Max Consecutive 1s in Array Java Solution

🚀 Day 92/100 ✅ 𝐏𝐫𝐨𝐛𝐥𝐞𝐦: 𝐓𝐡𝐢𝐫𝐝 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐍𝐮𝐦𝐛𝐞𝐫 Today’s problem was about finding the third distinct maximum number in an array. If it doesn’t exist, we return the maximum number instead. 💡 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬: Maintain three variables to track top 3 distinct values Handle 𝐝𝐮𝐩𝐥𝐢𝐜𝐚𝐭𝐞𝐬carefully Use 𝐋𝐨𝐧𝐠.𝐌𝐈𝐍_𝐕𝐀𝐋𝐔𝐄 to avoid edge case issues Single pass solution → 𝐎(𝐧) 𝐭𝐢𝐦𝐞 𝐜𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Instead of sorting, we efficiently track max1, max2, and max3 while iterating through the array. This improves performance and avoids unnecessary computations. #Day92 #100DaysOfCode #LeetCode #Java #DSA #CodingJourney #ProblemSolving #TechGrowth

🚀𝐃𝐚𝐲 88/100 – 𝐀𝐝𝐝 𝐁𝐢𝐧𝐚𝐫𝐲 Today’s problem was Add Binary — a great exercise to understand how addition works at the binary level. 🔍 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠: Just like decimal addition, binary addition also uses a carry, but with base 2. 💡 𝐂𝐨𝐫𝐞 𝐈𝐝𝐞𝐚: Traverse both strings from right to left Add digits along with carry Append result and update carry  𝐖𝐡𝐲 𝐢𝐭 𝐰𝐨𝐫𝐤𝐬? Binary addition rules: 0 + 0 = 0 1 + 0 = 1 1 + 1 = 0 (carry 1) ⚡ 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Use two pointers (end of both strings) Add digits + carry Append (sum % 2) Update carry (sum / 2) Reverse the result ⏱️ 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(𝐦𝐚𝐱(𝐧, 𝐦)) 📦 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(𝐧) #Day88 #100DaysOfCode #Java #DSA #LeetCode #Strings #CodingJourney

🚀Day 2/100 – LeetCode Journey Today’s problem: Two Sum II (Sorted Array) 💡Approach (Two Pointer Method) Since the array is sorted, I used two pointers: One at the start (i) One at the end (j) 👉Logic: If nums[i] + nums[j] == target → return answer ✅ If sum is greater → move j-- If sum is smaller → move i++ In this problem, we return 1-based index → so answer is (i + 1, j + 1) ⚡Time Complexity: O(n) → single pass using two pointers 🧠 Space Complexity: O(1) → no extra space used 🙌 Thanks to RAVI KUMAR Sir for guidance! . . #100DaysOfCode #LeetCode #DSA #Java

Many linked list problems are about pointer manipulation and maintaining order. 🚀 Day 113/365 — DSA Challenge Solved: Merge Two Sorted Lists Problem idea: We are given two sorted linked lists, and we need to merge them into a single sorted list. Efficient approach: Use a two-pointer technique to compare elements and build the merged list. Steps: 1. Create a dummy node to simplify result construction 2. Compare current nodes of both lists 3. Attach the smaller node to the result 4. Move the corresponding pointer forward 5. Once one list ends, attach the remaining part of the other list This ensures the final list remains sorted. ⏱ Time: O(n + m) 📦 Space: O(1) Day 113/365 complete. 💻 252 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #LinkedList #LeetCode #LearningInPublic

Day 66 — LeetCode Progress (Java) Problem: Find the Difference of Two Arrays Required: Given two integer arrays, return: Elements present in nums1 but not in nums2 Elements present in nums2 but not in nums1 Idea: Use sets to remove duplicates and quickly check membership. Approach: Convert both arrays into sets Iterate over nums1 set: Add elements not present in nums2 set to result1 Iterate over nums2 set: Add elements not present in nums1 set to result2 Return both lists Time Complexity: O(n + m) Space Complexity: O(n + m) #LeetCode #DSA #Java #HashSet #Arrays #ProblemSolving #CodingJourney

Linked list problems often become simpler when you break them into clear steps. 🚀 Day 111/365 — DSA Challenge Solved: Remove Nth Node From End of List Problem idea: We need to remove the nth node from the end of a linked list. Efficient approach: Convert the problem into finding the (size − n)th node from the start. Steps: 1. Traverse the list to calculate its size 2. If n equals size → remove the head 3. Otherwise, find the node just before the target 4. Update pointers to remove the node This simplifies the problem using basic traversal and pointer manipulation. ⏱ Time: O(n) 📦 Space: O(1) Day 111/365 complete. 💻 254 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #LinkedList #LeetCode #LearningInPublic

Day 69 - Remove Linked List Elements Working with linked lists to efficiently remove all nodes with a specific value using a dummy node approach. Time Complexity: O(n) Space Complexity: O(1) #Day69 #LeetCode #Java #LinkedList #DSA #ProblemSolving #CodingJourney

🚀 Day 21 of #128DaysOfCode 🔍 Key Learnings: Efficient searching in O(log n) Using low & high to narrow the range Finding correct position even if target is absent 🧠 Approach: Compare mid with target → move left/right → return index or insert position Consistency is key 🔥 #DSA #Java #CodingJourney #PlacementPreparation

💡 Day 48 of LeetCode Problem Solved! 🔧 🌟128. Longest Consecutive Sequence🌟 🔗 Solution Code: https://lnkd.in/gyN5ZJBr Task: Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence. You must write an algorithm that runs in O(n) time.   Example 1: Input: nums = [100,4,200,1,3,2] Output: 4 Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4. Example 2: Input: nums = [0,3,7,2,5,8,4,6,0,1] Output: 9 Example 3: Input: nums = [1,0,1,2] Output: 3 #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfChallenge #CodingJourney #KeepGrowing

Day 63 - Odd Even Linked List Today’s problem was about rearranging nodes based on their positions. Approach: • Maintain two pointers → odd and even • Traverse and rearrange links in-place • Keep track of even head to attach later • Merge odd list with even list at the end Key insight: 👉 Focus on node positions, not node values. Time Complexity: O(n) Space Complexity: O(1) #Day63 #LeetCode #Java #LinkedList #CodingPractice #TechJourney #DSA