Introduction

The Ransom Note problem involves checking whether one string can be constructed using characters from another string.

Given:

  • a string ransomNote
  • a string magazine

the task is to determine whether:

  • every character of ransomNote
    can be formed using characters from magazine

Each character in magazine can only be used once.

This problem helps in understanding:

  • hashing
  • frequency counting
  • string traversal
  • resource allocation logic

Example

Input:ransomNote = "code"
magazine = "ocedxyz"

Output:
True

Explanation:
All characters required for "code"
are present in magazine.

Input:
ransomNote = "hello"
magazine = "helo"

Output:
False

Explanation:
Character 'l' is required twice
but appears only once.

Constraints

1 <= ransomNote.length,magazine.length <= 10^5
ransomNote and magazine
contain lowercase English letters

Approach 1 : Brute Force

Explanation

The simplest way to solve this problem is:

  1. Traverse ransomNote characters
  2. Search matching character inside magazine
  3. Mark used characters

This approach works but repeated searching increases time complexity.

Steps

  1. Convert magazine into list.
  2. Traverse ransomNote characters.
  3. Search matching character in magazine.
  4. Mark character as used.
  5. If any character is missing:
    • return false
  6. Otherwise:
    • return true

Dry Run

Input:ransomNote = "code"magazine = "ocedxyz"
Check c:
Found in magazine
Mark as used

Check o:
Found in magazine

Check d:
Found in magazine

Check e:
Found in magazine

Final Result:
True

Brute Force Code

Complexity Analysis

Time Complexity: O(n²)Explanation:Repeated searching is performed
for every character.

Space Complexity: O(n)
Explanation:
Extra storage may be used for traversal.

Approach 2 : Optimized Solution (Hashing)

Explanation

The optimized solution uses frequency counting.

The idea is:

  1. Count magazine character frequencies
  2. Reduce frequencies using ransomNote
  3. If any frequency becomes negative:
    • construction is impossible

This avoids repeated searching.

Steps

  1. Create frequency hashmap.
  2. Count magazine characters.
  3. Traverse ransomNote.
  4. Reduce frequencies.
  5. If frequency becomes negative:
    • return false
  6. Otherwise:
    • return true

Dry Run

Input:ransomNote = "code"magazine = "ocedxyz"

Store frequencies:
o → 1
c → 1
e → 1
d → 1

Process ransomNote:
c → decrease frequency
o → decrease frequency
d → decrease frequency
e → decrease frequency
All frequencies valid

Final Result:
True

Optimized Code

Complexity Analysis

Time Complexity: O(n)Explanation:Characters are processed once.

Space Complexity: O(n)
Explanation:
Extra hashmap is used for frequencies.

Edge Cases

  1. ransomNote is empty
  2. magazine is empty
  3. Characters repeat multiple times
  4. All characters available
  5. Missing characters present

Why This Problem is Important

This problem helps in understanding:

  1. Hashing
  2. Frequency counting
  3. Resource allocation logic
  4. Efficient searching
  5. String traversal

It is one of the most common hashing interview problems.

Real-World Applications

Frequency validation concepts are used in:

  1. Inventory systems
  2. Resource allocation
  3. Text analytics
  4. Compiler design
  5. Data verification systems

Common Mistakes

  1. Forgetting repeated characters
  2. Incorrect frequency reduction
  3. Missing zero-frequency checks
  4. Improper hashmap updates

Interview Tips

Interviewers often expect:

  1. Hashing optimization
  2. Frequency counting explanation
  3. O(n) solution

Always explain why magazine frequencies decrease after character usage.

Related Questions

  1. Valid Anagram
  2. Frequency Count
  3. First Unique Character
  4. Single Number
  5. Character Frequency

Final Takeaway

The Ransom Note problem is a fundamental hashing problem that teaches frequency counting and efficient resource allocation techniques. Understanding this problem builds a strong foundation for advanced hashing and string-processing interview problems.