Two Sum Problem Solved on LeetCode with Java

✅ DSA Day 8 / 100 Solved Reverse an Array on HackerRank using Java. Problem: Given an array, return the array in reverse order. Logic : - Use the built-in method Collections.reverse() -This method reverses the elements of the list directly -Then return the reversed list This problem helped me understand how arrays/lists can be manipulated easily using Java utility methods. #DSA #100DaysOfCode #Java #HackerRank #LearningInPublic #BeginnerDSA

✅ DSA Day 7 / 100 Solved Median of Two Sorted Arrays on LeetCode using Java. Logic : - Combine both sorted arrays into one array -Sort the merged array - Find the middle index - If the total length is odd → median is the middle element -If even → median is the average of the two middle elements #DSA #100DaysOfCode #Java #LeetCode #BeginnerDSA #LearningInPublic #Consistency

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 41 of #100DaysOfLeetCode 💻✅ Solved #242. Valid Anagram problem in Java. Approach: • Converted both strings to lowercase to ensure case-insensitive comparison • Converted the strings into character arrays • Sorted both arrays using Arrays.sort() • Compared the sorted arrays using Arrays.equals() • If both arrays match, the strings are anagrams Performance: ✓ Runtime: 11 ms (Beats 28.32% submissions) ✓ Memory: 46.83 MB (Beats 8.28% submissions) Key Learning: ✓ Practiced string manipulation and character array handling in Java ✓ Understood how sorting helps in comparing character frequencies ✓ Strengthened problem-solving skills for string-based problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #ProblemSolving #CodingJourney #100DaysOfCode

🚀 Day 2/30 – Java DSA Challenge 🔎 Problem 17: 169. Majority Element (LeetCode – Easy) Solved the Majority Element problem today 🔥 🧠 Problem Statement Given an array of size n, find the element that appears more than ⌊ n / 2 ⌋ times. It is guaranteed that a majority element always exists. 💡 Approach Used – HashMap (Frequency Counting) ✔ Traverse the array ✔ Store frequency using HashMap<Integer, Integer> ✔ Track the element with maximum frequency ✔ Return the element whose count > n/2 ⏱ Time Complexity: O(n) 📦 Space Complexity: O(n) 📌 Key Learning Strengthened HashMap usage Understood frequency-based problems Learned how to track maximum occurrence 🔥 17 Problems Completed Step-by-step improving logic and confidence 💪 #Day2 #30DaysOfCode #Java #DSA #LeetCode #HashMap #ProblemSolving #CodingJourney #Consistency

Day 27 of #100DaysOfLeetCode 💻✅ Solved #83. Remove Duplicates from Sorted List on LeetCode using Java. Approach: • Utilized the fact that the linked list is already sorted • Traversed the list using a single pointer • Compared current node value with next node value • If duplicate found, skipped the next node by updating links • Continued traversal until reaching the end of the list Performance: ✓ Runtime: 0 ms (Beats 100% submissions) ✓ Memory: 45.30 MB (Beats 85.70% submissions) Key Learning: ✓ Understood how sorting simplifies duplicate removal logic ✓ Strengthened pointer manipulation skills in linked lists ✓ Learned efficient in-place modification without extra space Learning one problem every single day 🚀 #Java #LeetCode #DSA #LinkedList #ProblemSolving #CodingJourney #100DaysOfCode

🚀 DSA in Java – Day 72 ✅ LeetCode Problem Solved: Subsets II 🔁 Concept Used: Backtracking + Recursion 🧠 Key Focus: Handling duplicate elements correctly Today I solved Subsets II, where the main challenge was generating all unique subsets from an array that may contain duplicates. 🔑 What I learned: Why sorting the array is important before recursion How to skip duplicates using this condition: if (idx > i && nums[idx] == nums[idx - 1]) continue; Proper backtracking steps (add → recurse → remove) How recursion builds subsets step by step Learning something new every day and getting better at problem-solving 💪 #DSA #Java #LeetCode #Backtracking #Recursion #ProblemSolving #CodingJourney #DailyLearning #Consistency

🚀Java practice - Day 87 Completed! 👍 Problem: Sum of Squares of Special Elements Language: Java Today’s problem was about identifying special elements in a 1-indexed array. An element is considered special if its index divides the length of the array (n % i == 0). The task was to calculate the sum of the squares of such elements.✨ #Day87 #Java #LeetCode #Arrays #ProblemSolving #DailyCoding #Consistency #100DaysOfCode

Day 36 of #100DaysOfLeetCode 💻✅ Solved #111. Minimum Depth of Binary Tree on LeetCode using Java. Approach: • Used recursive approach to calculate minimum depth • Returned 0 when root is null (base case) • If one subtree is null, avoided taking minimum of 0 (handled edge case carefully) • Returned 1 + minimum depth of valid subtree • Ensured correct handling of skewed trees Performance: ✓ Runtime: 6 ms ✓ Memory: 82.20 MB Key Learning: ✓ Understood difference between minimum depth and maximum depth logic ✓ Learned importance of handling null subtree cases correctly ✓ Improved confidence in solving binary tree recursion problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTree #Recursion #TreeProblems #ProblemSolving #CodingJourney #100DaysOfCode

Day 43 of #100DaysOfLeetCode 💻✅ Solved #100. Same Tree problem in Java. Approach: • Used recursion to compare both binary trees simultaneously • If both nodes are null, returned true • If one node is null and the other is not, returned false • Checked whether the values of both nodes are equal • Recursively compared the left subtrees and right subtrees Performance: ✓ Runtime: 0 ms (Beats 100% submissions) 🚀 ✓ Memory: 43.10 MB (Beats 30.15% submissions) Key Learning: ✓ Practiced recursive traversal of binary trees ✓ Learned how to compare two trees node by node ✓ Strengthened understanding of tree recursion logic Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTrees #ProblemSolving #CodingJourney #100DaysOfCode