Picnic Technologies’ Post

🌟 Celebrating a Milestone: Mastering Dependency Lookup in Spring! Excited to share a significant milestone in my journey of mastering Spring Framework! 🚀💡 After diving deep into the concept of "dependency lookup in Spring," I'm thrilled to announce that I've completed this essential aspect of modern Java development. 🔍🌱 What is Dependency Lookup in Spring? Dependency lookup in Spring is the process of retrieving beans or dependencies from the Spring Inversion of Control (IoC) container when they are needed by a component. It's a fundamental concept that enables flexible and modular application development. Highlights of the Journey: 🚀 Understanding the Fundamentals: Grasping the core principles of dependency lookup laid a solid foundation for my understanding of Spring's powerful features. 🔗 Exploring Different Methods: I explored various techniques, including ApplicationContext.getBean(), @Autowired annotation, and more, to perform dependency lookup efficiently.

🚀 Leetcode Daily Challenge 📅 6th May 2024 Post Part-II (Intuition and Solution) Problem Post: https://lnkd.in/dMGHt4Zp Intuition: 💠The problems says "Remove every node which has a node with a greater value anywhere to the right side of it." , So the values greater from right side need to be return, also need to check is there any smaller numbers are in between and are connect if yes remove the entire nodes. 💠Intuition here is to make the list reverse and check the current value is less than the previous value if yes we need to remove the node else the next node will be the greater number and will go accordingly. 💠After remove the nodes makes the list again reverse, to get to the original form. Approach: 💠 Use reverse node logic in make link list reverse. 💠 Change head.next = null, and head = reversed link list. 💠 Now, prev = head, curr = head.next using while loop iterate over the link list. 💠 Inside while loop check current val is less than the previous val if yes curr = curr.next, means move to next node. 💠 In case the condition fails store current node's next node to a variable. 💠 And current node's next will point to previous node and previous node will be current and current is next node. #comment the Time Complexity and Space Complexity for both the approach. Also, Justify which approach do you thinks is best and why? This problem can be solved in different ways this is the most better way. When I say better it means this is clean and can be easily understandable. #Comment section is open for more better and optimized solution. #dailycodingchallenge #leetcode #problem #solution #FAANG #interviewquestions #softwareengineer #java #python #c #cpp #javascript #se #linkList

🚀 Excited to share my latest achievement on LeetCode! 🚀 Problem 19: Remove Nth Node From End of List Successfully tackled the challenge of removing the Nth node from the end of a linked list on LeetCode! Example: Input: head = [1, 2, 3, 4, 5], n = 2 Output: [1, 2, 3, 5] Mastered the task of removing the Nth node from the end of a linked list using Java. Implemented a robust solution by leveraging the two-pointer technique to efficiently locate and remove the desired node. Ensured efficiency by using a dummy node to handle edge cases smoothly, such as removing the head node. Optimized the solution to traverse the list only once, making it highly efficient for various input sizes. Thoroughly tested the solution with different test cases to ensure its accuracy and reliability. This problem deepened my understanding of linked list manipulation, two-pointer techniques, and efficient coding practices in Java. #LeetCode #Java #Algorithms #ProblemSolving #LinkedList #TwoPointers #Coding #DataStructures

🚀 Excited to share my latest achievement on LeetCode! 🚀 Problem 14: Longest Common Prefix Successfully tackled the challenge of finding the longest common prefix among an array of strings on LeetCode! Example: Input: strs = ["flower","flow","flight"] Output: "fl" Solution Highlights: Mastered the task of efficiently identifying the common prefix using Java. Implemented a concise and robust solution by iteratively comparing prefixes of the strings. Reduced the prefix length dynamically until a common prefix is found across all strings. Handled edge cases like arrays with a single string or no common prefix. Thoroughly tested the solution with various test cases to ensure its accuracy and reliability. This problem deepened my understanding of string manipulation techniques and iterative comparisons in Java. Tags: #LeetCode #Java #Algorithms #ProblemSolving #LongestCommonPrefix #StringManipulation

🌟 Day 10 of my LeetCode Challenge! 🚀 Today, I took on the task of finding the longest common prefix among an array of strings. The goal was to efficiently determine the common prefix shared by all strings in the array. Here's the Java solution I crafted to solve the challenge: public String longestCommonPrefix(String[] strs) {    if (strs == null || strs.length == 0) {        return "";    }    String prefix = strs[0];    for (int i = 1; i < strs.length; i++) {        while (strs[i].indexOf(prefix) != 0) {            prefix = prefix.substring(0, prefix.length() - 1);            if (prefix.isEmpty()) {                return "";            }        }    }    return prefix; } This solution efficiently identifies the longest common prefix among the array of strings by iteratively comparing prefixes and reducing them until a common prefix is found. The problem challenged me to think creatively about string manipulation and optimization techniques, and I'm thrilled to have found a robust solution! Stay tuned for more updates as I continue my 100-day LeetCode challenge! 🌟💻 #LeetCode #100DaysOfCode #ProblemSolving #CodingJourney

