Updated the ransom note program with solution and explaination

Author: Vitessse

Arrays & Strings/#383 - Ransom Note - Easy/Explaination.md

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+# Problem Title
+
+**Difficulty:** Easy   
+**Category:** Strings,Hash table
+**Leetcode Link:** [Problem Link]https://leetcode.com/problems/ransom-note/
+
+---
+
+## 📝 Introduction
+
+You are given two strings: ransomNote and magazine.
+You need to determine if you can construct the ransomNote using the letters from magazine.
+
+Each letter in magazine can only be used once in ransomNote.
+Return true if possible, otherwise return false.
+
+Constraints:
+
+1 <= ransomNote.length, magazine.length <= 10⁵
+
+ransomNote and magazine consist of lowercase English letters
+
+---
+
+## 💡 Approach & Key Insights
+
+The key idea is to count the frequency of each character in both ransomNote and magazine.
+Then, ensure that for each character required by the ransom note, the magazine has at least that many occurrences.
+
+A brute force approach would be inefficient.
+Instead, we can use a hash map (or Python’s collections.Counter) to store and compare character frequencies efficiently.
+
+---
+
+## 🛠️ Breakdown of Approaches
+
+### 1️⃣ Brute Force / Naive Approach
+
+-**Explanation:** For every character in ransomNote, loop through magazine and find a match, removing used characters.
+This leads to nested iterations and repeated string operations.
+
+- **Time Complexity:** *O(n^2) -  nested loops for character comparisons*
+- **Space Complexity:** *O(1) - no additional data structures used*
+- **Example/Dry Run:**
+
+Example input: ransomNote = "aa", magazine = "aab"
+Step 1 → Check for 'a' in magazine → Found
+Step 2 → Remove 'a' → magazine = "ab"
+Step 3 → Check for 'a' again → Found
+Step 4 → Return true
+
+
+### 2️⃣ Optimized Approach
+
+- **Explanation:** Use a hash map to count how many times each letter appears in magazine.
+Then, for each letter in ransomNote, check if it exists in the map with sufficient count. Decrease the count as we go
+
+- **Time Complexity:** *O(m + n) - where m = len(magazine), n = len(ransomNote)*
+- **Space Complexity:** *O(1)-because character set is fixed (26 lowercase letters)
+
+- **Example/Dry Run:**
+
+ransomNote = "aa", magazine = "aab"
+Step 1 → magazine_counter = {'a': 2, 'b': 1}
+Step 2 → Check 'a' in ransomNote → Yes (2 available) → decrement to 1
+Step 3 → Check 'a' again → Yes (1 available) → decrement to 0
+Step 4 → Return true
+
+
+### 3️⃣ Best / Final Optimized Approach (if applicable)
+
+- **Explanation:** * Use a hash map to count how many times each letter appears in magazine.
+Then, for each letter in ransomNote, check if it exists in the map with sufficient count. Decrease the count as we go*
+- **Time Complexity:** *O(m + n) - where m = len(magazine), n = len(ransomNote)*
+- **Space Complexity:** *O(1) - because character set is fixed (26 lowercase letters)*
+- **Example/Dry Run:**
+
+Example input: 
+ransomNote = "aa", magazine = "aab"
+Step 1 → magazine_counter = {'a': 2, 'b': 1}
+Step 2 → Check 'a' in ransomNote → Yes (2 available) → decrement to 1
+Step 3 → Check 'a' again → Yes (1 available) → decrement to 0
+Step 4 → Return true
+---
+
+## 📊 Complexity Analysis
+
+| Approach      | Time Complexity | Space Complexity |
+| ------------- | --------------- | ---------------- |
+| Brute Force   | O(n^2)          | O(1)             |
+| Optimized     | O(m + n)        | O(1)             |
+| Best Approach | O(m + n)        | O(1)             |
+
+---
+
+## 📉 Optimization Ideas
+
+Use collections.Counter in Python for cleaner syntax and built-in methods for frequency counting and subtraction.
+
+Early exit: if len(ransomNote) > len(magazine), return False immediately.
+
+---
+
+## 📌 Example Walkthroughs & Dry Runs
+
+Example:
+Input: ransomNote = "abc", magazine = "aabbcc"
+Process:
+1 → Count magazine: {'a': 2, 'b': 2, 'c': 2}
+2 → 'a' needed → OK
+3 → 'b' needed → OK
+4 → 'c' needed → OK
+Output: true
+
+Input: ransomNote = "aab", magazine = "abc"
+Process:
+1 → Count magazine: {'a': 1, 'b': 1, 'c': 1}
+2 → 'a' needed → Only 1 'a' available
+Output: false
+
+---
+
+## 🔗 Additional Resources
+
+- collections.Counter docs
+- Pythonic Solution Discussion
+
+
+---
+
+Author: Andrew
+Date: 13/06/2025

Arrays & Strings/#383 - Ransom Note - Easy/Solution.py

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+class Solution(object):
+    def canConstruct(self, ransomNote, magazine):
+        ransom_count = Counter(ransomNote)
+        magazine_count = Counter(magazine)
+
+        for letter, count in ransom_count.items():
+            if magazine_count[letter] < count:
+                return False
+    
+        return True

LeetCode-Solutions

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+Subproject commit dd7a2d5de50b2544eb285506601392340c529811