Why HashMap behaves weird: equals() and hashCode()

💡 HashMap vs HashSet — Same Roots, Different Purpose Ever wondered how HashMap and HashSet work internally in Java? Here’s the quick breakdown 👇 🔹 HashMap Stores data in key–value pairs. Uses hashing to compute an index in an internal array (called a bucket). If two keys produce the same hash, Java handles it using LinkedList or Balanced Tree (from Java 8) to avoid performance loss. Average lookup: O(1) ⚡ 🔹 HashSet Internally backed by a HashMap! It stores elements as keys in the map, with a dummy constant object as the value. Ensures no duplicates because keys in a HashMap are unique. 🔹 Core concept: Hashing + equals() + hashCode() = Fast lookups with uniqueness guarantee. 🧠 Mastering this helps in writing optimized code and debugging performance issues effectively. #Java #HashMap #HashSet #DataStructures #Coding #BackendDevelopment #Performance #SoftwareEngineering

🚀 Ever wondered why HashMap is a go-to for fast lookups in Java? Let's peel back the layers on its internal implementation – stuff most devs use daily but rarely dive into! 💻 At its core, HashMap is backed by an array of buckets (initially 16). When you put(key, value), it computes the key's hashCode(), mods it by the array length to find the bucket index, and stores an Entry (Node in Java terms) there. Collisions? Handled via chaining – each bucket can become a linked list of entries. But since Java 8, if a bucket's chain exceeds 8 nodes (TREEIFY_THRESHOLD), it morphs into a balanced red-black tree for O(log n) performance instead of O(n) worst-case! 🌳 Now, the fun part: resizing. When entries hit ~75% load factor (0.75 default), the map doubles its capacity and rehashes everything. This prevents performance degradation but can be costly in time and memory. What if it exceeds memory during this? Boom – OutOfMemoryError! 🛑 Java throws java.lang.OutOfMemoryError: Java heap space if the JVM can't allocate more heap for the new array. To mitigate, tune initial capacity or load factor via constructors, or bump -Xmx for more heap. Pro tip: Overriding hashCode() poorly can lead to uneven distribution and frequent resizes. Have you tuned a HashMap to avoid OOM in prod? Share your war stories below! 👇 #Java #HashMap #JVMInternals #ProgrammingTips #TechDeepDiveHave Check out this visual breakdown of HashMap's structure:

𝐉𝐚𝐯𝐚 𝐇𝐚𝐬𝐡𝐌𝐚𝐩 : 𝐒𝐢𝐦𝐩𝐥𝐞 𝐄𝐱𝐩𝐥𝐚𝐧𝐚𝐭𝐢𝐨𝐧 * HashMap is part of the java.util package and is one of the most commonly used data structures in Java. * It stores data in key-value pairs and provides very fast access, usually in O(1) time complexity. 𝐇𝐨𝐰 𝐈𝐭 𝐖𝐨𝐫𝐤𝐬 • When you store a key-value pair, Java uses the key’s hashCode() method to determine which bucket (index) to store it in. • If two keys produce the same hash value (a collision), HashMap uses equals() to compare the keys and decide where to store them. • From Java 8 onward, if too many collisions occur, the internal structure changes from a linked list to a balanced tree to improve performance. 𝐊𝐞𝐲 𝐏𝐨𝐢𝐧𝐭𝐬 • Stores key-value pairs • No duplicate keys, inserting a new value with an existing key replaces the old one • Not synchronized, use ConcurrentHashMap for thread-safe operations #Java #HashMap #DataStructures #Coding #JavaProgramming #SoftwareDevelopment #JavaTips #DeveloperLife

