1189. Maximum Number of Balloons - Explanation

1. Hash Map - I

class Solution:
    def maxNumberOfBalloons(self, text: str) -> int:
        countText = Counter(text)
        balloon = Counter("balloon")

        res = len(text)
        for c in balloon:
            res = min(res, countText[c] // balloon[c])
        return res

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$ since we have at most $26$ different characters.

2. Hash Map - II

class Solution:
    def maxNumberOfBalloons(self, text: str) -> int:
        mp = defaultdict(int)
        for c in text:
            if c in "balon":
                mp[c] += 1

        if len(mp) < 5:
            return 0

        mp['l'] //= 2
        mp['o'] //= 2
        return min(mp.values())

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$ since $balloon$ has $5$ different characters.