Incrementing Large Integer Array by One

🚀 Day 18/30– DSA Challenge 📌 LeetCode Problem – Search Insert Position 📝 Problem Statement Given a sorted array nums and a target value, return the index if found. If not found, return the index where it would be inserted in order. 📌 Example Input: nums = [1,3,5,6] target = 2 Output: 1 Explanation: 2 should be inserted at index 1. 💡 Key Insight This is a binary search variation. 👉 If element exists → return index 👉 If not → return the position where it should be And that position is exactly where left ends after the loop. 🚀 Algorithm 1️⃣ Initialize: left = 0 right = n - 1 2️⃣ While left <= right: Find mid If equal → return mid If smaller → move left Else → move right 3️⃣ If not found → return left ✅ Java Code (Optimal O(log n)) class Solution { public int searchInsert(int[] nums, int target) { int left = 0; int right = nums.length - 1; while (left <= right) { int mid = left + (right - left) / 2; if (nums[mid] == target) { return mid; } else if (nums[mid] < target) { left = mid + 1; } else { right = mid - 1; } } return left; } } ⏱ Complexity Time Complexity: O(log n) Space Complexity: O(1) 📊 Performance ⚡ Runtime: 0 ms (Beats 100%) 💾 Memory: ~44 MB 📚 Key Learnings – Day 18 ✔ Binary Search is more than just finding elements ✔ “Insert position” problems are common variations ✔ Final value of left is very important ✔ Always avoid overflow → use mid = left + (right - left)/2 Simple problem. Strong concept. Perfect execution. Day 18 completed. Consistency continues 💪🔥 #30DaysOfCode #DSA #Java #InterviewPreparation #ProblemSolving #CodingJourney #BinarySearch #LeetCode

🚀 Day 21/30 – DSA Challenge 📌 LeetCode Problem – Median of Two Sorted Arrays 📝 Problem Statement Given two sorted arrays nums1 and nums2, find the median of the combined array. 📌 Example Input: nums1 = [1,2] nums2 = [3,4] Output: 2.5 💡 My Approach Instead of using complex binary search, I followed a simple and reliable method: 👉 Merge both arrays 👉 Sort the merged array 👉 Find the median This approach is easy to understand and implement. 🚀 Algorithm 1️⃣ Create a new array of size n1 + n2 2️⃣ Copy elements of both arrays 3️⃣ Sort the merged array 4️⃣ If length is odd → return middle element 5️⃣ If even → return average of two middle elements ✅ Java Code import java.util.Arrays; class Solution { public double findMedianSortedArrays(int[] nums1, int[] nums2) { int n1 = nums1.length; int n2 = nums2.length; int[] merged = new int[n1 + n2]; for (int i = 0; i < n1; i++) { merged[i] = nums1[i]; } for (int i = 0; i < n2; i++) { merged[n1 + i] = nums2[i]; } Arrays.sort(merged); int n = merged.length; if (n % 2 == 1) { return merged[n / 2]; } else { return (merged[n / 2 - 1] + merged[n / 2]) / 2.0; } } } ⏱ Complexity Time Complexity: O((n + m) log(n + m)) Space Complexity: O(n + m) 📚 Key Learnings – Day 21 ✔ Simple solutions are often easiest to implement ✔ Always understand problem constraints ✔ This problem can be optimized further using binary search ✔ Multiple approaches exist — choose based on context Simple approach. Clear logic. Strong understanding. Day 21 completed. Consistency continues 💪🔥 #30DaysOfCode #DSA #Java #InterviewPreparation #ProblemSolving #CodingJourney #Arrays #LeetCode

