Implement strStr() LeetCode Challenge

Day 12/100 – LeetCode Challenge 🚀 Problem: Guess Number Higher or Lower Approach: Applied Binary Search Narrowed the search space based on feedback from the API Time Complexity: O(log n) Space Complexity: O(1) Key takeaway: Whenever the problem gives “higher or lower” feedback, binary search should be your first instinct. #LeetCode #100DaysOfCode #DSA #Java #ProblemSolving #InterviewPrep #100DaysOfLeetCode

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 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 #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 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 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 13/100 – LeetCode Challenge 🚀 Solved Count Binary Substrings using an optimized O(n) approach by grouping consecutive characters and comparing segment lengths. Key takeaway: Instead of generating all substrings, identify patterns and reduce the problem to counting adjacent groups. Consistency > Motivation 💪 #100DaysOfCode #DSA #Java #LeetCode

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

🚀 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 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