Maximum Number Of Balloons

---

Intuition

• Count the maximum number of times balloon amount of characters returns. This reduces to: Count the minimum amount of times each character exists in unison. Anything left over doesn’t count.
• l and o need to repeat twice, so we divide those by 2

Complexity

Runtime

$O(N)$, where N is the size of the text

Space

$O(1)$ due to constant size HashMap

Code

Hashmap

class Solution:
    def maxNumberOfBalloons(self, text: str) -> int:
        hm = {'b': 0, 'a': 0, 'l': 0, 'o': 0, 'n': 0}
        for c in text:
            if c in hm:
                hm[c] += 1
 
        hm['l'] //= 2
        hm['o'] //= 2
 
        return min(hm.values())

Cards

START Basic Front: Maximum Number Of Balloons Back: Count the minimum amount of times each character exists in unison. END

---

References

• Maximum Number Of Balloons