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Jack authored 8 years ago . add some solutions
# Time: O(n)
# Space: O(1)
# 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
class Solution(object):
def canConstruct(self, ransomNote, magazine):
"""
:type ransomNote: str
:type magazine: str
:rtype: bool
"""
counts = [0] * 26
letters = 0
for c in ransomNote:
if counts[ord(c) - ord('a')] == 0:
letters += 1
counts[ord(c) - ord('a')] += 1
for c in magazine:
counts[ord(c) - ord('a')] -= 1
if counts[ord(c) - ord('a')] == 0:
letters -= 1
if letters == 0:
break
return letters == 0
# Time: O(n)
# Space: O(1)
import collections
class Solution2(object):
def canConstruct(self, ransomNote, magazine):
"""
:type ransomNote: str
:type magazine: str
:rtype: bool
"""
return not collections.Counter(ransomNote) - collections.Counter(magazine)
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