Unveiling the Mysterious Ransom Note
Table of Contents
 Introduction
 Problem Description
 Example
 Approach
 Implementation
 Complexity Analysis
 Conclusion
 Pros and Cons
 FAQs
 References
Introduction
In this article, we will discuss a common question asked by Microsoft called "Ransom Note". We will explore the problem description, provide an example, discuss the approach to solve the problem, and provide an implementation in Java. Additionally, we will analyze the complexity of the solution and conclude with the pros and cons. Finally, we will address some frequently asked questions and provide references for further reading.
Problem Description
The problem involves constructing a ransom note from the letters available in a magazine. Given two strings, one representing the ransom note and the other representing the magazine, our task is to determine if the ransom note can be created using the letters from the magazine. It is important to note that each letter in the magazine can only be used once in the ransom note. The strings only contain lowercase letters.
Example
Let's consider an example to understand the problem better. Suppose we have the following strings:
Ransom Note: "aa"
Magazine: "ab"
In this case, we cannot create the ransom note "aa" using the available letters in the magazine "ab". Since we only have one "a" available in the magazine, we cannot construct the required two "a"s for the ransom note. Hence, the output for this example would be false.
Approach
To solve this problem, we can use a hashmap to keep track of the count of each character in the magazine. We will iterate through the magazine and populate the hashmap accordingly. Then, we can traverse the ransom note and check if we have enough available characters in the hashmap to construct the note.
Implementation
Let's begin by creating a hashmap to store the counts of each character in the magazine:
HashMap<Character, Integer> counts = new HashMap<>();
Next, we need to populate the hashmap by iterating through the magazine string:
for (char c : magazine.toCharArray()) {
counts.put(c, counts.getOrDefault(c, 0) + 1);
}
After populating the hashmap, we can traverse the ransom note and check if we have enough characters available:
for (char c : ransomNote.toCharArray()) {
if (!counts.containsKey(c)  counts.get(c) <= 0) {
return false;
}
counts.put(c, counts.get(c)  1);
}
If we successfully iterate through the entire ransom note without encountering any issues, we can return true.
Complexity Analysis
The runtime complexity of this solution is O(M + N), where M is the number of characters in the magazine and N is the number of characters in the ransom note. The space complexity is O(N) as we are storing the counts of characters in the ransom note.
Conclusion
In this article, we discussed the "Ransom Note" problem asked by Microsoft. We explored the problem description, provided an example, discussed the approach to solve the problem, and implemented the solution in Java. We also analyzed the complexity of the solution and concluded with the pros and cons. This problem highlights the importance of efficient manipulation of character counts to solve stringrelated problems.
Pros and Cons
Pros:
 Efficient solution using a hashmap for character count tracking
 Simple and easy to understand implementation
 Suitable for solving similar string manipulation problems
Cons:
 Relies on iterating through both the magazine and ransom note strings, which could be timeconsuming for large inputs
FAQs

What is the "Ransom Note" problem?
The "Ransom Note" problem involves constructing a ransom note using the characters available in a magazine, considering the constraints on the usage of each character in the magazine.

How can we solve the "Ransom Note" problem efficiently?
We can solve the problem efficiently by using a hashmap to track the count of each character in the magazine and then check if we have enough characters available while constructing the ransom note.

What is the complexity of the solution?
The runtime complexity of the solution is O(M + N), where M is the length of the magazine string and N is the length of the ransom note string. The space complexity is O(N).

Can this solution handle casesensitive characters?
No, the solution assumes that both the magazine and the ransom note strings only contain lowercase letters.
References
 LeetCode Problem  Ransom Note
 HashMap Java Documentation
 Character Class Java Documentation