Merge Sorted Arrays with Two Pointer Technique

🚀 Day 50 / 100 | Median of Two Sorted Arrays Intuition: We are given two sorted arrays and need to find the median of the combined numbers. Since both arrays are already sorted, we can merge them in sorted order. Once we have the merged array, finding the median becomes simple. If the total number of elements is odd, the median is the middle element. If it's even, the median is the average of the two middle elements. Approach: Use two pointers to traverse both arrays. Compare the elements and insert the smaller one into a new array. Continue this process until all elements are merged. Finally, calculate the median based on the length of the merged array. Complexity: Time Complexity: O(n + m) Space Complexity: O(n + m) #100DaysOfCode #Java #DSA #LeetCode #ProblemSolving

Day 16/100 – LeetCode Challenge Problem: Merge Sorted Array Today’s problem involved merging two sorted arrays into one sorted array. Approach: Created a temporary array of size m + n Used two pointers to compare elements from both arrays Inserted the smaller element into the new array Copied remaining elements if any array still had values Finally copied the merged result back into nums1 Complexity: Time: O(m + n) Space: O(m + n) Concepts Practiced: Two-pointer technique Array traversal Merging sorted arrays #100DaysOfCode #LeetCode #DSA #Java #Arrays #ProblemSolving #CodingJourney

🚀 Day 39 of #100DaysOfCode Solved 80. Remove Duplicates from Sorted Array II on LeetCode 🔢 🧠 Key Insight: The array is already sorted, and we need to ensure that each element appears at most twice, modifying the array in-place. ⚙️ Approach: 🔹Maintain a pointer i representing the position to place the next valid element 🔹Start iterating from index 2 🔹For each element nums[j], compare it with nums[i] 🔹If they are different, place the element at nums[i + 2] and move the pointer forward This ensures that no element appears more than twice while maintaining the sorted order. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #Arrays #TwoPointers #Java #ProblemSolving #InterviewPrep #LearningInPublic

𝐃𝐚𝐲 𝟓𝟒 – 𝐃𝐒𝐀 𝐉𝐨𝐮𝐫𝐧𝐞𝐲 | 𝐀𝐫𝐫𝐚𝐲𝐬 🚀 Today’s problem focused on finding the minimum element in a rotated sorted array using binary search. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐒𝐨𝐥𝐯𝐞𝐝 • Find Minimum in Rotated Sorted Array 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 • Used binary search to reduce the search space • Compared the middle element with the rightmost element Logic: • If nums[mid] > nums[right] → minimum lies in the right half • Else → minimum lies in the left half (including mid) • Continued until left meets right 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬 • Rotated arrays require a modified binary search • Comparing with boundary elements helps identify the sorted half • Binary search is not only for exact matches • Reducing the search space is the core idea 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 • Time: O(log n) • Space: O(1) 𝐓𝐚𝐤𝐞𝐚𝐰𝐚𝐲 Binary search is about asking the right question — not just finding the middle. 54 days consistent 🚀 On to Day 55. #DSA #Arrays #BinarySearch #LeetCode #Java #ProblemSolving #DailyCoding #LearningInPublic #SoftwareDeveloper

🚀 Day 34 of #100DaysOfCode Solved 852. Peak Index in a Mountain Array on LeetCode ⛰️📈 🧠 Key Insight: A mountain array strictly increases to a peak and then decreases. Instead of scanning the whole array, we can use Binary Search to efficiently find the peak. ⚙️ Approach: 🔹Use binary search with two pointers left and right 🔹Compare arr[mid] with arr[mid + 1] 🔹If arr[mid] > arr[mid + 1] → peak is on the left side (including mid) 🔹Otherwise → peak is on the right side 🔹Continue until left == right, which gives the peak index ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #BinarySearch #Arrays #Java #ProblemSolving #InterviewPrep #LearningInPublic

