequency of each letter in the word “balloon”.
- Calculate the maximum number of instances of “balloon” that can be formed using the counted frequencies.

#### Python Code Implementation

```
from collections import Counter
def maxNumberOfBalloons(text: str) -> int:
text_count = Counter(text)
balloon_count = Counter("balloon")
max_instances = float('inf')
for char, freq in balloon_count.items():
if char not in text_count:
return 0
max_instances = min(max_instances, text_count[char] // freq)
return max_instances
# Test the function
print(maxNumberOfBalloons("nlaebolko")) # Output should be 1
print(maxNumberOfBalloons("loonbalxballpoon")) # Output should be 2
```

#### Time Complexity

The time complexity for this approach is O(n), where n is the length of the `text`

string.

#### Space Complexity

The space complexity is O(1) because the dictionary size does not depend on the input size and is fixed at 26 for lowercase English letters.

### Approach 2: One-pass Counting and Checking

We can improve the algorithm by doing everything in a single pass. Instead of first counting all characters and then iterating through the “balloon” counts, we can calculate the maximum instances on the fly.

#### Algorithm

- Initialize a dictionary to keep track of the counts of the characters in “balloon”.
- Iterate through the
`text`

string, updating the count for each character that belongs to the word “balloon”. - After each update, calculate the maximum number of instances that can be formed with the current counts.

#### Python Code Implementation

```
def maxNumberOfBalloons(text: str) -> int:
balloon_dict = {'b': 0, 'a': 0, 'l': 0, 'o': 0, 'n': 0}
max_instances = float('inf')
for char in text:
if char in balloon_dict:
balloon_dict[char] += 1
# Calculate the maximum instances that can be formed with the current counts
max_instances = min(max_instances,
min(balloon_dict['b'],
balloon_dict['a'],
balloon_dict['l'] // 2,
balloon_dict['o'] // 2,
balloon_dict['n']))
return max_instances if max_instances != float('inf') else 0
# Test the function
print(maxNumberOfBalloons("nlaebolko")) # Output should be 1
print(maxNumberOfBalloons("loonbalxballpoon")) # Output should be 2
```

#### Time Complexity

The time complexity for this approach is O(n), which is the same as the previous approach.

#### Space Complexity

The space complexity remains O(1).

## Conclusion

The “Maximum Number of Balloons” problem offers an exciting challenge that tests your understanding of string manipulation and efficient counting. Both approaches offer similar time complexities of O(n) and constant space complexities, but the one-pass counting approach does everything in a single traversal of the input string, making it slightly more efficient in practice.