Ransom note¶
Time: O(N); Space: O(1); easy
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.
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
Notes: * You may assume that both strings contain only lowercase letters.
[1]:
class Solution1(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
[2]:
s = Solution1()
ransomNote = "a"
magazine = "b"
assert s.canConstruct(ransomNote, magazine) == False
ransomNote = "aa"
magazine = "ab"
assert s.canConstruct(ransomNote, magazine) == False
ransomNote = "aa"
magazine = "aab"
assert s.canConstruct(ransomNote, magazine) == True
[3]:
import collections
class Solution2(object):
"""
Time: O(N)
Space: O(1)
"""
def canConstruct(self, ransomNote, magazine):
"""
:type ransomNote: str
:type magazine: str
:rtype: bool
"""
return not collections.Counter(ransomNote) - collections.Counter(magazine)
[4]:
s = Solution2()
ransomNote = "a"
magazine = "b"
assert s.canConstruct(ransomNote, magazine) == False
ransomNote = "aa"
magazine = "ab"
assert s.canConstruct(ransomNote, magazine) == False
ransomNote = "aa"
magazine = "aab"
assert s.canConstruct(ransomNote, magazine) == True