Count the five required letters and take the minimum available copies, halving the counts for the two letters used twice.
Complexity
O(n) time and O(1) space.
09 Sept 2026•Easy•Go / Python / TypeScript
Maximum Number of Balloons
Count how many copies of the word balloon can be formed from a text's letters.
Solution files
Go•maximum-number-of-balloons/solution.go
package main
func maxNumberOfBalloons(text string) int {
count := map[rune]int{}
for _, character := range text {
count[character]++
}
values := []int{count['b'], count['a'], count['l'] / 2, count['o'] / 2, count['n']}
answer := values[0]
for _, value := range values[1:] {
if value < answer {
answer = value
}
}
return answer
}
Python•maximum-number-of-balloons/solution.py
from collections import Counter
class Solution:
def maxNumberOfBalloons(self, text: str) -> int:
count = Counter(text)
return min(count["b"], count["a"], count["l"] // 2, count["o"] // 2, count["n"])
TypeScript•maximum-number-of-balloons/solution.ts
function maxNumberOfBalloons(text: string): number {
const count = new Map<string, number>(); for (const character of text) count.set(character, (count.get(character) ?? 0) + 1);
return Math.min(count.get("b") ?? 0, count.get("a") ?? 0, Math.floor((count.get("l") ?? 0) / 2), Math.floor((count.get("o") ?? 0) / 2), count.get("n") ?? 0);
}