. 🚀 Day 27/30 – DSA Challenge 📌 LeetCode Problem – Middle of the Linked List 📝 Problem Statement Given the head of a singly linked list, return the middle node of the linked list. 👉 If there are two middle nodes, return the second middle node. 📌 Example Input: 1 → 2 → 3 → 4 → 5 Output: 3 📌 Even Case Input: 1 → 2 → 3 → 4 → 5 → 6 Output: 4 💡 My Approach (Counting Method) Instead of using two pointers, I used a two-step approach: 👉 First count total nodes 👉 Then move to the middle 🚀 Algorithm 1️⃣ Traverse the list and count nodes 2️⃣ Calculate middle index: mid = count / 2 3️⃣ Traverse again till mid position 4️⃣ Return that node ✅ Java Code class Solution { public ListNode middleNode(ListNode head) { int count = 0; ListNode temp = head; // Step 1: Count nodes while (temp != null) { count++; temp = temp.next; } // Step 2: Find middle index int mid = count / 2; // Step 3: Traverse to middle temp = head; for (int i = 0; i < mid; i++) { temp = temp.next; } return temp; } } ⏱ Complexity Time Complexity: O(n) (two traversals) Space Complexity: O(1) 📚 Key Learnings – Day 27 ✔ Problems can have multiple valid approaches ✔ Counting method is simple and intuitive ✔ Always think about optimizing further ✔ Same problem can be solved in one pass using fast & slow pointers Simple idea. Clear logic. Solid understanding. Day 27 completed. Consistency continues 💪🔥 #30DaysOfCode #DSA #Java #InterviewPreparation #ProblemSolving #CodingJourney #LinkedList #LeetCode

📘 DSA Journey — Day 31 Today’s focus: Binary Search for boundaries and square roots. Problems solved: • Sqrt(x) (LeetCode 69) • Search Insert Position (LeetCode 35) Concepts used: • Binary Search • Search space reduction • Boundary conditions Key takeaway: In Sqrt(x), the goal is to find the integer square root. Using binary search, we search in the range [1, x] and check mid * mid against x to narrow down the answer. This avoids linear iteration and achieves O(log n) time. In Search Insert Position, we use binary search to find either: • The exact position of the target, or • The correct index where it should be inserted The key idea is that when the target is not found, the final position of the left pointer gives the correct insertion index. These problems highlight how binary search is not just for finding elements, but also for determining positions and boundaries efficiently. Continuing to strengthen fundamentals and consistency in problem solving. #DSA #Java #LeetCode #CodingJourney

🧠 LeetCode POTD — The Biggest Hint Was Hidden in One Line 1855. Maximum Distance Between a Pair of Values Today's problem looked tricky at first considering the time complexity. We need to find indices (i, j) such that: 👉 i <= j 👉 nums1[i] <= nums2[j] And maximize: 👉 j - i ━━━━━━━━━━━━━━━━━━━ My first instinct? Try brute force combinations. Check many pairs. Maybe binary search. But I was missing the most important line in the question: Both arrays are non-increasing (sorted). That one detail changes everything. ━━━━━━━━━━━━━━━━━━━ 💡 Once I noticed that, the solution became a clean two-pointer approach. Start with: 👉 i = 0 in nums1 👉 j = 0 in nums2 Then: If nums1[i] <= nums2[j] 👉 This pair is valid 👉 Update answer with j - i 👉 Move j forward to try for a bigger distance Else: 👉 Current i is too large 👉 Move i forward Because arrays are sorted, once a pair fails, moving i is the only useful move. ━━━━━━━━━━━━━━━━━━━ 📌 What I liked about this problem: The challenge wasn't the coding. It was reading carefully enough to spot the property that unlocks the solution. ✅ Sometimes the answer is already in the problem statement. ✅ You just have to notice it. Curious — did anyone else miss the sorted hint at first? 👀 #LeetCode #TwoPointers #ProblemSolving #SoftwareEngineering #DSA #Java #c++ #SDE

🚀 Solved: Find Dominant Index (LeetCode) Just solved an interesting problem where the goal is to find whether the largest element in the array is at least twice as large as every other number. 💡 Approach: 1. First, traverse the array to find the maximum element and its index. 2. Then, iterate again to check if the max element is at least twice every other element. 3. If the condition fails for any element → return "-1". 4. Otherwise → return the index of the max element. 🧠 Key Insight: Instead of comparing all pairs, just track the maximum and validate it — keeps the solution clean and efficient. ⚡ Time Complexity: O(n) ⚡ Space Complexity: O(1) 💻 Code (Java): class Solution { public int dominantIndex(int[] nums) { int max = -1; int index = -1; // Step 1: find max and index for (int i = 0; i < nums.length; i++) { if (nums[i] > max) { max = nums[i]; index = i; } } // Step 2: check condition for (int i = 0; i < nums.length; i++) { if (i == index) continue; if (max < 2 * nums[i]) { return -1; } } return index; } } 🔥 Got 100% runtime and 99%+ memory efficiency! #LeetCode #DSA #Java #Coding #ProblemSolving #Algorithms

