Non-Overlapping Intervals Java Solution

Day 12/100 – LeetCode Challenge Problem: Middle of the Linked List Today’s problem focused on finding the middle node of a singly linked list. Approach: Used the Two-Pointer Technique (Slow & Fast pointers). slow moves one step at a time fast moves two steps at a time When fast reaches the end of the list, slow will be at the middle node Complexity: Time: O(n) Space: O(1) Concepts Practiced: Linked List traversal Two-pointer technique Efficient single-pass solution #100DaysOfCode #LeetCode #DSA #Java #LinkedList #ProblemSolving #CodingJourney

Day 52/100 Completed ✅ 🚀 Solved LeetCode – Merge Intervals (Java) ⚡ Implemented an interval merging approach by first sorting intervals based on their start times and then traversing them to combine overlapping ranges. 🧠 Used comparison logic to detect overlaps and update interval boundaries, ensuring that all overlapping intervals are merged efficiently. This approach avoids redundant checks and processes the intervals in a single pass after sorting. 💯 Strengthened understanding of interval problems, sorting strategies, and greedy traversal techniques commonly used in real-world scheduling and timeline merging scenarios. Profile: https://lnkd.in/gaJmKdrA #LeetCode #100DaysOfCode #Java #DSA #CodingPractice #ProblemSolving

Day 7 – #50DaysLeetCodeChallenge Today, I solved the Largest Number problem. Problem Statement: Given a list of non-negative integers, arrange them such that they form the largest possible number. Since the result can be very large, return it as a string. Example: - [10, 2] → "210" - [3, 30, 34, 5, 9] → "9534330" Approach I Used – Custom Sorting: This problem is not about normal sorting; it’s about deciding the correct order of numbers based on concatenation. 1. Convert all integers into strings. 2. Sort the array using a custom comparator. 3. For two numbers a and b, compare: "ab" vs "ba". If "ba" is larger, place b before a. 4. Join all strings to form the final result. #LeetCode #Java #DSA #CodingChallenge #ProblemSolving #SoftwareEngineering

Day 35 — LeetCode Progress (Java) Problem: Divide Players Into Teams of Equal Skill Required: Pair players so that each team has the same total skill. Return the sum of chemistry (product of paired skills) or -1 if not possible. Idea: Sort the array and use two pointers. If teams must have equal total skill, the smallest and largest values must pair together to maintain a constant sum. Approach: Sort the skill array. Set left = 0, right = n - 1. Compute target sum = skill[left] + skill[right]. While left < right: If skill[left] + skill[right] ≠ target, return -1. Add skill[left] × skill[right] to result. Move both pointers inward. Return total chemistry. Time Complexity: O(n log n) Space Complexity: O(1) (excluding sorting space) #LeetCode #DSA #Java #TwoPointers #Greedy #Sorting #CodingJourney #100DaysOfCode

🚀 Day 47 / 100 – LeetCode Challenge Solved Remove Nth Node From End of List (Linked List problem). 💡 Key Idea: Used the Two Pointer (Fast & Slow) technique. Move the fast pointer n steps ahead, then move both pointers together until fast reaches the end. This helps identify the node just before the one to remove. ⚡ Result: ✅ Runtime: 0 ms (Beats 100%) ✅ Testcases Passed: 208 / 208 📚 Concepts Practiced: Linked Lists Two Pointer Technique Dummy Node for edge cases Consistency > Perfection. One problem closer to the goal! 💪 #Day47 #100DaysOfCode #LeetCode #DSA #Java #CodingJourney #LinkedList

Day 22/30 – Remove Nth Node From End of List Solved a classic Linked List problem today. Approach: • Use Two Pointers (Fast & Slow) • Move fast pointer n+1 steps ahead • Move both pointers until fast reaches the end • Slow pointer will be right before the node to remove Time Complexity: O(n) Space Complexity: O(1) Key concept: Two-pointer technique with a dummy node. #30DaysOfDSA #Java #LinkedList #CodingChallenge

💻 LeetCode Practice – Day 15 Solved Valid Anagram (Problem 242) today. 📌 Problem: Given two strings s and t, return true if t is an anagram of s, otherwise return false. 🧠 Approach: • Check if both strings have the same length. • Use a frequency array to count characters. • Increment for s and decrement for t. • If all values become zero, the strings are anagrams. #LeetCode #Java #DSA #CodingPractice

🚀 Day 38 of #100DaysOfCode 🌱 Topic: Linked List / Recursion ✅ Problem Solved: LeetCode 24 – Swap Nodes in Pairs 🛠 Approach: Used a recursive approach to swap every two adjacent nodes in the linked list. If the list has 0 or 1 node, return it directly (base case). Store the second node (head.next) as a temporary node. Recursively swap the remaining list starting from temp.next. Adjust pointers so the second node becomes the new head of the pair. Connect the swapped pair with the recursively processed list. This swaps nodes without modifying their values, only changing pointers. #100DaysOfCode #Day38 #DSA #LinkedList #Recursion #LeetCode #Java #ProblemSolving #CodingJourney #Consistency

Day 35 - Min Stack Designed a stack that supports push, pop, top, and retrieving the minimum element in constant time. Used an auxiliary stack to track the minimum value at each stage, ensuring efficient updates during push and pop operations. Time Complexity: O(1) for all operations #Day35 #LeetCode #Java #Stack #DSA #ProblemSolving #CodingJourney

Solved LeetCode 2839 – Check if Strings Can be Made Equal With Operations I today. In simple terms, we are given two strings and allowed to swap characters only at positions that are 2 indices apart (like index 0 with 2, 1 with 3). So instead of trying all swaps, the idea is to separate characters at even and odd positions and check if both strings have the same characters in those positions. If the frequency of characters matches for even and odd indices separately, then the strings can be made equal. A simple yet interesting problem that improves thinking around patterns and constraints. #LeetCode #DSA #Java #ProblemSolving #CodingJourney