1189. Maximum Number of Balloons

Hash Table
String
Counting

TIME

O(n)

SPACE

O(1)

Intuition

Forming the word "balloon" repeatedly is limited by whichever required letter runs out first. So count the letters in text and see how many copies of "balloon" the supply supports. The word uses l and o twice each, so their available counts must be halved before comparing.

Approach

Count the frequency of each character in text.
"balloon" requires b:1, a:1, l:2, o:2, n:1. The number of words each letter can support is:
- count['b'], count['a'], count['n']
- count['l'] // 2, count['o'] // 2
The answer is the minimum of these five values (the bottleneck letter).

Complexity

Time: O(n), one pass to count
Space: O(1), fixed-size frequency map (26 letters)

Edge Cases

A required letter missing entirely -> minimum is 0
Only one of the doubled letters present once -> // 2 makes it 0
Extra irrelevant characters -> ignored, do not affect the count

Code

from collections import Counter


class Solution:
    def maxNumberOfBalloons(self, text: str) -> int:
        """
        1189. Maximum Number of Balloons
        Time: O(n)
        Space: O(1)
        """
        counts = Counter(text)
        return min(
            counts["b"],
            counts["a"],
            counts["l"] // 2,
            counts["o"] // 2,
            counts["n"],
        )

/**
 * 1189. Maximum Number of Balloons
 * Time: O(n)
 * Space: O(1)
 */
class Solution {
    public int maxNumberOfBalloons(String text) {
        int[] count = new int[26];
        for (int i = 0; i < text.length(); i++) {
            count[text.charAt(i) - 'a']++;
        }

        int answer = Math.min(count['b' - 'a'], count['a' - 'a']);
        answer = Math.min(answer, count['l' - 'a'] / 2);
        answer = Math.min(answer, count['o' - 'a'] / 2);
        answer = Math.min(answer, count['n' - 'a']);

        return answer;
    }
}