Maximum Number of Balloons

Maximum Number of Balloons

What is this problem about?

The Maximum Number of Balloons coding problem asks how many times you can form the word "balloon" using characters from a given string. You are given a string, and you need to count how many complete instances of the word "balloon" can be formed using its characters. Note that the letters 'l' and 'o' each appear twice in "balloon", so they need twice as many occurrences.

Why is this asked in interviews?

Tesla, Wayfair, and Amazon include this problem as a character frequency fundamentals check. It tests careful counting with attention to character multiplicity — a very common real-world pattern in string processing. Candidates who cleanly handle the "l appears twice, o appears twice" condition without special-casing demonstrate good attention to detail.

Algorithmic pattern used

Character frequency counting: Count occurrences of each character in the input string using a hash map or frequency array. The word "balloon" requires: b×1, a×1, l×2, o×2, n×1. For each required character, divide its available count by how many times it appears in "balloon". The minimum across all required characters is the answer.

Specifically: min(count['b'], count['a'], count['l']//2, count['o']//2, count['n']).

Example explanation

Input: "nlaebololl"

• Frequencies: n=1, l=3, a=1, e=1, b=1, o=2.
• b: min needs 1, have 1 → can form 1.
• a: needs 1, have 1 → 1.
• l: needs 2, have 3 → floor(3/2) = 1.
• o: needs 2, have 2 → 1.
• n: needs 1, have 1 → 1.
• Answer = min(1, 1, 1, 1, 1) = 1.

Input: "loonbalxballpoon"

• b=2, a=2, l=4, o=3, n=1 (ignoring x, p).
• l//2=2, o//2=1. Answer = min(2,2,2,1,1) = 1.

Common mistakes candidates make

• Treating 'l' and 'o' as needing only 1 occurrence: "balloon" has two l's and two o's. Dividing by 2 is essential.
• Counting characters not in "balloon": Extra characters don't help and shouldn't affect the count. Only track b, a, l, o, n.
• Using integer division inconsistently: Floor division (//) is required; floating point division may cause off-by-one errors.

Interview preparation tip

For the Hash Table Counting String interview pattern, generalize this approach: given a target word and a source string, "how many times can target be formed from source?" is always solved by character frequency counting + taking the minimum ratio. Practice this with other words and learn to handle repeated characters explicitly by dividing available count by required count.