Ransom Note in Python

Given two strings ransomNote and magazine, return true if ransomNote can be constructed from magazine and false otherwise.

Each letter in magazine can only be used once in ransomNote.

Example 1:

Input:  ransomNote = "aa", magazine = "abb"
Output:  false

Example 2:

Input:  ransomNote = "aa", magazine = "aab"
Output:  true

Note:

Your algorithm should run in O(n + m) time and use O(Σ) extra space.

Problem Definition

The problem requires us to determine if we can construct the string ransomNote using the characters from the string magazine. Each character in magazine can only be used once.

Input:

ransomNote: A string representing the ransom note.
magazine: A string representing the magazine text.

Output:

Return true if ransomNote can be constructed from magazine, otherwise return false.

Constraints:

Both strings consist of lowercase English letters.

Example:

Input: ransomNote = "aa", magazine = "abb"
Output: false

Input: ransomNote = "aa", magazine = "aab"
Output: true

Understanding the Problem

The core challenge is to ensure that every character in ransomNote appears in magazine at least as many times as it does in ransomNote. This problem is significant in scenarios where resource allocation or availability needs to be checked.

Common Pitfalls:

Assuming characters can be reused in magazine.
Not accounting for all characters in ransomNote.

Approach

To solve this problem, we can use a hash map (dictionary in Python) to count the occurrences of each character in both ransomNote and magazine. We then compare these counts to determine if the construction is possible.

Naive Solution:

A naive solution would involve iterating through each character in ransomNote and checking if it exists in magazine, removing it if found. This approach is not optimal as it involves multiple scans and deletions, leading to higher time complexity.

Optimized Solution:

We can optimize the solution by using two hash maps to store the frequency of each character in ransomNote and magazine. This allows us to perform the comparison in linear time.

Algorithm

Initialize two dictionaries to count the frequency of characters in ransomNote and magazine.
Iterate through each character in ransomNote and populate the first dictionary.
Iterate through each character in magazine and populate the second dictionary.
For each character in the first dictionary, check if the count in magazine is at least as much as in ransomNote.
If any character's count in magazine is less, return false.
If all checks pass, return true.

Code Implementation

def can_construct(ransomNote: str, magazine: str) -> bool:
    # Initialize dictionaries to count character frequencies
    ransom_freq = {}
    magazine_freq = {}
    
    # Populate ransom_freq with characters from ransomNote
    for char in ransomNote:
        if char in ransom_freq:
            ransom_freq[char] += 1
        else:
            ransom_freq[char] = 1
    
    # Populate magazine_freq with characters from magazine
    for char in magazine:
        if char in magazine_freq:
            magazine_freq[char] += 1
        else:
            magazine_freq[char] = 1
    
    # Check if magazine contains enough characters for ransomNote
    for char in ransom_freq:
        if ransom_freq[char] > magazine_freq.get(char, 0):
            return False
    
    return True

# Example usage
print(can_construct("aa", "abb"))  # Output: False
print(can_construct("aa", "aab"))  # Output: True

Complexity Analysis

The time complexity of this solution is O(n + m), where n is the length of ransomNote and m is the length of magazine. This is because we iterate through both strings once to populate the frequency dictionaries.

The space complexity is O(Σ), where Σ is the number of unique characters in the input strings. This is due to the storage required for the frequency dictionaries.

Edge Cases

Empty ransomNote: Should return true as no characters are needed.
Empty magazine: Should return false if ransomNote is not empty.
Characters in ransomNote not present in magazine: Should return false.

Testing

To test the solution comprehensively, consider the following test cases:

assert can_construct("", "") == True
assert can_construct("a", "") == False
assert can_construct("", "a") == True
assert can_construct("a", "a") == True
assert can_construct("aa", "aab") == True
assert can_construct("aa", "ab") == False

Thinking and Problem-Solving Tips

Break down the problem into smaller parts and solve each part individually.
Use hash maps or dictionaries to efficiently count and compare frequencies.
Practice similar problems to improve your understanding of frequency counting and hash maps.

Conclusion

Understanding and solving the "Ransom Note" problem helps in developing skills related to frequency counting and efficient use of hash maps. This problem is a good exercise in optimizing solutions and understanding time and space complexity.

Ransom Note in O(n + m) Time Complexity using Python