Maximum Number of Balloons
A easy-tier problem at 60% community acceptance, tagged with Hash Table, String, Counting. Reported in interviews at Wayfair and 1 others.
Maximum Number of Balloons is an easy problem that shows up in real OAs for Wayfair and Tesla. It looks trivial on the surface, but the acceptance rate sits just under 60%, which means a lot of people either overthink it or miss the character-frequency trap. The trick is dead simple once you see it, but if you blank during a live assessment, StealthCoder surfaces the solution invisible to your proctor. The problem tests whether you can count and compare frequencies without overthinking the data structure.
Companies that ask "Maximum Number of Balloons"
Maximum Number of Balloons is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Made for the engineer who has done the work but might still blank with a webcam pointed at him.
Get StealthCoderYou need to count how many times you can spell 'balloon' using characters from a given string. The naive approach is to loop through the string and build a frequency map, then check how many complete 'balloon' words you can form. The gotcha is that 'l' and 'o' appear twice in 'balloon' but the other letters once. Candidates often forget this asymmetry or try overly complex solutions. Hash Table and Counting are the right tools, but many people reach for something fancier. The algorithm is straightforward: count all character frequencies, then divide each required character count by its demand in 'balloon', taking the minimum. If you hit this in your OA and freeze on how to structure the comparison, StealthCoder gives you the working code in seconds.
Pattern tags
You know the problem.
Make sure you actually pass it.
Maximum Number of Balloons recycles across companies for a reason. It's easy-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Made for the engineer who has done the work but might still blank with a webcam pointed at him. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Maximum Number of Balloons interview FAQ
Is this problem actually easy or does the acceptance rate suggest it's a trap?+
The 60% acceptance rate is surprisingly low for an easy problem. Most misses come from forgetting that 'l' and 'o' each appear twice in 'balloon'. Once you count that correctly and divide frequencies by demand, the solution is trivial. The trap is overcomplication, not algorithmic difficulty.
Do I need a specific data structure or can I use basic counting?+
Hash Table (dictionary or map) is the cleanest approach for frequency counting, but you could also use a list or array if you map characters to indices. The data structure matters less than correctly tracking how many of each character you need versus have.
What's the common mistake candidates make on this one?+
Forgetting that 'l' and 'o' appear twice in the word 'balloon'. If you count them as single-frequency characters or don't divide by 2 when checking capacity, your answer will be wrong. The second mistake is writing overly complex code when a single loop and division are enough.
Is this still being asked at Wayfair and Tesla?+
Yes, it appears in their reported problem sets. Easy string and counting problems are common screening filters. It's the kind of problem you should solve in under 5 minutes in a live OA, so missing it costs credibility.
How does this relate to the other Counting and Hash Table problems I should know?+
This is entry-level for both topics. It teaches you character frequency counting, which is the foundation for anagram detection, word-pattern matching, and other string problems. Master the frequency map pattern here, then scale to harder problems using the same structure.
Want the actual problem statement? View "Maximum Number of Balloons" on LeetCode →