RansomNote
Problem
Solution
Track the number of times each letter appears in the magazine. Then, when the letter appears in the ransom note, decrement the count. If the count is ever negative, then the ransom note cannot be constructed from the magazine. A small optimization is to initialize the counter dictionary with all letters in the alphabet.
Code
https://github.com/GeorgeRPu/tech-interview-prep/blob/main/solutions/RansomNote.py
def canConstruct(ransomNote: str, magazine: str) -> bool:
"""Checks if ``ransomNote`` can be constructed from ``magazine`` by cutting
and gluing letters.
"""
letters = 'abcdefghijklmnopqrstuvwxyz'
counter = {letter: 0 for letter in letters}
for char in magazine:
counter[char] += 1
for char in ransomNote:
counter[char] -= 1
for char, count in counter.items():
if counter[char] < 0:
return False
return True
Test
>>> from RansomNote import canConstruct
>>> canConstruct("a", "b")
False
>>> canConstruct("aa", "ab")
False
>>> canConstruct("aa", "aab")
True
Functions
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Checks if |