🚀 Day 29 of #75daysofLeetCode 2095 – Delete the Middle Node of a Linked List Just solved an interesting linked list problem that perfectly demonstrates the power of the two-pointer technique (slow & fast pointers). 🔍 Problem Insight: Given a linked list, delete its middle node where the middle is defined as ⌊n/2⌋ (0-based indexing). 💡 Key Idea: Instead of calculating the length, we can efficiently find the middle using: 🐢 Slow pointer (1 step) ⚡ Fast pointer (2 steps) When the fast pointer reaches the end, the slow pointer will be at the middle node! 🛠 Approach: ✔ Handle edge case (single node → return null) ✔ Traverse using slow & fast pointers ✔ Keep track of previous node ✔ Remove the middle node in one pass ⏱ Time Complexity: O(n) 💾 Space Complexity: O(1) 🔥 Why this matters? This pattern is widely used in: Finding middle of linked list Detecting cycles Splitting lists Mastering this unlocks many problems! #LeetCode #DSA #LinkedList #Java #CodingInterview #ProblemSolving #TechLearning

🔥 Day 97 of #100DaysOfCode Today’s problem: LeetCode – Search in Rotated Sorted Array 🔄🔍 📌 Problem Summary You are given a rotated sorted array and a target. Your task is to return the index of the target if it exists; otherwise return -1. Example: Input: nums = [4,5,6,7,0,1,2], target = 0 Output: 4 🧠 Approach Used: Linear Search In this solution, we simply iterate through the array and check if the current element equals the target. ⚙️ Logic for(int i = 0; i < nums.length; i++){ if(nums[i] == target){ return i; } } return -1; 💡 Explanation Traverse the array from start to end If the target is found → return the index If the loop finishes → return -1 ⏱ Time Complexity: O(n) 💾 Space Complexity: O(1) 🚀 Performance Runtime: 0 ms Memory: 43.68 MB 🧠 Learning This problem can also be solved using Binary Search in O(log n) by identifying which half of the rotated array is sorted. But for smaller inputs or quick validation, linear scan works correctly and is simple to implement. Only 3 days left to complete the #100DaysOfCode challenge 🚀 Consistency is the real win here! On to Day 98 🔥 #100DaysOfCode #LeetCode #Java #DSA #CodingJourney #InterviewPrep

🚀 Day 38 of #100DaysOfCode Solved 154. Find Minimum in Rotated Sorted Array II on LeetCode 🔍 🧠 Key Insight: The array is sorted but rotated, and this version introduces duplicates, which makes the binary search logic slightly trickier. ⚙️ Approach: 🔹Use binary search with left and right pointers 🔹Compare nums[mid] with nums[right]: 🔹If nums[mid] > nums[right] → minimum lies in the right half 🔹If nums[mid] < nums[right] → minimum lies in the left half (including mid) 🔹If equal → safely shrink the search space by decrementing right This handles the ambiguity caused by duplicates. ⏱️ Time Complexity: Average: O(log n) Worst case: O(n) (when many duplicates exist) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #BinarySearch #Arrays #Java #ProblemSolving #InterviewPrep #LearningInPublic

🚀 Day 31 of #100DaysOfCode Solved 240. Search a 2D Matrix II on LeetCode 🔍📊 🧠 Key insight: In this matrix, each row is sorted left → right and each column is sorted top → bottom. Starting from the top-right corner lets us eliminate an entire row or column at each step. ⚙️ Approach: 🔹Start at top-right element 🔹If value equals target → return true 🔹If value is smaller → move down (increase row) 🔹If value is larger → move left (decrease column) 🔹Continue until the target is found or boundaries are crossed ⏱️ Time Complexity: O(m + n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #Matrix #TwoPointers #Java #ProblemSolving #InterviewPrep #LearningInPublic

🚀 Day 29 of #100DaysOfCode Solved 442. Find All Duplicates in an Array on LeetCode ✅ 🧠 Key Insight: Since numbers are in the range 1 → n, each number ideally belongs at index num - 1. By repeatedly swapping elements into their correct positions (cyclic sort), duplicates naturally reveal themselves. ⚙️ Approach Used: 🔹Place each element at its correct index (nums[i] → nums[nums[i] - 1]) 🔹Skip when the element is already in the correct position 🔹After rearrangement, any index i where nums[i] ≠ i + 1 is a duplicate ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) (excluding output list) #100DaysOfCode #LeetCode #DSA #Arrays #CyclicSort #Java #ProblemSolving #InterviewPrep #LearningInPublic