383 Ransom Note

Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.

Each letter in the magazine string can only be used once in your ransom note.

Note: You may assume that both strings contain only lowercase letters.

canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true

The Idea: A ransom note can be constructed if and only if it is a subset of the magazine.

Complexity: O(2r+m) where r is the size of the ransom note, and m is the size of the magazine. O(1) space

from collections import Counter

class Solution(object):
    def canConstruct(self, ransomNote, magazine):
        """
        :type ransomNote: str
        :type magazine: str
        :rtype: bool
        """
        return not Counter(ransomNote) - Counter(magazine)

The Idea: Same idea, but this time stop immediately if the amount of occurrences in the ransom note exceeds the amount of letters available in the magazine.

Complexity: same as before, but at least as fast by a constant factor

from collections import Counter, defaultdict

class Solution(object):
    def canConstruct(self, ransomNote, magazine):
        """
        :type ransomNote: str
        :type magazine: str
        :rtype: bool
        """
        if len(ransomNote) > len(magazine):
            return False
        hist_ransom, hist_mag = Counter(ransomNote), Counter(magazine)
        for key, val in hist_ransom.items():
            if key not in hist_mag or val > hist_mag[key]:
                return False
        return True