383.ransom-note

Statement

简体中文English

Metadata

• Link: 赎金信
• Difficulty: Easy
• Tag: 哈希表 字符串 计数

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

Metadata

• Link: Ransom Note
• Difficulty: Easy
• Tag: Hash Table String Counting

Given two strings ransomNote and magazine, return true if ransomNote can be constructed from magazine and false otherwise.

Each letter in magazine can only be used once in ransomNote.

 

Example 1:

Input: ransomNote = "a", magazine = "b"
Output: false

Example 2:

Input: ransomNote = "aa", magazine = "ab"
Output: false

Example 3:

Input: ransomNote = "aa", magazine = "aab"
Output: true

 

Constraints:

• 1 <= ransomNote.length, magazine.length <= 105
• ransomNote and magazine consist of lowercase English letters.

Solution

Python 3

from collections import Counter


class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        a = Counter(ransomNote)
        b = Counter(magazine)

        for k, v in a.items():
            if k not in b.keys() or v > b[k]:
                return False

        return True

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最后更新: October 11, 2023