How do you solve the 'Maximum Number of Balloons' problem?

Count the frequency of relevant characters ('b', 'a', 'l', 'o', 'n'), then determine how many times you can form 'balloon' based on the minimum count of the required characters.

What is the time complexity of the 'Maximum Number of Balloons' problem?

The time complexity is O(n), where n is the length of the input string, because you need to count the frequency of each character in the string.

What is the primary approach to solving the 'Maximum Number of Balloons' problem?

The primary approach is using a hash table to count character frequencies and then determining the maximum number of times you can form the word 'balloon'.

What are common mistakes in solving the 'Maximum Number of Balloons' problem?

Common mistakes include not accounting for the fact that 'l' and 'o' appear twice, or missing some characters in the frequency count.

How can GhostInterview help me with the 'Maximum Number of Balloons' problem?

GhostInterview provides a structured approach to practicing the key patterns for this problem, helping you master counting techniques and optimize solutions.

#1189

Easy

auto_awesome哈希·表·结合·string

LeetCode 题解工作台

“气球” 的最大数量

给你一个字符串 text ，你需要使用 text 中的字母来拼凑尽可能多的单词 "balloon"（气球）。字符串 text 中的每个字母最多只能被使用一次。请你返回最多可以拼凑出多少个单词 "balloon" 。示例 1：输入： text = "nlaebolko" 输出： 1 示例 2：…

哈希表字符串计数

题目描述

给你一个字符串 text，你需要使用 text 中的字母来拼凑尽可能多的单词 "balloon"（气球）。

字符串 text 中的每个字母最多只能被使用一次。请你返回最多可以拼凑出多少个单词 "balloon"。

示例 1：

输入：text = "nlaebolko"
输出：1

示例 2：

输入：text = "loonbalxballpoon"
输出：2

示例 3：

输入：text = "leetcode"
输出：0

提示：

1 <= text.length <= 10⁴
text 全部由小写英文字母组成

注意：本题与 2287. 重排字符形成目标字符串相同。

lightbulb

解题思路

方法一：计数

我们统计字符串 text 中每个字母出现的次数，然后将其中字母 'o' 和 'l' 的出现次数分别除以 2，这是因为单词 balloon 中字母 'o' 和 'l' 都出现了 2 次。

接着，我们遍历单词 balon 中的每个字母，统计每个字母在字符串 text 中出现的次数的最小值，这个最小值就是单词 balloon 在字符串 text 中出现的最大次数。

时间复杂度 $O(n)$ ，空间复杂度 $O(C)$ 。其中 $n$ 为字符串 text 的长度；而 $C$ 为字符集大小，本题中 $C = 26$ 。

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class Solution:
    def maxNumberOfBalloons(self, text: str) -> int:
        cnt = Counter(text)
        cnt['o'] >>= 1
        cnt['l'] >>= 1
        return min(cnt[c] for c in 'balon')

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
The candidate identifies the need for frequency counting of characters relevant to 'balloon'.
question_mark
The candidate efficiently computes the minimum number of 'balloon' instances from the character frequencies.
question_mark
The candidate optimizes space by using a fixed-size hash table for character counts.

warning

常见陷阱

外企场景

error
Ignoring the fact that 'l' and 'o' appear twice in the word 'balloon'.
error
Not updating the result if some characters are missing or not sufficient to form a 'balloon'.
error
Using unnecessary data structures or overly complex approaches when a simple hash table suffices.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the string contains extra characters not relevant to the word 'balloon'?
arrow_right_alt
What if the input string is much shorter than the word 'balloon'?
arrow_right_alt
How would you approach the problem if it were required to form multiple words instead of just one?

help

常见问题

外企场景

继续练习

#1160 拼写单词 #1079 活字印刷 #1347 制造字母异位词的最小步骤数