Exploring the Stack class in Java with key methods and examples.

Day 80 of #100DaysOfCode Solved Frequency Sort in Java 🔠 Approach Today's problem was to sort an array based on the frequency of its numbers. My initial solution was accepted, but it exposed a major efficiency gap! My strategy involved: Sorting the array first. Using nested loops to count the frequency of each number and store it in a HashMap. (Implied) Using the counts to perform the final custom sort. The core issue was the counting method: running a full loop inside another full loop to get the frequency. This quadratic $O(N^2)$ counting completely tanked the performance. ✅ Runtime: 29 ms (Beats 12.02%) ✅ Memory: 45.26 MB (Beats 5.86%) Another great lesson in algorithmic complexity! The difference between $O(N^2)$ and the optimal $O(N \log N)$ for this problem is massive. Time to refactor and implement the fast, single-pass $O(N)$ frequency count using getOrDefault! 💪 #Java #100DaysOfCode #LeetCode #CodingChallenge #Algorithms #Arrays #HashMap #Optimization #ProblemSolving

Day 28/100 – #100DaysOfCode 🚀 | #Java #LeetCode #DynamicProgramming ✅ Problem Solved: Scramble String 🔀 🧩 Problem Summary: Given two strings s1 and s2, determine whether one is a scrambled version of the other. A string is scrambled by recursively dividing it into two non-empty substrings and swapping them. 💡 Approach Used: Implemented Recursion + Memoization using a HashMap for overlapping subproblems. Used character frequency checks to prune unnecessary recursion calls. Used Java’s BiFunction with inline helper logic for recursion. ⚙️ Time Complexity: O(n⁴) (due to substring operations and recursion) 📦 Space Complexity: O(n²) ✨ Takeaway: Even complex recursive problems can be optimized efficiently with Memoization and early pruning. 🚀 #Java #LeetCode #DynamicProgramming #Recursion #ProblemSolving #100DaysOfCode #CodingChallenge

LeetCode Question #199 — Binary Tree Right Side View Thrilled to share another Accepted Solution (100% Runtime 🚀) on LeetCode! This problem focuses on Binary Trees — specifically, how to capture the view of a tree from its right side 🌳➡️. I implemented a Modified Pre-Order Traversal (Root → Right → Left) in Java, ensuring that the first node at each level (from the right) gets recorded. 💡 Key Idea: Use recursion with a level tracker — when the current level equals the list size, it means we’re seeing the rightmost node for the first time. Here’s the performance snapshot: ⚙️ Runtime: 0 ms (Beats 100% of Java submissions) 💾 Memory: 42.4 MB Every problem like this sharpens my understanding of tree traversal patterns and depth-first search optimization. #LeetCode #Java #DataStructures #BinaryTree #ProblemSolving #CodingJourney #DSA #100PercentRuntime

Today I worked on the problem "Convert Sorted List to Binary Search Tree" in Java — a perfect blend of Linked List and Tree logic! 🌳 🔍 Key Concepts Covered: Finding the middle element of a Linked List using the slow & fast pointer approach Recursively constructing a height-balanced BST Strengthening understanding of Divide and Conquer strategies This problem really sharpened my thinking around recursive structures and pointer management — small yet powerful steps toward writing more efficient and elegant Java code. 💻✨ 💬 What’s your favorite data structure to work with — Linked Lists or Trees? #LeetCode #Java #DataStructures #Algorithms #CodingJourney #LearningEveryday #ProblemSolving #SoftwareEngineering

🚀 Day 55 of #100DaysOfCode Challenge Problem: LeetCode #219 – Contains Duplicate II Language: Java ☕ Today I solved an interesting array problem that checks whether a duplicate element exists within a given distance k in the array. 💡 Logic: Use a HashMap to remember the last index of each number. If the same number appears again, check if the difference between indices ≤ k. If yes → return true, else keep checking. 💻 Code: import java.util.HashMap; public class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { if (map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) { return true; } map.put(nums[i], i); } return false; } } 🧠 Example: Input: [1,2,3,1], k = 3 Output: true ✅ (same number 1 appears within distance 3) ⚙️ Key Concepts: HashMap for quick lookup Difference check using indices Time Complexity → O(n) Space Complexity → O(n) 💬 Every day’s problem teaches me something new — today it was about using maps smartly to track elements efficiently. On to the next challenge! 💪 #Day55 #LeetCode #Java #100DaysOfCode #CodingJourney #ProblemSolving #HashMap #DSA

