Intuition¶

The word balloon needs one b, one a, one n, and two each of l and o. Each complete word consumes those letters from text, so the answer is limited by whichever required letter runs out first.

Count how many times each needed letter appears, halve the counts for l and o, then take the minimum.

Approach: Frequency Count¶

Count character frequencies in text.
For l and o, divide their counts by 2 because each balloon uses two of each.
Return the minimum among the counts for b, a, l, o, and n.

Complexity¶

Time complexity: O(n), where n is the length of text.
Space complexity: O(1) — at most 26 letters (Go) or 5 tracked letters (Rust).

Code¶

Go¶

func maxNumberOfBalloons(text string) int {
    freq := [26]int{}
    for _, ch := range text {
        freq[ch-'a']++
    }
    return min(freq['a'-'a'], freq['b'-'a'], freq['l'-'a']/2, freq['o'-'a']/2, freq['n'-'a'])
}

Rust¶

impl Solution {
    pub fn max_number_of_balloons(text: String) -> i32 {
        let mut freq = [0; 5];
        for ch in text.as_bytes() {
            match ch {
                b'a' => freq[0] += 1,
                b'b' => freq[1] += 1,
                b'l' => freq[2] += 1,
                b'o' => freq[3] += 1,
                b'n' => freq[4] += 1,
                _ => {},
            }
        }
        freq[2] /= 2;
        freq[3] /= 2;
        *freq.iter().min().unwrap_or(&0)
    }
}