Ransom Note

title: Ransom Note tags: hash-frequency-counter

The basic building block used is using hash as frequency counter.

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def can_spell?(magazine, note)
  letters = Hash.new(0)
  
  for c in magazine
    letters[c] += 1
  end
  
  for c in note.chars
    if letters[c] < 1
      return false
    end

    letters[c] -= 1
  end
  
  return true
end

p can_spell?(['a', 'b', 'c', 'd', 'e', 'f'], 'bed')
p can_spell?(['a', 'b', 'c', 'd', 'e', 'f'], 'bat')

The time complexity is O(N) + O(M), where it takes O(N) time to populate the frequency of the characters in the magazine and O(M) to check if each character in the ransom note is found in the magazine or not. The space complexity is O(1) because we only have 26 characters to store in the hash table.

You can solve this problem by keeping track of the counts of each character in the magazine string and then checking if there are enough characters to construct the ransomNote string.

Here’s the code:

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class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        char_count = {}  # Dictionary to store the counts of characters in magazine

        # Count the occurrences of each character in magazine
        for char in magazine:
            char_count[char] = char_count.get(char, 0) + 1

        # Check if there are enough characters in magazine to construct ransomNote
        for char in ransomNote:
            if char_count.get(char, 0) <= 0: # If the character is not enough
                return False
            char_count[char] -= 1 # Reduce the count for the character

        return True

Explanation:

  1. First, we count the occurrences of each character in the magazine string.
  2. Then, we iterate through the ransomNote string and check if there are enough occurrences of each character in the magazine string.
  3. If we find a character that is not enough, we return False.
  4. If we can construct the ransomNote string with the characters from the magazine string, we return True.

This solution has O(n) time complexity, where n is the length of the magazine string, as we iterate through the characters of both the ransomNote and magazine strings.

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class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        st1, st2 = Counter(ransomNote), Counter(magazine)
        if st1 & st2 == st1:
            return True

        return False

Problem Classification

This problem can be classified into the following categories:

  1. String Manipulation: The problem involves the handling of strings, checking the characters in one string against another.

  2. Frequency Counting: The solution requires counting the frequency of each letter in both strings and then comparing those frequencies. This technique is a common pattern for problems involving arrays or strings.

  3. Data Structures: As the problem involves counting the frequency of characters, data structures like arrays (for fixed size character set) or dictionaries / hashmaps (for arbitrary size character set) are usually used to keep track of these frequencies.

  4. Comparative Analysis: The solution to this problem involves comparing elements (in this case, letters and their counts) between two collections (in this case, strings).

  5. Availability Checking: This problem requires us to check if the characters needed for the ‘ransomNote’ are available in the ‘magazine’.

Language Agnostic Coding Drills

  1. Understanding the Problem Statement: The first step in solving any problem is understanding the problem statement. In this case, you are given two strings: ransomNote and magazine. Your task is to determine if it is possible to construct ransomNote by using letters from magazine, with the constraint that each letter in magazine can only be used once.

  2. String Manipulation and Character Counting: Understanding how to work with strings and how to count characters within strings is crucial for this problem. You need to know how to iterate over the characters in a string, and how to count the occurrences of each character.

  3. Working with Data Structures: In order to solve this problem efficiently, a data structure that supports quick lookup, insertion, and deletion operations, like a hashtable, can be used. In this case, the Python Counter from the collections module serves as such a data structure. This allows us to count the occurrences of each character in both ransomNote and magazine.

  4. Comparative Analysis: Once you have the counts of characters in both strings, the next step is to compare these counts to check if magazine contains enough of each character to construct ransomNote.

  5. Boolean Logic and Conditional Statements: The final step is to use conditional statements to return the appropriate boolean value. If every character in ransomNote is present in magazine with an equal or greater count, return True. Otherwise, return False.

By combining these individual drills or components, you’d be able to solve the problem. Your solution would involve parsing the strings, counting character occurrences, comparing these counts, and finally making a decision based on this comparison.

Targeted Drills in Python

  1. Understanding the Problem Statement: There isn’t a specific code drill for this step as it is about comprehending the problem statement.

  2. String Manipulation and Character Counting:

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def count_chars(string):
    char_counts = {}
    for char in string:
        if char in char_counts:
            char_counts[char] += 1
        else:
            char_counts[char] = 1
    return char_counts

print(count_chars("hello"))  # Expected output: {'h': 1, 'e': 1, 'l': 2, 'o': 1}

  1. Working with Data Structures: This is covered by the code drill above as well. The Counter class in Python essentially automates the process of counting characters in a string.

  2. Comparative Analysis:

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def compare_counts(count1, count2):
    for char in count1:
        if char not in count2 or count2[char] < count1[char]:
            return False
    return True

print(compare_counts({'a': 1, 'b': 2}, {'a': 1, 'b': 2, 'c': 3}))  # Expected output: True
print(compare_counts({'a': 1, 'b': 2}, {'a': 1, 'b': 1, 'c': 3}))  # Expected output: False
  1. Boolean Logic and Conditional Statements: Again, this is covered by the above drill. The function compare_counts already returns a boolean value based on a condition.

The final drill which combines all the previous drills would be the actual function that solves the problem:

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from collections import Counter

def can_construct(ransomNote, magazine):
    ransomNote_counts = Counter(ransomNote)
    magazine_counts = Counter(magazine)
    return not ransomNote_counts - magazine_counts

print(can_construct("a", "b"))  # Expected output: False
print(can_construct("aa", "aab"))  # Expected output: True

This solution uses Python’s Counter class to count characters in the strings, then subtracts the two Counter objects. If there are any characters left in ransomNote_counts after the subtraction, it means magazine did not have enough of those characters, so False is returned. If there are no characters left, True is returned.