Maximum number of balloons

November 5, 2022

array-and-hashmap

Problem URL: Maximum number of balloons

We will count all the characters, then we will return the minimum of the count of b, a, l, o, and n, as the number of l and o are counted twice, we will divide the count by 2.

class Solution:
    def maxNumberOfBalloons(self, text: str) -> int:
        count = collections.Counter(text)
        return min([count['b'], count['a'], count['l']//2, count['o']//2, count['n']])

Time Complexity: O(n)
Space Complexity: O(n)