🚀 Leetcode Daily Challenge 📅 1st May 2024 Post Part-II (Intuition and Solution) Problem Post: https://lnkd.in/dDrf4Ptr Intuition: 💠Here we need to find the fist occurrence of the letter in `word` string as we found we will reverse the string from the current index to 0 and concatenate the rest of the string. Approach: 💠Declare a variable `index` with -1 , iterate over the word string check each character matches with `ch` variable if yes update the variable `index` with the i. 💠After the loop if we get index as `-1` that means the character is not found in the string and can return s. 💠If we found the character in `ch` variable we need to slice the word from 0 to `index` + 1 and then split in to array use reverse() to reverse and join() to join the split array as string. 💠 Then concatenate the rest of the slice with the new string and return, thus the problem solved. #comment the Time Complexity and Space Complexity for this approach. This problem can be solved in different ways this is the most better way. When I say better it means this is clean and can be easily understandable. #Comment section is open for more better and optimized solution. #dailycodingchallenge #leetcode #problem #solution #FAANG #interviewquestions #softwareengineer #java #python #c #cpp #javascript

🚀 Day 22 of #30DaysOfSpringBoot Unveiling the @Component Annotation in Spring Boot 🔍 Today, let’s explore the @Component annotation in Spring Boot, one of the core annotations that makes dependency injection seamless and helps Spring manage the objects in your application. 💡 🔹 What is @Component? @Component is an annotation that marks a Java class as a Spring-managed bean. It’s a general-purpose stereotype annotation, indicating that an annotated class is a component, which will be automatically detected and registered by Spring during component scanning. 🔹 Why Use @Component? -> Automatic Bean Detection: With @Component, you don’t need to explicitly declare beans in your configuration file. Spring automatically detects and registers them in the application context. -> Loose Coupling: It helps in creating loosely coupled components by allowing Spring to inject dependencies into the classes marked with @Component. -> Code Simplification: Simplifies the wiring of components, services, and repositories without manual bean registration. 🔹 When to Use @Component? When you have a general-purpose class that performs a specific function and should be managed by Spring (like utility or helper classes). Use @Component for basic classes, and for more specific use cases, Spring provides specialized annotations like @Service, @Repository, and @Controller. 🔔 Follow along and join me on this 30-day journey to master Spring Boot. 💻 Checkout the amazing Discord community for coders :https://lnkd.in/gVfSqKFd #SpringBoot #Java #SpringFramework #BackendDevelopment #TechChallenge #Day22 #ComponentAnnotation #JavaDeveloper #SpringAnnotations

Day 14 of coding consistency challenge ✨ 🎉 Excited to share that I’ve successfully solved a problem on LeetCode with a runtime of **0 ms** and 100% efficiency! 🚀 The challenge was to implement the `removeElement` function in Java, which involves removing all instances of a specific value from an array **in-place** while maintaining O(n) time complexity. Here’s a snapshot of the solution and performance stats. 🌟 👨💻 **Key Highlights:** - **Language:** Java - **Optimization:** Leveraged two-pointer technique for space efficiency. - **Runtime:** 0 ms - **Memory:** 42.24 MB Each solved problem is another step forward in strengthening problem-solving skills and preparing for larger challenges ahead. If you’re also exploring algorithms or data structures, let’s connect and grow together. 🙌 #Java #CodingChallenge #ProblemSolving #LeetCode #CodingJourney

🚀 Leetcode Daily Challenge 📅 30th May 2024 Post Part-I (Problem Statement) Problem Statement: Given an array of integers arr. We want to select three indices i, j and k where (0 <= i < j <= k < arr.length). Let's define a and b as follows: 1. a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1] 2. b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k] Note that ^ denotes the bitwise-xor operation. Return the number of triplets (i, j and k) Where a == b. Examples: Input: arr = [2,3,1,6,7] Output: 4 Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4) #dailycodingchallenge #leetcode #problemoftheday #programmers #FAANG #interviewquestions #softwareengineer #java #python #c #javascript

30 Days Javascript Challenge by Chai Aur Code [Update: Day 16-22] In the past week, I’ve continued my coding challenge and delved into several key topics: 1. Learned how recursion works. 2. Studied basic data structures. 3. Practiced using regular expressions. 4. Explored local and session storage. 5. Solved easy and medium LeetCode problems. Progressing well and excited to learn more! GitHub Link - https://lnkd.in/dNV9gsTb #ChaiAurCode #Javascript