Day 10- What I Learned In a Day(JAVA) In the real world, input is not inbuilt-the user has to provide it. Today, I learned how to take user input in Java and understood how the Scanner class works. I learned: ✔ Why Java does not automatically take input ✔ How System.in reads from the keyboard ✔ How nextLine() reads full user input ✔ How to create a Scanner object import java.util.Scanner; class ClassName { public static void main(String[] args) { Scanner sc = new Scanner(System.in); // input here sc.close(); } } and also using the new tool(VSCODE) practiced 👇 #Java #LearningJourney #CodingDaily #JavaDeveloper

Day 14/100 – LeetCode Challenge 🚀 Problem: #169 Majority Element   Difficulty: Easy   Language: Java   Approach: Sorting + Middle Element   Time Complexity: O(n log n)   Space Complexity: O(1) 🔍 Key Insight: The majority element appears **more than ⌊n / 2⌋ times** in the array. If we **sort the array**, the majority element must occupy the **middle position** because it appears more than half of the time. Therefore, the element at index **n/2** will always be the majority element. 🧠 Solution Brief: First sorted the array using `Arrays.sort()`. Since the majority element appears more than half of the array length, it will always be positioned at the middle index. Finally returned `nums[nums.length / 2]` as the majority element. 📌 What I Learned: Understanding problem constraints can simplify the solution significantly. Sometimes a simple observation (like majority occupying the middle after sorting) can avoid more complex implementations. #LeetCode #Day14 #100DaysOfCode #Java #DSA #Arrays #ProblemSolving #CodingJourney

🚀 Container With Most Water | Java | Two Pointer Approach I solved the “Container With Most Water” problem using an optimized Two Pointer technique in Java. The goal is to find two lines that together with the x-axis form a container, such that the container holds the maximum amount of water. 🧠 Approach: Start with two pointers at both ends of the array. Calculate the width between them. The height is determined by the smaller of the two values. Update the maximum area. Move the pointer pointing to the smaller height inward. This greedy strategy works because the area depends on both width and minimum height, and moving the smaller height gives a chance to find a larger area. ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) Practicing DSA problems daily to improve logical thinking and optimization skills. #Java #DSA #TwoPointer #ProblemSolving #CodingJourney #LeetCode #DataStructures

Day 33 of my DSA Journey Today I solved the problem “Reverse Vowels of a String” using Java. In this problem, the goal is to reverse only the vowels in a string while keeping all other characters in their original positions. To solve this efficiently, I used the Two Pointer approach. Key idea: • Convert the string into a character array. • Use two pointers (left and right). • Move the left pointer forward until a vowel is found. • Move the right pointer backward until a vowel is found. • Swap the vowels. • Continue until the pointers meet. This approach helps solve the problem in O(n) time complexity with O(1) extra space. Concepts practiced today: • Two Pointer Technique • String manipulation in Java • Character array operations Consistent practice is helping me strengthen my problem-solving skills step by step. #Day33 #DSA #Java #CodingJourney #ProblemSolving #LeetCode

🚀 Day 76 of #100DaysOfCode Today I solved a string manipulation problem: Clear Digits 🧠 Problem: Given a string, whenever a digit appears, remove the previous character. Digits act like a backspace operation. 💡 Key Insight: Instead of modifying the string directly (since Strings are immutable in Java), I used a StringBuilder, which works like a stack: If character → append (push) If digit → delete last character (pop) 📌 What I Learned: StringBuilder is powerful for string modifications Always handle edge cases (like digit at the start) Many string problems are actually stack problems in disguise ⏱ Complexity: Time: O(n) Space: O(n) #Java #DSA #LeetCode #CodingJourney #100DaysOfCode #ProblemSolving

🚀 Day 17 of #100DaysOfCode Solved the Concatenation of Consecutive Binary Numbers problem today using Bit Manipulation + Modulo Arithmetic in Java. Instead of converting numbers to binary strings, we shift the current result left by the number of bits in i and add i. We increase the bit count only when i is a power of 2 using: (i & (i - 1)) == 0 #Java #DSA #BitManipulation #LeetCode #CodingJourney #PlacementPrep

Day 3/50 | #50DaysOfCode 📍 Platform: LeetCode 💻 Language: Java ✅ 2520. Count the Digits That Divide a Number (Easy) Today’s problem focused on digit extraction and divisibility checks. It helped strengthen my understanding of number manipulation and conditional logic. 🔎 Approach: Extract each digit using modulus (%) Check if the digit divides the original number using modulo If divisible, increment the counter Continue until all digits are processed Return the final count 📌 Example: Input: num = 1248 Output: 4 Explanation: All digits (1, 2, 4, 8) divide 1248 evenly This problem improved my understanding of loops, digit handling, and divisibility logic in Java. #DSA #LeetCode #Java #CodingJourney #ProblemSolving #Consistency #LearningJourney #50DaysOfCode #LinkedIn

Day 3/50 | #50DaysOfCode 📍 Platform: LeetCode 💻 Language: Java ✅ 977. Squares of a Sorted Array (Easy) Today’s problem focused on array manipulation and sorting logic. It helped me understand how negative numbers affect order after squaring and how to maintain sorted order efficiently. 🔎 Approach: Traverse the array and calculate the square of each element Store the squared values in a new array Sort the new array in non-decreasing order Return the sorted squared array 📌 Example: Input: nums = [-4, -1, 0, 3, 10] Output: [0, 1, 9, 16, 100] This problem strengthened my understanding of arrays, sorting, and handling negative values in Java. #DSA #LeetCode #Java #CodingJourney #ProblemSolving #Consistency #LearningJourney #50DaysOfCode #LinkedIn

Day 24 of #100DaysOfLeetCode 💻✅ Solved #35. Search Insert Position on LeetCode using Java. Approach: • Applied Binary Search to achieve O(log n) time complexity • Initialized two pointers: left and right • Calculated mid using left + (right - left) / 2 to avoid overflow • Compared mid element with target to adjust search space • If target not found, returned left as the correct insert position Performance: ✓ Runtime: 0 ms (Beats 100% submissions) ✓ Memory: 43.2 MB Key Learning: ✓ Strengthened understanding of Binary Search pattern ✓ Learned how to determine insert position when element is absent ✓ Improved confidence in handling sorted array problems efficiently Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #ProblemSolving #CodingJourney #100DaysOfCode

LeetCode Problem || Find Unique Binary String (1980)🚀 Today I solved the problem "Find Unique Binary String" using Java. 🔹 Problem: We are given an array of n binary strings, each of length n. The goal is to return a binary string of length n that does not exist in the array. 🔹 Approach (Diagonal Flip Technique): The idea is simple but powerful: Traverse the array using index i. Look at the i-th character of the i-th string (nums[i][i]). Flip the bit (0 → 1, 1 → 0). Append it to a result string. 💡 Time Complexity: O(n) Practicing problems like this strengthens logical thinking and problem-solving skills. #LeetCode #Java #CodingPractice #ProblemSolving #DSA