Binary Search Solution for Guess Number Higher or Lower

Day 13/100 – LeetCode Challenge 🚀 Problem: Binary Search Approach: Maintained left and right pointers Compared middle element with target Reduced the search space by half each step Time Complexity: O(log n) Space Complexity: O(1) Key takeaway: Binary search is the foundation for solving problems with logarithmic efficiency. #LeetCode #100DaysOfCode #DSA #Java #ProblemSolving #InterviewPrep #100DaysOfLeetCode

🚀 Day 13 of #100DaysOfCode Solved Remove Linked List Elements on LeetCode 🔗 🧠 Key insight: While traversing a linked list, careful pointer updates are needed—especially when the head node itself matches the value to be removed. ⚙️ Approach: 🔹Handle cases where the head contains the target value 🔹Traverse the list using a pointer 🔹Skip nodes whose value matches the given target by adjusting next pointers ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

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 #42/100 Days of Code🔥 Solved #LeetCode 61: 𝐑𝐨𝐭𝐚𝐭𝐞 𝐋𝐢𝐬𝐭 𝐭𝐨𝐝𝐚𝐲  Focused on understanding linked list manipulation, edge cases, and optimizing the approach to 𝐎(𝐧) 𝐭𝐢𝐦𝐞 & 𝐎(1) 𝐬𝐩𝐚𝐜𝐞. Key takeaways: Always handle 𝐤 % 𝐥𝐞𝐧𝐠𝐭𝐡 in rotation problems Turning the list into a circular linked list simplifies logic #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #Consistency #LearningJourney

Day 30/100 – LeetCode Challenge 🚀 Problem: Count Binary Substrings Approach: Counted consecutive groups of 0’s and 1’s For every adjacent pair of groups, added min(prevGroup, currGroup) to the result Avoided generating substrings explicitly Time Complexity: O(n) Space Complexity: O(1) Key takeaway: Many substring problems don’t require generating substrings — they can be solved by analyzing group patterns and transitions. #LeetCode #100DaysOfCode #DSA #Java #Strings #ProblemSolving

🚀 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 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 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 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