Java Increment Operator Challenge

Most people solve this problem by checking every possible pair. It works… but it doesn’t scale. The key insight is simple: The container is always limited by the shorter line. Once you understand that, you don’t need to check everything. You can eliminate unnecessary pairs and reach an optimal solution in O(n). This is where the Two Pointer approach becomes powerful. Small shift in thinking → big improvement in efficiency. Swipe to see how it works → #DSA #CodingInterview #Java #TwoPointers #ProblemSolving #SoftwareEngineering #LeetCode

#Day95 of #100DaysOfCode Extended the previous number-to-words program to handle multiple digits. The program takes a number as input and prints each digit in its word form. Focused on improving logic by combining loops, string conversion, and conditional handling. #Java #MiniProject #100DaysOfCode

🚀 Day 21 of #128DaysOfCode 🔍 Key Learnings: Efficient searching in O(log n) Using low & high to narrow the range Finding correct position even if target is absent 🧠 Approach: Compare mid with target → move left/right → return index or insert position Consistency is key 🔥 #DSA #Java #CodingJourney #PlacementPreparation

🚀 𝐃𝐚𝐲 89/100 – 𝐈𝐬 𝐒𝐮𝐛𝐬𝐞𝐪𝐮𝐞𝐧𝐜𝐞 Today’s problem was 𝐈𝐬 𝐒𝐮𝐛𝐬𝐞𝐪𝐮𝐞𝐧𝐜𝐞 — a simple yet fundamental problem to understand two-pointer technique. 🔍 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠: We can efficiently check if one string is a subsequence of another using two pointers. 💡 𝐂𝐨𝐫𝐞 𝐈𝐝𝐞𝐚: Traverse both strings Move pointer of s only when characters match Always move pointer of t  𝐖𝐡𝐲 𝐢𝐭 𝐰𝐨𝐫𝐤𝐬? We try to match characters of s in order within t without disturbing sequence. ⚡ 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Initialize two pointers i (for s) and j (for t) If characters match → move both Else → move only j If i reaches end of s → subsequence exists ⏱️ 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(𝐧) 📦 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(1) #Day89 #100DaysOfCode #Java #DSA #LeetCode #TwoPointers #CodingJourney

🚀 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 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 42 of consistency 💻🔥 Solved Add Two Numbers using linked lists — a classic problem that really tests understanding of pointers, carry handling, and edge cases. Key takeaways: Handling carry efficiently is crucial Dummy nodes make list problems much cleaner Iterative approach keeps space optimal Not the fastest runtime yet, but improvement is a process — optimizing step by step. Consistency > Perfection. #Day42 #LeetCode #DSA #Java #CodingJourney #ProblemSolving #100DaysOfCode

Day 75 – Linked List Cycle Detection Solved a problem to detect whether a linked list contains a cycle using an efficient two-pointer approach. Key Learnings: Applied Floyd’s Cycle Detection Algorithm Understood how different pointer speeds help detect loops Achieved optimal solution with O(1) space and O(n) time complexity #DSA #Java #LinkedList #TwoPointers #ProblemSolving #CodingPractice

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 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