Problem Description

You are given a string text and need to determine how many times you can form the word "balloon" using the characters from text.

The word "balloon" contains:

• 1 'b'
• 1 'a'
• 2 'l's
• 2 'o's
• 1 'n'

Each character in text can only be used once. Your task is to find the maximum number of complete instances of "balloon" that can be formed.

For example, if text = "nlaebolko", you can form exactly 1 "balloon" because even though you have enough of most letters, you only have 1 'b', 1 'a', and 1 'n'.

The solution counts the frequency of each character in text, then adjusts for the fact that "balloon" needs 2 'l's and 2 'o's by dividing their counts by 2. Finally, it finds the minimum count among all required letters ('b', 'a', 'l', 'o', 'n'), which determines the maximum number of "balloon" instances possible.

Intuition

Think of this problem like a recipe. To make one "balloon", we need specific ingredients: 1 'b', 1 'a', 2 'l's, 2 'o's, and 1 'n'. The question becomes: given our available letters in text, how many complete sets of these ingredients can we make?

The key insight is that we're limited by whichever letter we have the least of (relative to what we need). It's like making sandwiches - if you have 100 slices of bread but only 2 slices of cheese, you can only make 2 cheese sandwiches.

First, we count how many of each letter we have in text. But here's the clever part: since "balloon" needs 2 'l's and 2 'o's, having 4 'l's is really like having enough for 2 balloons, not 4. So we divide the counts of 'l' and 'o' by 2 to normalize them to the same scale as the other letters.

After this adjustment, each letter's count represents how many "balloon"s that particular letter could support. The minimum among these counts is our answer because we can't make more "balloon"s than what our scarcest letter allows.

For instance, if after adjustment we have: 'b':3, 'a':5, 'l':2, 'o':4, 'n':1, then we can only make 1 "balloon" because 'n' is our bottleneck.

Solution Approach

The implementation uses a frequency counting approach with a clever normalization step:

1. Count character frequencies: We use Python's Counter from the collections module to count the frequency of each character in text. This gives us a dictionary-like object where keys are characters and values are their counts.

2. Normalize for double letters: Since "balloon" contains 2 'l's and 2 'o's, we need to adjust their counts. The code uses the right shift operator >>= to divide by 2:

   • cnt['o'] >>= 1 divides the count of 'o' by 2
   • cnt['l'] >>= 1 divides the count of 'l' by 2

   This normalization ensures that if we have 4 'o's, we can only make 2 "balloon"s from them, not 4.

3. Find the minimum: We iterate through each character in 'balon' (note it's 'balon' not 'balloon' to avoid counting 'l' and 'o' twice) and find the minimum count using min(cnt[c] for c in 'balon'). This minimum represents the bottleneck - the maximum number of complete "balloon"s we can form.

The algorithm is efficient with O(n) time complexity where n is the length of the input string (for counting characters), and O(1) space complexity since we're only storing counts for at most 26 lowercase letters.

The beauty of this solution lies in its simplicity - by normalizing the counts of repeated letters and finding the minimum, we directly get the answer without complex logic or multiple passes through the data.

Example Walkthrough

Let's walk through the solution with text = "loonbalxballpoon".

Step 1: Count character frequencies First, we count how many of each character we have:

• 'l': 4 occurrences
• 'o': 4 occurrences
• 'n': 2 occurrences
• 'b': 2 occurrences
• 'a': 2 occurrences
• 'x': 1 occurrence (not needed for "balloon")
• 'p': 1 occurrence (not needed for "balloon")

Step 2: Normalize for double letters Since "balloon" needs 2 'l's and 2 'o's per instance, we divide their counts by 2:

• 'l': 4 ÷ 2 = 2 (meaning we have enough 'l's for 2 balloons)
• 'o': 4 ÷ 2 = 2 (meaning we have enough 'o's for 2 balloons)
• 'n': 2 (enough for 2 balloons)
• 'b': 2 (enough for 2 balloons)
• 'a': 2 (enough for 2 balloons)

Step 3: Find the minimum After normalization, all required letters have a count of 2:

• 'b': 2
• 'a': 2
• 'l': 2 (normalized)
• 'o': 2 (normalized)
• 'n': 2

The minimum is 2, so we can form exactly 2 instances of "balloon".

To verify: From "loonbalxballpoon", we can extract:

• First "balloon": b-a-l-l-o-o-n (using letters at various positions)
• Second "balloon": b-a-l-l-o-o-n (using the remaining letters)

All letters are accounted for, confirming our answer of 2.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string text. This is because:

• Creating the Counter object requires iterating through all characters in text once, which takes O(n) time
• The bit shift operations (>>= 1) for 'o' and 'l' are O(1) operations
• Finding the minimum among 5 specific characters ('b', 'a', 'l', 'o', 'n') takes O(5) = O(1) time

The space complexity is O(C), where C is the size of the character set. In this problem, C = 26 (for lowercase English letters). The Counter object stores at most C distinct characters from the input string, resulting in O(C) space usage. Since C is a constant (26), this can also be expressed as O(1) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Missing Characters Causing KeyError

The most common pitfall in this solution is that if any required character ('b', 'a', 'l', 'o', 'n') is missing from the input text, accessing char_count[c] will raise a KeyError since Counter doesn't automatically return 0 for missing keys when using bracket notation.

Example of the issue:

text = "xyz"  # No characters from 'balloon'
char_count = Counter(text)
# This will crash with KeyError: 'b'
return min(char_count[c] for c in 'balon')  

Solution: Use the .get() method with a default value of 0, or check if the character exists first:

def maxNumberOfBalloons(self, text: str) -> int:
    char_count = Counter(text)
    char_count['o'] //= 2
    char_count['l'] //= 2
  
    # Safe approach using .get() with default value
    return min(char_count.get(c, 0) for c in 'balon')

2. Integer Division vs Right Shift Confusion

While the problem description mentions using the right shift operator >>=, the actual code uses floor division //=. Both achieve the same result for positive integers (dividing by 2), but mixing them up or using regular division /= would cause issues.

Incorrect approach:

char_count['o'] /= 2  # Results in float, not integer
char_count['l'] /= 2  # Will cause issues in min() comparison

Correct approaches (both work):

# Floor division
char_count['o'] //= 2
char_count['l'] //= 2

# OR right shift (equivalent for dividing by 2)
char_count['o'] >>= 1
char_count['l'] >>= 1

3. Forgetting to Handle Empty String

While Counter handles empty strings gracefully, the logic should still account for edge cases where the input is empty or contains no valid characters.

Robust solution combining all fixes:

def maxNumberOfBalloons(self, text: str) -> int:
    if not text:
        return 0
  
    char_count = Counter(text)
  
    # Check if all required characters exist
    required = {'b', 'a', 'l', 'o', 'n'}
    if not required.issubset(char_count.keys()):
        return 0
  
    char_count['o'] //= 2
    char_count['l'] //= 2
  
    return min(char_count[c] for c in 'balon')