🚀 LeetCode Practice Update: Concatenation of Array Solved an interesting array problem today that reinforces indexing fundamentals. 🔍 Problem: Given an array nums of size n, create a new array ans of size 2n such that: ans[i] = nums[i] ans[i + n] = nums[i] 💡 My Approach: ✔️ Created a new array of size 2 * n ✔️ Used a single loop to fill both halves of the array ✔️ Leveraged index shifting (i + n) for duplication 👨💻 Code (Java): class Solution { public int[] getConcatenation(int[] nums) { int[] ans = new int[nums.length * 2]; int n = nums.length; for(int i = 0; i < n; i++) { ans[i] = nums[i]; ans[i + n] = nums[i]; } return ans; } } ⚡ Key Learning: Using index manipulation helps avoid extra loops and keeps the solution efficient. 📊 Complexity: ⏱ Time: O(n) 💾 Space: O(n) 📌 Small problem, big clarity boost! #LeetCode #Java #DSA #CodingJourney #ProblemSolving #Developers

🚀 Day 41/100 – LeetCode DSA Practice 📌 Problem: 561. Array Partition Today’s challenge was about grouping elements into pairs such that the sum of the minimum values of each pair is maximized. 💡 My Approach: Initially thought about checking all possible pair combinations ❌ (too complex) Then realized this is a problem Sorted the array in ascending order Paired adjacent elements Added elements at even indices (0, 2, 4...) to get the maximum sum 🧠 Key Insight: After sorting, pairing neighbors ensures that the minimum in each pair is as large as possible, which maximizes the total sum. 💻 Code (Java): import java.util.Arrays; class Solution { public int arrayPairSum(int[] nums) { Arrays.sort(nums); int sum = 0; for(int i = 0; i < nums.length; i += 2) { sum += nums[i]; } return sum; } } 📚 What I Learned Today: Importance of recognizing patterns Sorting can simplify complex pairing problems Always look for pattern instead of brute force 🔥 Small insight, big impact! #Day41 #100DaysOfCode #LeetCode #DSA #GreedyAlgorithm #CodingJourney

🚀 DAY 94/150 — CONNECTING TREE LEVELS! 🌳🔗 Day 94 of my 150 Days DSA Challenge in Java and today I solved a very interesting problem that focuses on level connections in Binary Trees 💻🧠 📌 Problem Solved: Populating Next Right Pointers in Each Node 📌 LeetCode: #116 📌 Difficulty: Medium 🔹 Problem Insight The task is to connect each node’s next pointer to its right node on the same level. If there is no right node → set next = null. 👉 This problem is based on a perfect binary tree, which allows some optimizations. 🔹 Approaches Used ✅ Approach 1: BFS (Level Order Traversal) • Use a queue to traverse level by level • Connect nodes within the same level using a prev pointer ✔️ Easy to understand ❌ Uses extra space (queue) ✅ Approach 2: Optimized (Constant Space) • Use already established next pointers • Connect: left → right right → next.left ✔️ No extra space (O(1)) ✔️ More optimized and elegant ⏱ Complexity Time Complexity: O(n) Space Complexity: • BFS → O(n) • Optimized → O(1) 🧠 What I Learned • Same problem can have multiple approaches (basic → optimized) • Perfect binary tree structure helps reduce complexity • Using existing pointers smartly can eliminate extra space 💡 Key Takeaway This problem taught me: How to connect nodes level-wise Difference between BFS vs pointer-based optimization Writing space-efficient solutions 🌱 Learning Insight Now I’m focusing more on: 👉 Moving from brute-force → optimized solutions 🚀 ✅ Day 94 completed 🚀 56 days to go 🔗 Java Solution on GitHub: 👉 https://lnkd.in/gDWj7K5Q 💡 Optimization is about using what you already have. #DSAChallenge #Java #LeetCode #BinaryTree #BFS #DFS #Pointers #150DaysOfCode #ProblemSolving #CodingJourney #InterviewPrep #LearningInPublic