Java LeetCode Solution: Find Missing Array Elements

🚀 Day 46/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2315. Count Asterisks Used a boolean flag approach to track whether the current position is inside a pair of '|' and count only the valid '*' outside those sections. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) Strengthening understanding of state-based string traversal and conditional counting. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 57/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 242. Valid Anagram Used a HashMap frequency counting approach (Unicode-safe using code points) to compare character frequencies of both strings. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) Strengthening understanding of string frequency analysis and Unicode handling. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency 

🚀 Day 65/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2535. Difference Between Element Sum and Digit Sum of an Array Used a single loop + digit extraction approach to calculate both element sum and digit sum efficiently. ⏱️ Time Complexity: O(n × d) 📦 Space Complexity: O(1) Strengthening understanding of number manipulation and digit-based operations. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 58/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 278. First Bad Version Used binary search on answer space to efficiently find the first bad version while minimizing API calls. ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) Strengthening understanding of binary search optimization and decision-based problems. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

Day 88 - LeetCode Journey Solved LeetCode 24: Swap Nodes in Pairs in Java ✅ This problem is a great exercise in pointer manipulation in linked lists. Instead of changing values, I swapped the actual nodes by carefully adjusting the next pointers. Using a dummy node made the process much cleaner and helped handle edge cases smoothly. Step by step, I swapped pairs while moving forward in the list. Key takeaways: • Deep understanding of pointer manipulation • Importance of dummy node in linked list problems • Clean handling of edge cases • Iterative approach for swapping nodes ✅ All test cases passed ⚡ Efficient O(n) time and O(1) space Linked list problems like this really sharpen your fundamentals 🔥 #LeetCode #DSA #Java #LinkedList #Pointers #ProblemSolving #CodingJourney #InterviewPrep #Consistency #100DaysOfCode

🚀 Day 48/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1534. Count Good Triplets Used a brute-force triple nested loop to check all possible triplets and validate the given conditions using absolute differences. ⏱️ Time Complexity: O(n³) 📦 Space Complexity: O(1) Strengthening understanding of nested loop patterns and condition-based filtering. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

Day 62 of #100DaysOfLeetCode 💻✅ Solved #219. Contains Duplicate II problem in Java. Approach: • Used two nested loops to compare elements within range k • For each element, checked next k elements • If any duplicate is found within distance k, returned true • If no such pair exists, returned false Key Learning: ✓ Practiced checking duplicates within a given range ✓ Strengthened understanding of array traversal with conditions ✓ Learned the importance of optimizing nested loop solutions Learning one problem every single day 🚀 #Java #LeetCode #DSA #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

🚀 Day 69/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2103. Rings and Rods Used a 2D boolean array to track presence of colors (R, G, B) on each rod and counted rods containing all three. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) Strengthening understanding of mapping, indexing, and state tracking techniques. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

Solved a problem where we need to check if two strings can be made equal using a special operation. The rule is: you can swap characters only if the distance between their positions is even. So basically, characters at even indices can only swap among themselves, and same for odd indices. Idea: Instead of actually swapping, I just counted characters separately for even and odd positions in both strings. If both match, then it’s possible — otherwise not. Simple concept, but interesting twist! 😊 #LeetCode #DSA #Java #ProblemSolving #CodingJourney

🚀 Day 54/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1656. Design an Ordered Stream Used an array + pointer approach to store values by index and return consecutive chunks by moving a pointer forward whenever elements are available. ⏱️ Time Complexity: O(n) (amortized across all insert calls) 📦 Space Complexity: O(n) Strengthening understanding of design problems and pointer-based simulation techniques. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency