"Day 75: Subarray Sum Equals K in Java with Brute Force"

💻 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 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 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 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

💡 LeetCode 3110 – Score of a String 💡 Today, I solved LeetCode Problem #3110: Score of a String, a neat little challenge that emphasizes understanding of ASCII values and absolute differences in Java strings. 🔠✨ 🧩 Problem Overview: You’re given a string s. The score of the string is defined as the sum of the absolute differences between the ASCII values of consecutive characters. Return the total score. 👉 Example: Input → "hello" Output → 13 Explanation → |'e'-'h'| + |'l'-'e'| + |'l'-'l'| + |'o'-'l'| = 3 + 7 + 0 + 3 = 13 💡 Approach: 1️⃣ Initialize a variable sum to store the total score. 2️⃣ Traverse the string from index 1 to end. 3️⃣ For each pair of consecutive characters, find their ASCII difference using Math.abs(). 4️⃣ Add the difference to sum and return it. ⚙️ Complexity Analysis: ✅ Time Complexity: O(n) — Single traversal of the string. ✅ Space Complexity: O(1) — Constant extra space. ✨ Key Takeaways: Strengthened understanding of character encoding and ASCII values. Reinforced clean looping logic and mathematical precision in Java. A great example of how simplicity can elegantly solve a real problem. 🌱 Reflection: Even the smallest coding challenges help sharpen attention to detail — from understanding data types to leveraging built-in functions effectively. Consistency is what transforms learning into mastery. 🚀 #LeetCode #3110 #Java #StringManipulation #ASCII #ProblemSolving #DSA #CodingJourney #CleanCode #AlgorithmicThinking #ConsistencyIsKey

🚀 Day 60 of #100DaysOfCode 🚀 Today, I solved LeetCode Problem #137 – Single Number II 🧩 📘 Problem Statement: Given an integer array where every element appears three times except for one that appears exactly once, find that single element and return it. The challenge: solve it with O(n) time and O(1) space complexity. 💻 Language: Java ⚡ Runtime: 0 ms — Beats 100.00% 📉 Memory: 45.54 MB — Beats 56.30% 🧠 Concept Used: 🔹 Bit Manipulation — A powerful yet tricky technique that leverages binary operations to track occurrences of bits efficiently. 🔹 Instead of using extra space or hash maps, we track bits that appear once and twice using two integer variables: ✅ once → bits that appeared once ✅ twice → bits that appeared twice The trick lies in updating them using XOR (^) and NOT (~) to "cancel out" bits appearing three times. 🧩 Approach Summary: 1️⃣ Initialize two variables once = 0 and twice = 0. 2️⃣ For every number in the array: - Update once and twice based on how many times a bit has appeared. 3️⃣ After processing all numbers, once holds the value of the single element. ✅ Complexity: Time: O(n) Space: O(1) ✨ Takeaway: This problem teaches how bitwise logic can replace traditional data structures, achieving the same result with constant space — a crucial optimization in system-level programming and interviews. #100DaysOfCode #LeetCode #Java #BitManipulation #ProblemSolving #CleanCode #DataStructures #Algorithms #TechLearning #CodingChallenge #SoftwareEngineering #InterviewPreparation

💡 #GFG160 Day 30 — Search in Rotated Sorted Array Today’s challenge focused on finding a target element in a sorted and rotated array — one of the most elegant uses of binary search. 🧩 Problem: Given a rotated sorted array, find the index of the target element. If it’s not present, return -1. Example: arr = [4,5,6,7,0,1,2], target = 0 → Output: 4 ⚙️ Approach / Intuition: Even after rotation, one half of the array is always sorted. So in each step: Find the mid element. Check which half (left or right) is sorted. Decide where to move next based on the target’s range. This approach eliminates half of the search space every iteration — pure binary search efficiency. 🧩 Dry Run (Example): arr = [4,5,6,7,0,1,2], target = 0 mid = 7 → left half sorted target not in left → move right mid = 1 → right half sorted target in left → move left mid = 0 → found ✅ ⏱ Time Complexity: O(log N) 💾 Space Complexity: O(1) 💻 Language: Java 📚 Key takeaway: Binary Search is not just for plain sorted arrays — the trick lies in identifying which half is sorted each time! #GeeksforGeeks #Java #BinarySearch #CodingChallenge #DSA #ProblemSolving #LearningEveryday

Day 38/100 – #100DaysOfCode 🚀 | #Java #LeetCode #StringParsing #Validation ✅ Problem Solved: Valid Number 🧩 Problem Summary: Given a string, determine whether it represents a valid number. This includes handling cases like: Integers Decimals Scientific notation (e / E) Sign symbols (+ / -) Trailing and leading spaces 💡 Approach Used: This is a string interpretation + state validation problem. I validated the number by checking: Only one decimal point allowed. Only one e or E, and it must split the number into valid components. Sign characters allowed only at the start or just after e/E. At least one digit must appear appropriately. Instead of using regex or parsing libraries, I implemented logical rules step-by-step to ensure strong input handling. ⚙️ Time Complexity: O(n) 📦 Space Complexity: O(1) ✨ Takeaway: Even simple inputs can require careful rule-based parsing — validating strings is more about logic than syntax. #Java #LeetCode #Strings #Parser #StateMachine #ProblemSolving #100DaysOfCode #CodingChallenge

Day 41 of #100DaysOfCode Problem: 3228. Maximum Number of Operations to Move Ones to the End Difficulty: Medium Language: Java Status: Solved Problem Summary: Given a binary string s, you can repeatedly choose an index i such that: s[i] == '1' and s[i + 1] == '0', and move that '1' to the right until it either reaches the end of the string or sits just before another '1'. You must find the maximum number of such operations possible. Key Idea: Every '0' that appears after a '1' contributes to new operations. Each '0' can “absorb” all previous '1's — since each of them can eventually be moved right past it. Hence, for every '0' that immediately follows a '1', we can add the total count of '1's so far to our result. Algorithm Steps: Initialize counters: ones = 0, res = 0. Traverse the string: When you see '1', increment ones. When you see '0' after a '1', add ones to res. Return res. Time Complexity: O(n) Space Complexity: O(1) Learning Takeaway: Great example of prefix accumulation logic — counting how many prior elements influence the current one. Efficiently transforms a greedy movement problem into a simple linear counting one. #Day41 #100DaysOfCode #LeetCode #StringManipulation #Greedy #Java #Algorithms #CodingChallenge #ProblemSolving #DSA

🚀 Day 29 of #100DaysOfCode – LeetCode Problem #643: Maximum Average Subarray I 🧩 Problem: Given an integer array nums and an integer k, find the contiguous subarray of length k that has the maximum average value. 💡 Approach: This problem is a great example of the Sliding Window Technique 🪟 Calculate the sum of the first k elements. Then slide the window by one element at a time — subtract the element that goes out, add the new one coming in. Keep track of the maximum sum seen. Return the maximum average (maxSum / k). 💻 Java Code: class Solution { public double findMaxAverage(int[] nums, int k) { double sum = 0; for (int i = 0; i < k; i++) { sum += nums[i]; } double max = sum; for (int i = k; i < nums.length; i++) { sum += nums[i] - nums[i - k]; max = Math.max(max, sum); } return max / k; } } ⏱️ Complexity: Time: O(n) Space: O(1) ✅ Result: Accepted (Runtime: 0 ms) This one reinforces how sliding window optimizations can drastically simplify what looks like a brute-force problem!