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
Last modified 3yr ago