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

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

Clarifying Questions

  1. Case Sensitivity: Should the comparison be case-sensitive? (Assume yes for this problem unless otherwise stated)
  2. Non-alphabetic characters: Can we assume only alphabetic characters in both strings?
  3. Input Length: Is there an upper bound on the length of the input strings?

Assuming the answers align with common problem constraints:

  1. Yes, the comparison is case-sensitive.
  2. Yes, we are only dealing with lower-case English letters.
  3. Inputs can be reasonably large, up to tens of thousands of characters.

Strategy

  1. Frequency Counts: We’ll use a histogram (or frequency count) approach. This involves counting the occurrences of each character in both ransomNote and magazine.
  2. Comparison: For each unique character in ransomNote, verify that the corresponding count in magazine is at least as large.
  3. Return Result: Return true if all characters in ransomNote are sufficiently available in magazine, otherwise return false.

Code

def canConstruct(ransomNote, magazine):
    from collections import Counter

    ransom_counter = Counter(ransomNote)
    magazine_counter = Counter(magazine)

    for char, count in ransom_counter.items():
        if magazine_counter[char] < count:
            return False

    return True

Time Complexity

  1. Counting Frequencies: Creating the frequency histograms for both strings takes (O(n + m)) time, where (n) is the length of ransomNote and (m) is the length of magazine.
  2. Comparing Counts: Checking the counts involves iterating over the unique characters in ransomNote, which is (O(k)) where (k) is the number of unique characters in ransomNote.

Overall Time Complexity: (O(n + m))

This approach should be efficient for the input sizes generally considered in these types of problems.

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