100 Days of Code: Solved LeetCode 61 - Rotating a Linked List

Day 20/100 – LeetCode Challenge 🚀 Problem: Subsets (Power Set) Approach: Started with an empty subset For each number in the array, created new subsets by adding the number to all existing subsets Appended the newly formed subsets to the result list Time Complexity: O(n × 2ⁿ) Space Complexity: O(n × 2ⁿ) (to store all subsets) Key takeaway: Many combinatorial problems can be solved by building results incrementally, expanding previously generated subsets. #LeetCode #100DaysOfCode #DSA #Java #ProblemSolving #Backtracking #100DaysOfLeetCode

🚀 Day 18 of #100DaysOfCode Solved Remove Duplicates from Sorted List II on LeetCode 🔗 🧠 Key insight: When duplicates appear in a sorted linked list, we need to remove all occurrences, not just one. Using a dummy (sentinel) node makes it easier to handle cases where duplicates start at the head. ⚙️ Approach: 🔹Create a dummy node pointing to the head 🔹Traverse the list with two pointers 🔹If a duplicate sequence is found, skip the entire block 🔹Otherwise, move forward normally ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 36/100 – LeetCode Challenge 🚀 Problem: 3Sum Closest Approach: Sorted the array Fixed one element and applied the two-pointer technique Tracked the closest sum by comparing absolute differences Returned immediately if an exact match was found Time Complexity: O(n²) Space Complexity: O(1) Key takeaway: Many optimization problems are variations of classic patterns. Understanding 3Sum deeply makes solving 3Sum Closest straightforward. #LeetCode #100DaysOfCode #DSA #Java #TwoPointers #ProblemSolving #InterviewPrep

Day 31/100 – LeetCode Challenge 🚀 Problem: Roman to Integer Approach: Mapped Roman symbols to their integer values Traversed the string once If the current value was less than the next value, subtracted it Otherwise, added it to the total Time Complexity: O(n) Space Complexity: O(1) Key takeaway: When dealing with special formatting rules (like subtraction cases), comparing the current and next element often simplifies the logic. #LeetCode #100DaysOfCode #DSA #Java #Strings #ProblemSolving

🚀 Day 19 of #100DaysOfCode Solved 2. Add Two Numbers on LeetCode ➕🔗 (in-place linked list approach) 🧠 Key insight: Instead of creating a new list, we can reuse the longer linked list and perform addition in place, handling carry carefully across nodes. ⚙️ Approach: 🔹Calculate the size of both linked lists 🔹Use the longer list as the result list 🔹Traverse both lists, adding values with carry 🔹Continue processing remaining nodes of the longer list 🔹Append a new node if a carry remains at the end ⏱️ Time Complexity: O(max(n, m)) 📦 Space Complexity: O(1) (no extra list created) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 34/100 – LeetCode Challenge 🚀 Problem: Container With Most Water Approach: Used two pointers (left & right) Calculated area using min(height[left], height[right]) × width Moved the pointer pointing to the smaller height Updated maximum area during traversal Time Complexity: O(n) Space Complexity: O(1) Key takeaway: When maximizing area between two boundaries, moving the smaller boundary is the only way to potentially increase the result. #LeetCode #100DaysOfCode #DSA #Java #TwoPointers #ProblemSolving

🚀 Day 15 of #100DaysOfCode Solved Reverse Linked List on LeetCode 🔄🔗 🧠 Key insight: Reversing a linked list is all about careful pointer manipulation. By tracking prev, current, and next, we can reverse links in-place without extra memory. ⚙️ Approach: 🔹Initialize prev = null, current = head 🔹Store next node before breaking the link 🔹Reverse current.next to point to prev 🔹Move pointers forward until the list ends ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

🚀 Day 14 of #100DaysOfCode Solved Merge Two Sorted Lists on LeetCode 🔗🔀 🧠 Key insight: Using a dummy (sentinel) node simplifies pointer handling and avoids edge cases when building the merged list. ⚙️ Approach: 🔹Initialize a dummy node to act as the start of the merged list 🔹Compare nodes from both lists one by one 🔹Attach the smaller node and move the pointer forward 🔹Append remaining nodes once one list is exhausted ⏱️ Time Complexity: O(n + m) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 40 of 100 Days of Code 🚀 Solved Maximum Consecutive Ones today. 🔹 Problem: Find the longest streak of 1’s in a binary array. 🔹 Approach: Single pass with running counter. Key idea: - Increment count when we see 1 - Reset when we see 0 - Track maximum during traversal ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) Small problems strengthen logical thinking and loop control. #LeetCode #DSA #Java #100DaysOfCode #ProblemSolving

Day 18/100 – LeetCode Challenge 🚀 Problem: Merge Sorted Array Approach: Used three pointers starting from the end of the arrays Compared elements from nums1 and nums2 Placed the larger element at the last available position in nums1 Time Complexity: O(m + n) Space Complexity: O(1) (in-place) Key takeaway: When merging arrays in-place, working from the end avoids unnecessary shifting of elements. #LeetCode #100DaysOfCode #DSA #Java #ProblemSolving #InterviewPrep #100DaysOfLeetCode