Ransom Note

Clarify the problem:

• The problem requires determining if a ransom note can be constructed from a given magazine.
• We need to implement a function that takes two strings, ransomNote and magazine, as input and returns a boolean indicating whether the ransom note can be constructed from the magazine.

Analyze the problem:

• Input: Two strings, ransomNote and magazine, consisting of only lowercase English letters.
• Output: A boolean value indicating whether the ransom note can be constructed from the magazine.
• Constraints:
  • The length of ransomNote and magazine is at most 10^3.
  • The characters in magazine can be used to construct the ransomNote.

Design an algorithm:

• We can solve this problem by counting the frequency of characters in both the ransomNote and magazine.
• We can use an array or a dictionary to store the character frequencies.
• First, we iterate through the magazine and count the frequency of each character.
• Then, we iterate through the ransomNote and decrement the corresponding frequency for each character.
• If, at any point, the frequency of a character in the ransomNote becomes negative or the character is not present in the magazine, we return False.
• If we successfully iterate through the ransomNote without encountering any issues, we return True.

Explain your approach:

• We will implement a function called canConstruct that takes two strings, ransomNote and magazine, as input.
• We initialize an array freq of size 26 to store the frequency of characters (assuming lowercase English letters).
• We iterate through the magazine and increment the frequency of each character using its ASCII value as the index in the freq array.
• Next, we iterate through the ransomNote and decrement the frequency of each character.
• If, at any point, the frequency becomes negative or the character is not present in the magazine, we return False.
• Finally, if we successfully iterate through the ransomNote without any issues, we return True.

Write clean and readable code:

python

def canConstruct(ransomNote, magazine):
    freq = [0] * 26

    # Count the frequency of characters in magazine
    for ch in magazine:
        freq[ord(ch) - ord('a')] += 1

    # Check if ransomNote can be constructed
    for ch in ransomNote:
        freq[ord(ch) - ord('a')] -= 1
        if freq[ord(ch) - ord('a')] < 0:
            return False

    return True

Test your code:

python

# Test case 1: Example case
ransomNote = "a"
magazine = "b"
# The ransomNote cannot be constructed from the magazine.
print(canConstruct(ransomNote, magazine))  # Output: False

# Test case 2: Example case
ransomNote = "aa"
magazine = "ab"
# The ransomNote cannot be constructed from the magazine.
print(canConstruct(ransomNote, magazine))  # Output: False

# Test case 3: Additional case
ransomNote = "aa"
magazine = "aab"
# The ransomNote can be constructed from the magazine by using one 'a'.
print(canConstruct(ransomNote, magazine))  # Output: True

# Test case 4: Additional case
ransomNote = "abc"
magazine = "abcdefg"
# The ransomNote can be constructed from the magazine by using 'a', 'b', and 'c'.
print(canConstruct(ransomNote, magazine))  # Output: True

Optimize if necessary:

• The current approach has a time complexity of O(m + n), where m is the length of ransomNote and n is the length of magazine. This is because we iterate through both strings once.
• The space complexity is O(1) because we use a fixed-size array freq of size 26 to store the character frequencies.

Handle error cases:

• We need to consider the case where either ransomNote or magazine is an empty string. In such cases, we can return False since an empty string cannot be constructed from any other string.

Discuss complexity analysis:

• Let m and n be the lengths of ransomNote and magazine, respectively.
• The time complexity of the solution is O(m + n) since we iterate through both strings once.
• The space complexity is O(1) because we use a fixed-size array of size 26 to store the character frequencies, which is independent of the input size.