💡 What I Learned Today: equals() and hashCode() in Java Today I explored how equals() and hashCode() work together — a small concept that plays a huge role in collections like HashMap and HashSet. Here’s what I learned 👇 🔹 equals() → compares two objects for logical equality. 🔹 hashCode() → returns an integer (hash value) used by hash-based collections. ✅ Key rule: If two objects are equal according to equals(), they must have the same hash code. Otherwise, collections like HashMap or HashSet may not work correctly. Example: - If you override equals() in your class but forget to override hashCode(), you might end up with duplicate entries in a HashSet or missing keys in a HashMap. Understanding this relationship ensures your objects behave correctly in all Java collections — a simple but powerful concept! #Java #HashCode #Equals #Collections #JavaDeveloper #CodingTips #LearningJourney

💻 Day 52 of #100DaysOfCodeChallenge Today’s problem: LeetCode 203 – Remove Linked List Elements 🧩 I worked on a Linked List manipulation problem where the goal is to remove all nodes that match a given value. This problem helped strengthen my understanding of pointer handling and dummy node usage in linked lists. 🧠 What I Learned: Handling edge cases where the head itself might need to be removed. Using a dummy node before the head to simplify pointer operations. How current and next pointers work together to modify a linked list safely. 🧩 Java Solution: class Solution { public ListNode removeElements(ListNode head, int val) { ListNode dummy = new ListNode(0); dummy.next = head; ListNode current = dummy; while (current.next != null) { if (current.next.val == val) current.next = current.next.next; else current = current.next; } return dummy.next; } } ⏱️ Complexity: Time: O(n) Space: O(1) Every day, I’m getting more confident with data structures and linked list traversal logic. On to the next challenge 🚀 #Day52 #LeetCode #Java #100DaysOfCode #CodingChallenge #LinkedList #ProblemSolving

⚙️ equals() and hashCode() — The Silent Duo That Holds Java Collections Together 🧠 You use them every day — maybe without realizing it. But these two methods literally decide how your Java objects behave in collections. Let’s decode them 👇 --- 🔹 1️⃣ The Golden Rule If two objects are equal, they must have the same hash code. But having the same hash code ❌ doesn’t mean they’re equal. It’s like house numbers — two houses can’t share one number, but two houses in different cities can both have “21A” 🏠 --- 🔹 2️⃣ How Collections Use Them Take a HashMap or HashSet: When you add an object, Java: 1️⃣ Uses hashCode() to find the right bucket 🪣 2️⃣ Uses equals() to find the exact match inside that bucket If you break the contract (override one but not the other)… 💥 your map can lose entries or fail to find keys that actually exist. --- 🔹 3️⃣ A Quick Example class Employee { String id; String name; public boolean equals(Object o) { return o instanceof Employee && id.equals(((Employee) o).id); } public int hashCode() { return id.hashCode(); } } Here, two employees with the same ID are considered equal — and because hashCode aligns with that logic, HashMap works perfectly. ✅ --- ⚡ The Takeaway Always override equals() and hashCode() together Use IDEs or Lombok (@EqualsAndHashCode) to generate them safely Test equality logic for correctness, not convenience They may look small — but they hold the Java Collection Framework together 🔗 --- 💬 Your turn: Have you ever faced a bug because of incorrect equals() or hashCode() implementation? 👇 Drop your story — I bet every dev has one! #Java #Collections #HashMap #CleanCode #BackendDevelopment #SoftwareEngineering #JavaDeveloper

💭 Ever thought about what really happens when you do map.put("key", "value"); Seems simple, right? But under the hood — it’s a masterpiece of engineering. ⚙️ --- 🔹 1. It all starts with hashCode() Java calls hashCode() on your key and uses it to find the bucket where the key-value pair will live. Good hash function = fewer collisions = faster lookups. 🚀 --- 🔹 2. Then comes equals() If two keys land in the same bucket (collision), Java compares them using equals() to find the exact match. That’s why both hashCode() and equals() must be consistent — break that rule and your map breaks too 😬 --- 🔹 3. What happens after Java 8? Before Java 8 — collisions were stored in linked lists. Now, if too many keys end up in one bucket, it’s converted to a balanced tree (Red-Black Tree) 🌳 → Faster lookups, even with bad hash distributions. --- ⚡ The takeaway Every HashMap operation hides a brilliant balance of speed, memory, and engineering simplicity. Knowing how it works helps you debug smarter and write more predictable code. --- 💬 Your turn: Did you know about HashMap’s internal switch from LinkedList to Tree after Java 8? Or was this new to you? 👇 #Java #HashMap #DataStructures #BackendDevelopment #CodeInternals #JavaDeveloper

