Solving String Problem with ASCII Values and Sum Difference Logic

🚀 Day 56/100 Today’s problem: String Compression (In-place) 💡 Key Learnings: - Learned how to use Two Pointers (Read & Write) efficiently - Understood in-place modification without extra space - Handled edge cases like counts greater than 9 (e.g., 12 → '1','2') - Improved thinking around grouping consecutive characters 🧠 Approach: Used a read pointer to traverse the array and a write pointer to update the compressed result directly in the same array. ⚡ Time Complexity: O(n) 📦 Space Complexity: O(1) Consistency is the real game 💯 #Day56 #Java #DSA #CodingJourney #LeetCode #ProblemSolving #100DaysOfCode

🚀 Day 64/100 Today’s problem: Find all strings that are substrings of another word 🧠 What I learned: - How to compare strings using nested loops - Using ".contains()" to check substrings efficiently - Importance of breaking early to optimize performance - Strengthening problem-solving with brute-force approach 💡 Key Insight: Sometimes simple solutions (O(n²)) are enough when constraints are small. No need to overcomplicate! 🔁 Consistency > Perfection #Day64 #DSA #Java #CodingJourney #Consistency #KeepLearning #100DaysOfCode

🚀 Day 46/100 Today I worked on an interesting problem: Shuffle String 💡 Problem Understanding: Given a string and an indices array, the task is to rearrange the characters so that each character moves to its specified position, forming a new string. 🧠 Approach: Used a simple mapping idea — place each character directly at its correct index. ✨ Key Learnings: - Mapping-based problems become easy with a clear approach - Direct placement helps keep the solution efficient - Avoiding unnecessary loops improves performance 📈 Learning something new every day and getting better step by step! #Day46 #DSA #CodingJourney #ProblemSolving #Java #100DaysOfCode

🚀 Day 67 of #LeetCode Challenge ✅ Problem Solved: Check If Two String Arrays are Equivalent 💡 What I learned today: • Learned how to compare two string arrays without joining them • Understood how to traverse multiple strings using pointers • Improved handling of indices across arrays and strings • Realized the importance of edge cases to avoid runtime errors 🧠 Approach: • Used four pointers to track positions in both arrays and strings • Compared characters one by one • Moved to the next string when current string ends • Ensured both arrays are fully traversed at the end 📊 Key Takeaway: Efficient solutions avoid extra space — comparing character by character is better than building new strings 🔥 Consistency + small improvements every day = big progress #Day67 #LeetCode #CodingJourney #DSA #Java #ProblemSolving #Consistency

Day 33/50 🚀 — Valid Palindrome (Two Pointer Approach) Today’s problem was a great mix of string manipulation + two pointers. 🔹 Ignored non-alphanumeric characters 🔹 Handled case-insensitivity 🔹 Compared characters from both ends efficiently Key insight: Instead of preprocessing the string, we can optimize in-place using two pointers, skipping unwanted characters on the go. 💡 This improves both readability and performance. Performance: ⚡ Runtime: 2 ms (99%+) 📦 Memory: Efficient #Day33 #LeetCode #TwoPointers #DSA #Java #CodingJourney #50DaysOfCode #ProblemSolving

🚀 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

𝐃𝐚𝐲 87/100 – 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 🚀 Problem: 228. 𝐒𝐮𝐦𝐦𝐚𝐫𝐲 𝐑𝐚𝐧𝐠𝐞𝐬  Today I solved a problem where we need to summarize consecutive numbers in a sorted unique array into ranges. 🔑 𝐈𝐝𝐞𝐚: Traverse the array and keep extending the range while consecutive numbers continue. Once the sequence breaks, close the range and store it. 💡 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Start with the first element as start Move forward while nums[i] + 1 == nums[i+1] If range exists → "start->end" Else → single number "start" 𝐊𝐞𝐲 𝐈𝐧𝐬𝐢𝐠𝐡𝐭: Efficient single pass solution (O(n)) by grouping consecutive elements on the fly. #LeetCode #Java #ProblemSolving #DSA #100DaysOfCode #CodingJourney

🚀 Day 27/100 Days of Code Challenge Today’s problem: Find Peak Element (Leetcode 162) 🔍 What I learned: How to find a peak element efficiently without scanning the entire array Using Binary Search to reduce time complexity from O(n) to O(log n) Understanding how the “slope” of elements helps decide the search direction 🧠 Key Idea: Instead of checking every element, compare the middle element with its neighbor: If nums[mid] < nums[mid + 1] → move right Else → move left ✅ Example: Input: [1, 2, 3, 1] Output: 2 (index of peak element 3) Consistency is key 🔑 — improving problem-solving skills one day at a time! 💪 #Day27 #100DaysOfCode #LeetCode #BinarySearch #DSA #Java #CodingJourney

🚀 Day 54 of my #100DaysOfCode Journey Today, I solved LeetCode – Sort Array By Parity II Problem Insight: Rearrange the array such that even-indexed positions have even numbers and odd-indexed positions have odd numbers. Approach: • Created a new result array of same size • Used two pointers: evenplace = 0 and oddplace = 1 • Traversed the array once • Placed even numbers at even indices and odd numbers at odd indices • Incremented pointers by 2 to maintain correct positions Time Complexity: O(n) | Space Complexity: O(n) Key Takeaway: Using two pointers for index placement makes the solution clean, efficient, and avoids unnecessary swaps. #DSA #Java #LeetCode #100DaysOfCode #CodingJourney

LeetCode — Problem 189 | Day 3 💡 Problem: Rotate Array Given an array, rotate it to the right by k steps. 🧠 My Approach: - Used reverse technique for in-place rotation - First reversed the entire array - Then reversed first k elements - Finally reversed remaining elements - Handled k using k = k % n This problem gave a good understanding of: ✔️ Array manipulation ✔️ In-place optimization (O(1) space) ✔️ Reverse logic #LeetCode #DSA #Java #CodingJourney