Java LeetCode Solution: Difference Between Element Sum and Digit Sum

🚀 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 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 47/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 448. Find All Numbers Disappeared in an Array Used an in-place marking technique by treating indices as a hash map and marking visited numbers as negative to identify missing elements. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) (excluding output list) Strengthening understanding of array manipulation and in-place hashing tricks. 💪 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 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

🚀 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 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 52/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 3174. Clear Digits Used a stack-like approach with StringBuilder to remove the closest non-digit character whenever a digit is encountered. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) Strengthening understanding of string processing and stack-based logic simulation. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🔥 Day 58/100 Today’s problem: Reverse Integer (LeetCode) 💡 What I learned: - How to reverse digits using modulo (%) and division (/) - Handling negative numbers automatically - Most important: dealing with 32-bit integer overflow without using long - Learned to check overflow before updating the result ⚡ Key Concept: Before doing "rev = rev * 10 + digit", always check: 👉 If value exceeds "Integer.MAX_VALUE" or "Integer.MIN_VALUE", return 0 🧠 Takeaway: Small problems can hide tricky edge cases. Overflow handling is the real challenge here! 💻 Language used: Java #Day58 #100DaysOfCode #Java #DSA #CodingJourney #LeetCode