𝗛𝗼𝘄 𝗛𝗮𝘀𝗵𝗠𝗮𝗽 𝗪𝗼𝗿𝗸𝘀 𝗜𝗻𝘁𝗲𝗿𝗻𝗮𝗹𝗹𝘆 𝗶𝗻 𝗝𝗮𝘃𝗮? I recently watched a video from Concepts & Coding by Shrayansh Jain — he explained the internal working of HashMap in a very simple way. Here’s what I learned👇 🎥 Video link:https://lnkd.in/gmQ3qe8a 1️⃣ When we create a HashMap without a size, Java creates an internal array of 16 buckets by default. Each bucket is a place where data entries are stored. 2️⃣ When we call put(key, value) — Java first calculates the hash code of the key. Then it spreads the hash evenly using (hash >>> 16). After that, it finds the bucket index using (n - 1) & hash — this is faster than using %, because the array size is always a power of 2. Finally, the entry is stored as a Node (which keeps hash, key, value, and next). 3️⃣ If the same key is added again — Java checks hashCode() to find the right bucket. Then it uses equals() to check if the key already exists. If it matches, the value is updated. 4️⃣ If there’s a collision (different key but same bucket): The new entry is added to the linked list at that bucket. If the list size becomes 8 or more (and total size ≥ 64), it converts into a Red-Black Tree for faster lookups. 5️⃣ When the map fills up beyond 75% of its capacity (load factor = 0.75) — Java performs rehashing: doubles the array size and re-inserts all entries, because their new index changes. 6️⃣ Time complexity: Average → O(1) Worst case → O(log n) (if tree used) or O(n) (in linked list) 💡 HashMap looks simple, but it’s one of the most efficient and well-designed data structures in Java. #Java #HashMap #DataStructure #CollectionsFramework #BackendDevelopment #CodingConcepts #CleanCode #techieanky #javainterviewquestion #grow #linkedin

LeetCode Question #199 — Binary Tree Right Side View Thrilled to share another Accepted Solution (100% Runtime 🚀) on LeetCode! This problem focuses on Binary Trees — specifically, how to capture the view of a tree from its right side 🌳➡️. I implemented a Modified Pre-Order Traversal (Root → Right → Left) in Java, ensuring that the first node at each level (from the right) gets recorded. 💡 Key Idea: Use recursion with a level tracker — when the current level equals the list size, it means we’re seeing the rightmost node for the first time. Here’s the performance snapshot: ⚙️ Runtime: 0 ms (Beats 100% of Java submissions) 💾 Memory: 42.4 MB Every problem like this sharpens my understanding of tree traversal patterns and depth-first search optimization. #LeetCode #Java #DataStructures #BinaryTree #ProblemSolving #CodingJourney #DSA #100PercentRuntime

Day 39/100 – #100DaysOfCode 🚀 | #Java #LeetCode #HashMap ✅ Problem Solved: Minimum Index Sum of Two Lists 🧩 Problem Summary: Given two string arrays, find the common strings with the smallest index sum (index in list1 + index in list2). If multiple answers exist, return all of them. 💡 Approach Used: ➤ Stored the elements of the first list in a HashMap with their indices. ➤ Iterated the second list and checked for matches. ➤ Tracked the minimum index sum and collected all strings that match that sum. This avoids nested loops and improves efficiency. ⚙️ Time Complexity: O(n + m) 📦 Space Complexity: O(n) ✨ Takeaway: HashMaps continue to be one of the most effective tools for reducing time complexity through direct lookups 🔍⚡ #Java #LeetCode #HashMap #DataStructures #ProblemSolving #100DaysOfCode #CodingJourney