Valid Number Parsing in Java: A Logic-Based Approach

💡 Day 31/100 Days Today’s problem was “Valid Palindrome” 🪞 The challenge was to check if a string reads the same forward and backward after removing all non-alphanumeric characters and ignoring case differences. This helped strengthen my understanding of string manipulation and the two-pointer technique in Java. 🧠 Key Features: Cleaned the string using regex to remove unwanted characters. Applied two-pointer logic to compare characters from both ends. Simple, efficient, and elegant — a great example of how logic matters more than length of code. ⚙️ Time Complexity: O(n) → Each character checked once. 💾 Space Complexity: O(n) → For the cleaned string (or O(1) in in-place approach). ✨ Takeaway: This problem reinforced how small optimizations — like using pointers instead of extra data structures — make algorithms more efficient. Every problem is a step closer to cleaner and smarter code! #Day31Of100 #DSA #Java #ProblemSolving #TwoPointer #CodingJourney #100DaysOfCode #LearningEveryday

Day 47/100 – #100DaysOfCode 🚀 | #Java #Arrays #InPlaceAlgorithms ✅ Problem Solved: First Missing Positive (LeetCode 41) 🧩 Problem Summary: Given an unsorted array, find the smallest missing positive integer. The solution must run in O(n) time and constant extra space. 💡 Approach Used: This is not a direct sorting or hashing problem — the key is index positioning. Logic: The answer must lie within 1 to n+1. Place each positive number x at index x-1 if possible. Then scan to find the first index where nums[i] != i+1. Steps: Iterate and swap values into their correct position. Ignore values ≤0 or >n. Final scan to find the first missing positive. ⚙️ Time Complexity: O(n) 📦 Space Complexity: O(1) ✨ Takeaway: This problem teaches how index-based hashing can eliminate the need for extra space — very common in interview-style optimization problems. #Java #LeetCode #Algorithm #Array #ProblemSolving #100DaysOfCode #CodingChallenge

Day 88 of #100DaysOfCode Solved Decrypt String from Alphabet to Integer Mapping in Java 🔡 Approach Today's challenge involved mapping a string of digits to lowercase English characters, where single digits '1' through '9' map to 'a' through 'i', and two digits followed by a '#' (e.g., '10#', '11#') map to 'j' through 'z'. I used an iterative approach with a while loop and an index i to traverse the input string. I checked ahead to see if a two-digit mapping existed by looking for a '#' at s.charAt(i + 2). If a '#' was found, I parsed the two-digit number (e.g., "10", "11") using s.substring(i, i + 2), converted it to an integer, and then mapped it to the correct character by adding it to the ASCII value of 'a' and subtracting one. I then advanced the index i by 3. If no '#' was found, it meant the current character s.charAt(i) was a single-digit mapping. I converted this digit to an integer and mapped it to a character similarly, then advanced the index i by 1. This approach processes the string from left to right, correctly handling the two-digit encodings first, which resulted in a quick runtime that beat 79.30% of other submissions. #Java #100DaysOfCode #LeetCode #CodingChallenge #Algorithms #StringManipulation #ProblemSolving

💻 Day 75 of #100DaysOfCodingChallenge ✨ Problem: 560. Subarray Sum Equals K 📚 Category: Array Manipulation | Subarray Problems 💡 Approach: Brute Force 🧠 Language: Java 🔍 Problem Understanding: We’re given an integer array nums and an integer k. We need to count how many subarrays (continuous parts of the array) have a sum equal to k. Example: Input: nums = [1, 2, 3], k = 3 Output: 2 Explanation: [1,2] and [3] are the subarrays with sum 3. 🧠 Brute Force Approach: In the brute-force method, we check every possible subarray using two loops: 1️⃣ Start from each element as a subarray beginning. 2️⃣ Keep adding elements until the end, and check if the sum equals k. 3️⃣ If yes, increment the count. Although it’s not the most optimized, this approach helps build a strong foundation in understanding subarray logic. ⚙️ Complexity: Time Complexity: O(n²) Space Complexity: O(1) 🌱 Learning: This problem taught me how to approach subarray-based questions step-by-step — starting from brute force before moving toward optimized solutions like prefix sums + hashmaps. Building clarity in fundamentals always makes optimization easier later 🚀 #Day75 #100DaysOfCode #LeetCode #Java #ProblemSolving #CodingJourney #WomenInTech #DSA #ArrayManipulation

💻 Day 4 of #100DaysOfCode Challenge Topic: Binary Tree Traversals 🌳 Problems Solved: 🔹 94. Binary Tree Inorder Traversal 🔹 144. Binary Tree Preorder Traversal 🔹 145. Binary Tree Postorder Traversal Concept Recap: Today, I explored the three fundamental depth-first traversal techniques used in binary trees: ✅ Inorder (Left → Root → Right) – Produces a sorted order for BSTs. ✅ Preorder (Root → Left → Right) – Useful for creating a copy of the tree or serialization. ✅ Postorder (Left → Right → Root) – Ideal for deleting trees or evaluating expressions. Each traversal follows a recursive approach to explore nodes systematically. Implementing them helped me strengthen my understanding of recursion and how stack frames manage function calls behind the scenes. Key Learnings: 🧠 Understood how traversal order impacts output sequence. ⚙️ Practiced recursive depth-first traversal logic in Java. 🌱 Improved code readability by modularizing recursive functions. #LeetCode #DataStructures #BinaryTree #Recursion #Java #CodingChallenge #100DaysChallenge #Day4

