What is the primary pattern for solving the Ransom Note problem?

The primary pattern involves using a hash table or frequency map to count occurrences of characters in both the ransomNote and the magazine, then comparing these counts to ensure the ransomNote can be constructed.

What is the time complexity of the Ransom Note problem?

The time complexity is O(n), where n is the length of the longer string (either ransomNote or magazine). This is because each string is processed once to count letter frequencies.

What if the ransomNote has more characters than the magazine?

If the ransomNote contains more characters than the magazine can supply, the answer should be false. The hash table comparison will show insufficient occurrences of required characters.

How does early termination improve the solution?

Early termination helps by stopping further processing once it's clear that the ransomNote cannot be fully constructed from the magazine, thus saving time and reducing unnecessary computation.

Can the Ransom Note problem be solved using brute force?

While a brute-force solution could work, it would be inefficient. Using hash tables or frequency maps to count characters is a more optimal approach with O(n) time complexity.

#383

Easy

auto_awesome哈希·表·结合·string

LeetCode 题解工作台

赎金信

给你两个字符串： ransomNote 和 magazine ，判断 ransomNote 能不能由 magazine 里面的字符构成。如果可以，返回 true ；否则返回 false 。 magazine 中的每个字符只能在 ransomNote 中使用一次。示例 1：输入： ransomN…

哈希表字符串计数

题目描述

给你两个字符串：ransomNote 和 magazine ，判断 ransomNote 能不能由 magazine 里面的字符构成。

如果可以，返回 true ；否则返回 false 。

magazine 中的每个字符只能在 ransomNote 中使用一次。

示例 1：

输入：ransomNote = "a", magazine = "b"
输出：false

示例 2：

输入：ransomNote = "aa", magazine = "ab"
输出：false

示例 3：

输入：ransomNote = "aa", magazine = "aab"
输出：true

提示：

1 <= ransomNote.length, magazine.length <= 10⁵
ransomNote 和 magazine 由小写英文字母组成

lightbulb

解题思路

方法一：哈希表或数组

我们可以用一个哈希表或长度为 $26$ 的数组 $cnt$ 记录字符串 magazine 中所有字符出现的次数。然后遍历字符串 ransomNote，对于其中的每个字符 $c$ ，我们将其从 $cnt$ 的次数减 $1$ ，如果减 $1$ 之后的次数小于 $0$ ，说明 $c$ 在 magazine 中出现的次数不够，因此无法构成 ransomNote，返回 $false$ 即可。

否则，遍历结束后，说明 ransomNote 中的每个字符都可以在 magazine 中找到对应的字符，因此返回 $true$ 。

时间复杂度 $O(m + n)$ ，空间复杂度 $O(C)$ 。其中 $m$ 和 $n$ 分别为字符串 ransomNote 和 magazine 的长度；而 $C$ 为字符集的大小，本题中 $C = 26$ 。

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class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        cnt = Counter(magazine)
        for c in ransomNote:
            cnt[c] -= 1
            if cnt[c] < 0:
                return False
        return True

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
Checks for efficient handling of string operations and hash table manipulation.
question_mark
Evaluates understanding of counting characters and early termination optimization.
question_mark
Tests ability to correctly implement hash table usage and manage frequency comparisons.

warning

常见陷阱

外企场景

error
Incorrectly assuming that characters can be reused from the magazine when they can only be used once.
error
Failing to account for the scenario where a character in the ransomNote does not appear enough times in the magazine.
error
Overcomplicating the solution by using unnecessary data structures or algorithms, instead of leveraging a simple hash table or frequency map.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the magazine and ransomNote contain uppercase letters? Modify the hash table to account for both cases.
arrow_right_alt
Extend the problem to handle more complex scenarios like punctuations or numbers. How would the approach change?
arrow_right_alt
Consider the impact on performance if the input strings were significantly larger (e.g., with lengths up to 10^6).

help

常见问题

外企场景

继续练习

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