Reverse Vowels of a String in Java with Two Pointers and HashSet

🎉🎉 Day-38 of #60DaysOfDSA 🎉🎉 Problem: Row with Maximum 1s 👉 Problem Statement: Given a 2D binary matrix (rows sorted), find the index of the row having the maximum number of 1s. 👉 Approach Used: ✔️ Traverse each row ✔️ Find first occurrence of 1 ✔️ Count number of 1s → (m - j) ✔️ Track row with maximum count 👉Time Complexity: O(n * m) #geekstreak60 #dsa #java #problemsolving

🚀 Day 83 of #100DaysOfCode Solved 3217. Delete Nodes from Linked List Present in Array on LeetCode 🔗 🧠 Key Insight: We need to remove all nodes from the linked list whose values exist in a given array 👉 Fast lookup is required → use a HashSet ⚙️ Approach (HashSet + Traversal): 1️⃣ Store all array elements in a HashSet 🔹 set.add(nums[i]) 2️⃣ Use a dummy node before head 🔹 Helps handle deletion at head easily 3️⃣ Traverse the list: 🔹 If node.next.val exists in set → delete it 👉 node.next = node.next.next 🔹 Else → move forward 4️⃣ Return dummy.next ⏱️ Time Complexity: O(n + m) 📦 Space Complexity: O(m) #100DaysOfCode #LeetCode #DSA #LinkedList #HashSet #TwoPointers #Java #InterviewPrep #CodingJourney

Day 69/75 — Reverse String Simple and clean two-pointer problem. Approach: • Use two pointers (start & end) • Swap characters • Move pointers inward Key idea: while (i < j) {  char temp = s[i];  s[i] = s[j];  s[j] = temp;  i++;  j--; } Time Complexity: O(n) Space Complexity: O(1) Basic problem, but strengthens in-place thinking and pointer control. 69/75 🔁 #Day69 #DSA #TwoPointers #Strings #Java #LeetCode

Day 41 Today’s problem: Anti-Diagonals of a Matrix The trick was noticing that elements on the same anti-diagonal have row + col = constant. Looped over all possible sums of indices and collected elements in order. Clean, simple, and efficient solution Example: Input: [[1,2],[3,4]] → Output: [1,2,3,4] Nice problem to see patterns in 2D arrays #geekstreak #Day41 #GeeksforGeeks #DSA #Java #ProblemSolving #CodingJourney #Consistency

Day 73/100 🚀 | #100DaysOfDSA Solved LeetCode 219 – Contains Duplicate II today. The problem was to check if there are two equal elements in the array such that their indices differ by at most k. Approach: Used a sliding window + HashSet. • Iterated through the array while maintaining a window of size k • For each element: Checked if it already exists in the set → if yes, duplicate found within range → return true Added the current element to the set • If the set size exceeds k, removed the element that goes out of the window (nums[i - k]) This ensures we only track elements within the allowed index distance. Time Complexity: O(n) Space Complexity: O(k) Key takeaway: When constraints involve a fixed index range (k), think of using a sliding window with HashSet to efficiently track elements. #100DaysOfDSA #LeetCode #DSA #Java #SlidingWindow #HashSet #ProblemSolving #Consistency

Day 71 - Delete Middle Node Handling linked list traversal to identify and remove the middle node efficiently. Approach: • Traverse once to count nodes • Find middle index using n/2 • Traverse again to reach previous node • Update pointers to remove middle Key Insight: Careful handling needed when list size is small (edge cases) Time Complexity: O(n) Space Complexity: O(1) #Day71 #LeetCode #Java #CodingPractice #TechJourney #DSA #LinkedList

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

Day 61 - Add Two Numbers Worked on adding two numbers represented as linked lists. Approach: • Traverse both lists simultaneously • Add corresponding digits along with carry • Store result in a new linked list • Continue until both lists and carry are processed Key insight: Treat the linked list like a number stored in reverse order. Time Complexity: O(max(n, m)) #Day61 #LeetCode #Java #LinkedList #CodingPractice #DSA

🚀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-32 of #60DaysOfDSA 🎉🎉 Problem: K-th Element of Two Sorted Arrays 👉 Problem Statement: Given two sorted arrays, find the element that would appear at the k-th position in the merged sorted array. 👉 Approach Used (Two Pointer / Merge Logic): ✔️ Use two pointers i and j for both arrays ✔️ Traverse like merge step of merge sort ✔️ Keep counting elements until count == k ✔️ Return the k-th element 👉 Time Complexity : O(n + m) #dsa #problemsolving #java #geekstreak60