🚀 Day 55 of #100DaysOfCode Challenge Problem: LeetCode #219 – Contains Duplicate II Language: Java ☕ Today I solved an interesting array problem that checks whether a duplicate element exists within a given distance k in the array. 💡 Logic: Use a HashMap to remember the last index of each number. If the same number appears again, check if the difference between indices ≤ k. If yes → return true, else keep checking. 💻 Code: import java.util.HashMap; public class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { if (map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) { return true; } map.put(nums[i], i); } return false; } } 🧠 Example: Input: [1,2,3,1], k = 3 Output: true ✅ (same number 1 appears within distance 3) ⚙️ Key Concepts: HashMap for quick lookup Difference check using indices Time Complexity → O(n) Space Complexity → O(n) 💬 Every day’s problem teaches me something new — today it was about using maps smartly to track elements efficiently. On to the next challenge! 💪 #Day55 #LeetCode #Java #100DaysOfCode #CodingJourney #ProblemSolving #HashMap #DSA

🚀 Day 55 of #100DaysOfCode Challenge Problem: LeetCode #219 – Contains Duplicate II Language: Java ☕ Today I solved an interesting array problem that checks whether a duplicate element exists within a given distance k in the array. 💡 Logic: Use a HashMap to remember the last index of each number. If the same number appears again, check if the difference between indices ≤ k. If yes → return true, else keep checking. 💻 Code: import java.util.HashMap; public class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { if (map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) { return true; } map.put(nums[i], i); } return false; } } 🧠 Example: Input: [1,2,3,1], k = 3 Output: true ✅ (same number 1 appears within distance 3) ⚙️ Key Concepts: HashMap for quick lookup Difference check using indices Time Complexity → O(n) Space Complexity → O(n) 💬 Every day’s problem teaches me something new — today it was about using maps smartly to track elements efficiently. On to the next challenge! 💪 #Day55 #LeetCode #Java #100DaysOfCode #CodingJourney #ProblemSolving #HashMap #DSA

✅ Day 33 of #100DaysOfCode Challenge 📘 LeetCode Problem 110: Balanced Binary Tree 🧩 Problem Statement:  Given a binary tree, determine if it is height-balanced. A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than 1. Example: Input: root = [3,9,20,null,null,15,7] Output: true Explanation:  Both left and right subtrees of every node differ in height by at most 1. 💡 Approach (Simple & Efficient): Use a recursive function to calculate the height of each subtree. If any subtree is unbalanced, return -1. Otherwise, return the height of the tree. If the final result is -1 → tree is not balanced. 💻 Java Code: class Solution {   public boolean isBalanced(TreeNode root) {     return checkHeight(root) != -1;   }   private int checkHeight(TreeNode node) {     if (node == null) return 0;     int leftHeight = checkHeight(node.left);     if (leftHeight == -1) return -1;     int rightHeight = checkHeight(node.right);     if (rightHeight == -1) return -1;     if (Math.abs(leftHeight - rightHeight) > 1) return -1;     return 1 + Math.max(leftHeight, rightHeight);   } } ⚙ Complexity: ⏱ Time: O(n) → Each node visited once 💾 Space: O(h) → Recursion stack (h = tree height) 🌱 Another step towards mastering Binary Trees!  #Day33 #100DaysOfCode #LeetCode #Java #DSA #BinaryTree #BalancedTree #Recursion #CodingChallenge

Day 45/100 – #100DaysOfCode 🚀 | #Java #Hashing #SlidingWindow ✅ Problem Solved: Contains Duplicate III (LeetCode 220) 🧩 Problem Summary: Given an integer array and integers k and t, determine if there exist two distinct indices i and j such that: |i - j| ≤ k |nums[i] - nums[j]| ≤ t 💡 Approach Used: Used a Sliding Window + TreeSet to maintain numbers in a sorted structure. Steps: Traverse the array and maintain a window of at most size k. For each element x, find if there exists another element in the set such that |x - y| ≤ t. This is done using ceiling() to find the closest value ≥ x. Insert the element in TreeSet and remove the element that slides out. ⚙️ Time Complexity: O(n log k) 📦 Space Complexity: O(k) ✨ Takeaway: This problem teaches how ordered data structures like TreeSet help efficiently handle range queries in sliding window scenarios. #Java #TreeSet #SlidingWindow #LeetCode #ProblemSolving #100DaysOfCode #CodingChallenge

📌 Day 16/100 - Reverse String (LeetCode 344) 🔹 Problem: Reverse a given string in-place — meaning you must modify the original array of characters without using extra space. 🔹 Approach: Used the two-pointer technique — one starting at the beginning and one at the end of the array. Swap characters at both pointers, then move them closer until they meet. Efficient, clean, and runs in linear time without additional memory allocation. 🔹 Key Learnings: In-place algorithms optimize space complexity significantly. The two-pointer pattern is a versatile tool for many array and string problems. Understanding mutable vs immutable structures in Java is crucial for memory efficiency. Sometimes, the simplest logic beats the most complex one. 🧠 “True efficiency lies in simplicity, not complexity.” #100DaysOfCode #LeetCode #Java #ProblemSolving #DSA #CodingJourney #TwoPointers