🚀 LeetCode #344 — Reverse String (Two Pointer Approach) Solved this classic problem today! The task is simple: reverse a string in-place using O(1) extra memory — perfect for practicing the two-pointer technique. 💡 Idea: Use two pointers — one at the start and one at the end. Swap characters while moving both pointers toward the center. #LeetCode #Java #TwoPointer #ProblemSolving #CodingJourney
Reversed a string in-place with two pointers
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💡 LeetCode #344 — Reverse String Today I solved LeetCode Problem 344: Reverse String 🔁 Problem Summary: Write a function that reverses a string. The input string is given as a character array s, and you must reverse it in-place (without using extra space). Key Idea: Use the two-pointer technique 👈👉 Initialize one pointer at the start (left) and another at the end (right). Swap the characters at both pointers and move them toward the center. #LeetCode #Java #DSA #TwoPointers #CodingJourney #ProblemSolving
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Day 54/100 – #100DaysOfCode 🚀 | #Java #SlidingWindow #TwoPointers ✅ Problem Solved: Count the Number of Substrings With Dominant Ones (LeetCode) 🧩 Problem Summary: You’re given a binary string. A substring is dominant if: number of 1s > number of 0s × k (where k is given). Return the total count of such substrings. 💡 Approach Used: Used an optimized Sliding Window + Two Pointers approach: Traverse with a right pointer. Adjust the left pointer whenever the substring stops being dominant. All valid windows contribute (left + 1) substrings. This avoids brute-force O(N²) and makes the solution efficient. ⚙️ Time Complexity: O(N) 📦 Space Complexity: O(1) ✨ Takeaway: Sliding window transforms heavy substring problems into clean linear-time solutions. #Java #LeetCode #SlidingWindow #TwoPointers #ProblemSolving #CodingChallenge #100DaysOfCode
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🚀 Day 57/100 of #100DaysOfLeetCode Today’s problem was 1323. Maximum 69 Number 💡 The task was to maximize a given positive integer (consisting only of digits 6 and 9) by changing at most one digit — specifically, changing a 6 to a 9 to get the largest possible number. 🔑 Key Takeaways: Learned how to efficiently manipulate numbers as strings. #LeetCode #Day57 #Java #100DaysOfCode #ProblemSolving #CodingChallenge #LearningEveryday
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#100DaysOfCode – Day 67 Reverse Words in a String Problem: – Reverse Words in a String Task: – Given a string, reverse the order of the words and remove extra spaces. Example: Input: s = " the sky is blue " → Output: "blue is sky the" My Approach: Used trim() to remove leading and trailing spaces. Split the string using split("\\s+") to handle multiple spaces. Reversed the array and joined the words with a single space. Time Complexity: O(N) | Space Complexity: O(N) Even simple string problems can teach the importance of clean code and efficient use of built-in methods. #takeUforward #100DaysOfCode #Java #ProblemSolving #LeetCode #GeeksForGeeks #CodeNewbie #StringManipulation
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💻 Day 45 of #LeetCode Journey 🚀 ✅ Problem: Count and Say 📘 Language: Java 🔹 Status: Accepted (30/30 test cases passed) 🔹 Runtime: 4 ms | Beats 55.38% 🔹 Memory: 41.86 MB 🧠 Concept: This problem is about generating the n-th term in the “count and say” sequence. Each term is built by describing the previous term — count the number of digits and say them in order. 🧩 Approach: Start with "1". For each iteration, use a StringBuilder to construct the next sequence. Track consecutive digits using a counter. Append the count and digit when the sequence changes. 💡 Efficient string manipulation and iteration give optimal performance. 🔥 Every solved problem builds confidence. One step closer to mastering patterns in strings! #Day45 #LeetCode #Java #CodingChallenge #ProblemSolving #CountAndSay #50DaysOfCode
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💡 LeetCode #58 — Length of Last Word Today I solved LeetCode Problem 58: Length of Last Word 🧠 Problem Summary: Given a string s consisting of words and spaces, return the length of the last word in the string. A word is defined as a sequence of non-space characters. Key Idea: We can traverse the string from end to start — Skip trailing spaces. Count the number of characters until the next space (which marks the last word). This avoids unnecessary splitting and saves extra space. #LeetCode #Java #DSA #Strings #ProblemSolving #CodingChallenge
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Day 40/100 – #100DaysOfCode 🚀 | #Java #LeetCode #TwoPointers #DynamicProgramming ✅ Problem Solved: Longest Palindromic Substring 🟣 🧩 Problem Summary: Given a string, return the longest palindromic substring within it. The substring must read the same forward and backward. 💡 Approach Used: Expand Around Center Technique Every palindrome is centered around: A single character (odd length) Or between two characters (even length) For each index: Expand left and right while characters match. Track the longest substring during expansions. This avoids unnecessary recomputation and works efficiently. ⚙️ Time Complexity: O(n²) 📦 Space Complexity: O(1) ✨ Takeaway: Sometimes, the cleanest approach comes from observing patterns — in this case, symmetry around a center 🎯 #Java #LeetCode #TwoPointers #ExpandAroundCenter #ProblemSolving #100DaysOfCode #CodingJourney
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Day 29/100 – #100DaysOfCode 🚀 | #Java #LeetCode #BinaryTree ✅ Problem Solved: Verify Preorder Serialization of a Binary Tree 🌲 🧩 Problem Summary: Given a string representing a preorder serialization of a binary tree, determine if it’s valid. Example: "9,3,4,#,#,1,#,#,2,#,6,#,#" → ✅ valid "1,#" → ❌ invalid 💡 Approach Used: Used the slot-counting method 🧠 Each node uses one slot and non-null nodes create two new slots. Process the preorder string, ensuring slots never go negative and exactly zero remain at the end. ⚙️ Time Complexity: O(n) 📦 Space Complexity: O(1) ✨ Takeaway: This problem teaches how binary tree structure can be validated with simple counting logic — no need to rebuild the tree! 🌱 #Java #LeetCode #BinaryTree #100DaysOfCode #ProblemSolving #CodingChallenge
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🎯 LeetCode 394 – Decode String Today I solved a really interesting Stack + String based problem! 💡 Problem Summary: We are given an encoded string containing patterns like 3[a2[c]] where: Numbers represent how many times the substring should repeat. Nested patterns are allowed. The goal is to decode the string correctly. 🧠 Approach: I used two stacks: One to store counts (how many times to repeat) One to store previously formed strings We iterate through the string: Build numbers when digits appear When [ appears, push state to stacks When ] appears, pop and reconstruct substring Else, just append characters normally ⏱️ Time Complexity: O(n) 💾 Space Complexity: O(n) #LeetCode #Java #DSA #Stack #StringManipulation #CodingJourney #LearnEveryday #100DaysOfCode 🚀
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🎯 Day 62 of #100DaysOfLeetCode Today I solved the “Merge Two Sorted Lists” problem using Java This problem is a great exercise in understanding how to work with Linked Lists and pointer manipulation. The goal is to merge two sorted linked lists into one sorted list — without using extra space for another list. Key Concepts Practiced: Linked List traversal Dummy node technique Efficient merging using pointers Handling null edge cases Approach Summary: Create a dummy node to simplify edge cases. Use two pointers to traverse both lists. Compare values and link nodes in sorted order. Attach any remaining nodes at the end. #100DaysOfCode #LeetCode #Java #CodingChallenge #DataStructures #LinkedList #ProblemSolving #SoftwareEngineering
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