🧑‍💻 Interviewer Guide: Ransom Note (UMPIRE Method)

https://leetcode.com/problems/ransom-note/

📌 Problem Statement

Given two strings ransomNote and magazine,
Return True if ransomNote can be constructed using the letters from magazine, False otherwise.

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

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🧭 U — Understand the Problem

Interviewer Prompts & Possible Answers:

• "Can you restate the problem in your own words?"
  → "We need to check if we have enough of each letter in the magazine to build the ransom note, where each letter can only be used once."

• "What does 'used once' mean exactly?"
  → "If magazine has two 'a's, we can use at most two 'a's in the ransom note. Each letter in magazine is consumed when used."

• "What if ransomNote is longer than magazine?"
  → "Then it's definitely impossible to construct it, so we'd return False."

• "Can the same letter appear multiple times?"
  → "Yes, we need to count how many times each letter appears in both strings."

• "What if ransomNote is empty?"
  → "An empty ransom note can always be constructed from any magazine, so return True."

• "Are the strings case-sensitive?"
  → "No, the constraints specify only lowercase English letters."

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🧪 M — Match with Patterns

Interviewer Prompts & Possible Answers:

• "What data structure would help track letter counts?"
  → "A hash map or Counter would be perfect for counting character frequencies."

• "Have you seen similar counting problems before?"
  → "Yes, like checking if one string is an anagram of another, or finding character frequencies."

• "What's the key insight here?"
  → "We need to ensure that for every character in ransomNote, magazine has at least that many occurrences of that character."

• "Could you solve this without extra space?"
  → "We could sort both strings and compare, but that would be less efficient. Or we could use a fixed-size array since we only have 26 lowercase letters."

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🛠️ P — Plan the Solution

Interviewer Prompts & Possible Answers:

• "What's the first thing you'd check?"
  → "If ransomNote is longer than magazine, we can return False immediately since we don't have enough letters."

• "How would you count the letters?"
  → "Create a frequency map of letters in magazine, then iterate through ransomNote and check if each letter is available."

• "What do you do when you find a letter in ransomNote?"
  → "Check if it exists in the magazine's frequency map with a count greater than 0. If yes, decrement the count. If no, return False."

• "Do you need to count ransomNote's letters separately?"
  → "Not necessarily. We can count magazine letters first, then iterate through ransomNote and decrement counts as we go."

• "Could you outline the algorithm step by step?"
  → "1. Quick length check. 2. Count all letters in magazine using a hash map. 3. For each letter in ransomNote, check if it's available in magazine map. 4. If available, decrement count. If not available or count is 0, return False. 5. If we complete the loop, return True."

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⌨️ I — Implement

Hash Map Solution

def canConstruct(ransomNote: str, magazine: str) -> bool:
    # Quick length check
    if len(ransomNote) > len(magazine):
        return False
    
    # Count letters in magazine
    letter_count = {}
    for char in magazine:
        letter_count[char] = letter_count.get(char, 0) + 1
    
    # Check if ransomNote can be constructed
    for char in ransomNote:
        if char not in letter_count or letter_count[char] == 0:
            return False
        letter_count[char] -= 1
    
    return True

Using Counter from collections

from collections import Counter

def canConstruct(ransomNote: str, magazine: str) -> bool:
    if len(ransomNote) > len(magazine):
        return False
    
    magazine_count = Counter(magazine)
    ransom_count = Counter(ransomNote)
    
    # Check if magazine has enough of each letter
    for char, count in ransom_count.items():
        if magazine_count[char] < count:
            return False
    
    return True

Counter Subtraction Approach

from collections import Counter

def canConstruct(ransomNote: str, magazine: str) -> bool:
    ransom_count = Counter(ransomNote)
    magazine_count = Counter(magazine)
    
    # Subtract and check if all counts are non-positive
    ransom_count.subtract(magazine_count)
    
    return all(count <= 0 for count in ransom_count.values())

Array-Based Solution (Fixed Space)

def canConstruct(ransomNote: str, magazine: str) -> bool:
    if len(ransomNote) > len(magazine):
        return False
    
    # Use array for 26 lowercase letters
    letter_count = [0] * 26
    
    # Count letters in magazine
    for char in magazine:
        letter_count[ord(char) - ord('a')] += 1
    
    # Check ransomNote
    for char in ransomNote:
        index = ord(char) - ord('a')
        if letter_count[index] == 0:
            return False
        letter_count[index] -= 1
    
    return True

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🔁 R — Review

Interviewer Prompts & Possible Answers:

• "What edge cases should we test?"
  → "Empty ransomNote, empty magazine, ransomNote longer than magazine, single character matches and mismatches, all letters the same, ransomNote requiring multiple of same letter."

• "Can you trace through the example 'aa' and 'aab'?"
  → "Count magazine: {'a': 2, 'b': 1}. Check ransomNote: first 'a' → count becomes {'a': 1, 'b': 1}, second 'a' → count becomes {'a': 0, 'b': 1}. Both found, return True."

• "What about 'aa' and 'ab'?"
  → "Count magazine: {'a': 1, 'b': 1}. Check ransomNote: first 'a' → count becomes {'a': 0, 'b': 1}, second 'a' → count is 0, return False."

• "Why do the length check first?"
  → "It's a quick optimization. If ransomNote is longer, we know immediately it's impossible without processing any characters."

Test Cases:

def test_canConstruct():
    # Basic failures
    assert canConstruct("a", "b") == False
    assert canConstruct("aa", "ab") == False
    
    # Basic success
    assert canConstruct("aa", "aab") == True
    
    # Empty ransomNote
    assert canConstruct("", "abc") == True
    
    # Empty magazine
    assert canConstruct("a", "") == False
    
    # Exact match
    assert canConstruct("abc", "abc") == True
    
    # Magazine has extras
    assert canConstruct("abc", "aabbcc") == True
    
    # Multiple same letters
    assert canConstruct("aaa", "aaaa") == True
    assert canConstruct("aaaa", "aaa") == False
    
    # Different letters entirely
    assert canConstruct("abc", "def") == False
    
    # Longer ransom note
    assert canConstruct("abcdef", "abc") == False

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🧹 E — Evaluate & Optimize

Interviewer Prompts & Possible Answers:

• "What's the time and space complexity?"
  → "Time: O(m + n) where m is magazine length and n is ransomNote length. Space: O(1) if using array approach (fixed 26 letters), or O(k) where k is number of unique characters in magazine for hash map approach."

• "Which approach is better - hash map or array?"
  → "Array is slightly more space-efficient since it's always 26 elements, while hash map only stores characters that appear. For small alphabets like lowercase English letters, array is great. Hash map is more flexible for larger character sets."

• "Could we optimize the Counter subtraction approach?"
  → "The basic hash map approach that iterates once through magazine and once through ransomNote is already optimal and more intuitive than Counter subtraction."

• "Is there a way to avoid counting all magazine letters?"
  → "Not really if we want O(1) lookup. We could count only when needed, but that would make us scan magazine multiple times, resulting in worse performance overall."

• "How does this compare to sorting both strings?"
  → "Sorting would be O(n log n + m log m) which is worse than our O(n + m) solution. Our approach is optimal."

• "What if the strings were very large?"
  → "Our algorithm scales linearly, which is optimal since we need to examine every character at least once. The constant space array approach would be preferable for memory efficiency."

• "Could we stop early in any case?"
  → "Yes, we already do - as soon as we find a character in ransomNote that's not available in magazine, we return False immediately."