✅Day 42 : Leetcode 154 - Find Minimum in Rotated Sorted Array-2 #60DayOfLeetcodeChallenge 🧩 Problem Statement You are given an array nums that is sorted in ascending order and then rotated between 1 and n times. The array may contain duplicates. Your task is to find and return the minimum element in the rotated sorted array. You must minimize the number of overall operations as much as possible. 💡 My Approach I used a modified binary search technique to handle both rotation and duplicates. Initialize two pointers — low = 0 and high = n - 1. Calculate the middle index mid = low + (high - low) / 2. Update the answer as ans = min(ans, nums[mid]). If nums[low] == nums[mid] && nums[mid] == nums[high], move both low++ and high-- to skip duplicates. If the left half is sorted (nums[low] <= nums[mid]), update the answer and move to the right half (low = mid + 1). Otherwise, move to the left half (high = mid - 1). Continue until low > high. This efficiently finds the minimum even when duplicates exist. ⏱️ Time Complexity Worst Case: O(n) — when many duplicates exist. Average Case: O(log n) — behaves like binary search when duplicates are few. #BinarySearch #LeetCode #RotatedArray #Java #DSA #ProblemSolving #CodingPractice #LeetCodeHard

✅ Day 35 of #100DaysOfCode Challenge 📘 LeetCode Problem 112: Path Sum 🧩 Problem Statement:  Given the root of a binary tree and an integer targetSum, return true if there exists a root-to-leaf path whose sum of node values equals targetSum. Example: Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22   Output: true Explanation:  Path 5 → 4 → 11 → 2 gives the sum = 22 ✅ 💡 Simple Approach: If the tree is empty → return false If it’s a leaf node → check if its value equals targetSum Otherwise → subtract node value and check left and right recursively 💻 Easiest Java Code: class Solution {   public boolean hasPathSum(TreeNode root, int sum) {     if (root == null) return false;     if (root.left == null && root.right == null && root.val == sum)       return true;     sum = sum - root.val;     return hasPathSum(root.left, sum) || hasPathSum(root.right, sum);   } } ⚙ Complexity: ⏱ Time: O(n) → visit each node once 💾 Space: O(h) → recursion stack (h = height of tree) 🌿 Small code, big concept — recursion makes trees easy 🌱  #Day35 #100DaysOfCode #LeetCode #Java #DSA #BinaryTree #Recursion #CodingChallenge

💻 Day 53 of #100DaysOfCode 💻 Today, I solved LeetCode Problem #709 – To Lower Case 🔠 This problem focuses on string manipulation and character encoding (ASCII values) — one of the simplest yet fundamental concepts in text processing. 💡 Key Learnings: Reinforced understanding of ASCII value ranges for uppercase and lowercase alphabets. Practiced efficient string traversal and conditional conversion using StringBuilder in Java. Time complexity: O(n) — iterates through the string once. Space complexity: O(n) — for the output string. 💻 Language: Java ⚡ Runtime: 0 ms — Beats 100% 🚀 📉 Memory: 41.78 MB — Beats 43.48% 🧠 Approach: 1️⃣ Iterate through each character in the string. 2️⃣ Check if it’s an uppercase letter (A–Z). 3️⃣ Convert it to lowercase by adding 32 (based on ASCII values). 4️⃣ Append to the final string using StringBuilder. ✨ Takeaway: Sometimes, understanding the basics like ASCII values can help you write your own efficient methods — without relying on built-in functions! #100DaysOfCode #LeetCode #Java #CodingChallenge #ProblemSolving #Algorithms #StringManipulation #SoftwareEngineering #LearningEveryday #CleanCode

✅ Day 35 of #100DaysOfCode Challenge 📘 LeetCode Problem 112: Path Sum 🧩 Problem Statement: Given the root of a binary tree and an integer targetSum, return true if there exists a root-to-leaf path whose sum of node values equals targetSum. Example: Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true Explanation: Path 5 → 4 → 11 → 2 gives the sum = 22 ✅ 💡 Simple Approach: If the tree is empty → return false If it’s a leaf node → check if its value equals targetSum Otherwise → subtract node value and check left and right recursively 💻 Easiest Java Code: class Solution { public boolean hasPathSum(TreeNode root, int sum) { if (root == null) return false; if (root.left == null && root.right == null && root.val == sum) return true; sum = sum - root.val; return hasPathSum(root.left, sum) || hasPathSum(root.right, sum); } } ⚙️ Complexity: ⏱️ Time: O(n) → visit each node once 💾 Space: O(h) → recursion stack (h = height of tree) 🌿 Small code, big concept — recursion makes trees easy 🌱 #Day35 #100DaysOfCode #LeetCode #Java #DSA #BinaryTree #Recursion #CodingChallenge

✅ Day 35 of #100DaysOfCode Challenge 📘 LeetCode Problem 112: Path Sum 🧩 Problem Statement: Given the root of a binary tree and an integer targetSum, return true if there exists a root-to-leaf path whose sum of node values equals targetSum. Example: Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true Explanation: Path 5 → 4 → 11 → 2 gives the sum = 22 ✅ 💡 Simple Approach: If the tree is empty → return false If it’s a leaf node → check if its value equals targetSum Otherwise → subtract node value and check left and right recursively 💻 Easiest Java Code: class Solution { public boolean hasPathSum(TreeNode root, int sum) { if (root == null) return false; if (root.left == null && root.right == null && root.val == sum) return true; sum = sum - root.val; return hasPathSum(root.left, sum) || hasPathSum(root.right, sum); } } ⚙️ Complexity: ⏱️ Time: O(n) → visit each node once 💾 Space: O(h) → recursion stack (h = height of tree) 🌿 Small code, big concept — recursion makes trees easy 🌱 #Day35 #100DaysOfCode #LeetCode #Java #DSA #BinaryTree #Recursion #CodingChallenge