Algorithms Problem Solving: Ransom Note

This post is part of the Algorithms Problem Solving series. And it was originally published on TK's blog.

Problem description

This is the Ransom Note problem. The description looks like this:

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.

Examples

# ransomNote = "a", magazine = "b"
# => false

# ransomNote = "aa", magazine = "ab"
# => false

# ransomNote = "aa", magazine = "aab"
# => true

Solution

def can_construct(ransom_note, magazine):
    if ransom_note == '':
        return True

    if magazine == '':
        return False

    letters_counter = {}

    for letter in magazine:
        if letter in letters_counter:
            letters_counter[letter] += 1
        else:
            letters_counter[letter] = 1

    for char in ransom_note:
        if char not in letters_counter or letters_counter[char] == 0:
            return False

        letters_counter[char] -= 1

    return True

Resources

• Learning Python: From Zero to Hero
• Algorithms Problem Solving Series
• Big-O Notation For Coding Interviews and Beyond
• Learn Python from Scratch
• Learn Object-Oriented Programming in Python
• Data Structures in Python: An Interview Refresher
• Data Structures and Algorithms in Python
• Data Structures for Coding Interviews in Python
• One Month Python Course

Comments

[Jing Xue]

Another solution:

    def can_construct(ransom_note, magazine):
        for c in ransom_note:
            prev_len = len(magazine)
            magazine = magazine.replace(c, '', 1)
            if len(magazine) == prev_len:
                return False
        return True

[TK]

Great, Jing!

This was very similar with my first solution. But I wanted to try a different solution without using the replace method. Just to challenge me.

[Ryan Palo]

Have you heard about the collections module in the standard library? It's got a class called Counter available that is really useful in situations like this. Check out the docs

from collections import Counter

def can_build_ransom_note(message, magazine):
    needed_letters = Counter(message)
    available_letters = Counter(magazine)

    for letter, count in needed_letters.items():
        if available_letters[letter] < count:
            return False

    return True

[TK]

I didn't know about this Counter class. It is awesome! But my idea with this whole algorithm series is to try to implement the algorithm without help from built-in methods or classes. Just the default data structures.

Thanks for the note! I really liked this class.

[Ryan Palo]

That makes sense. Thanks for sharing!

[Amr Monier]

this my solution, i didn't want to match any character again if i already know that it's n't in the ransom note.
i think this would be a good solution in case the magazines string is huge.
but i have a question for huge strings (would it be better if we sorted the strings first ?)

class RansomNote:
    def __init__(self, note: str, mags: str) -> bool:
        self.note = note
        self.mags = mags
        if self.checkValidInput():
            if(self.canBeConstructed()):
                print('Message Can Constructed')
            else:
                print('Can\'t be constructed')
        else:
            print('Can\'t be constructed')

    def checkValidInput(self) -> bool:
        if(len(self.mags) < len(self.note)):
            return False
        return True

    def isInRansom(self, letter) -> bool:
        pos = self.note.find(letter)
        if pos != -1:
            self.note = self.note[:pos] + self.note[pos+1:]
            return True
        else:
            return False

    def canBeConstructed(self) -> bool:
        for letter in self.mags:
            if len(self.note) == 0:
                return True
            else:
                if(not self.isInRansom(letter)):
                    self.mags = self.mags.replace(letter, '')

        return False