Invert Binary Tree Problem - Java Solution

#day333 of #1001daysofcode problem statement (0257): Binary Tree Paths 💡 Approach: Used DFS recursion to explore all root-to-leaf paths in the binary tree. While traversing, I kept building the path string. When a leaf node is reached, the complete path is added to the result list. Example path format: 1->2->5 ⏱ Time Complexity: O(n) 🧠 Space Complexity: O(h) — recursion stack (h = height of the tree) Consistency in solving one problem every day 📈 #1001DaysOfCode #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode

#day334 of #1001daysofcode problem statement (1009): Complement of Base 10 Integer 💡 Approach: To find the complement of a number, I converted it into binary and flipped every bit (0 → 1 and 1 → 0). After flipping the bits, the resulting binary string was converted back to decimal. Example: 5 → Binary: 101 Complement: 010 → 2 ⏱ Time Complexity: O(log n) 🧠 Space Complexity: O(log n) #1001DaysOfCode #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode

Day 65 — LeetCode Progress (Java) Problem: Find All Numbers Disappeared in an Array Required: Given an array of size n containing numbers in the range [1, n], return all the numbers that are missing from the array. Idea: Compare the expected range [1…n] with the actual elements to identify missing values. Approach: Initialize a set containing all numbers from 1 to n. Traverse the array: Remove each element from the set The remaining elements in the set are the missing numbers. Time Complexity: O(n) Space Complexity: O(n) #LeetCode #DSA #Java #HashSet #Arrays #Algorithms #CodingJourney #100DaysOfCode

𝗬𝗼𝘂 𝗵𝗮𝘃𝗲 𝗯𝗲𝗲𝗻 𝘂𝘀𝗶𝗻𝗴 𝗕𝗶𝗻𝗮𝗿𝘆 𝗦𝗲𝗮𝗿𝗰𝗵 𝘄𝗿𝗼𝗻𝗴. It is not just for sorted arrays. The real definition — "Eliminate HALF the search space with each decision." That one shift in thinking unlocks 20+ LeetCode problems. Swipe to see the template, live code, and 7 problems you can now solve with one pattern. 👇 Save this. 🔖 #DSA #BinarySearch #Java #LeetCode #CodingInterview #DebugWithPurpose

Day 59 - Merge Two Sorted Lists Worked on merging two sorted linked lists into one sorted list. Approach: • Used a dummy node to simplify pointer handling • Compared nodes from both lists and attached the smaller one • Appended remaining elements after traversal A classic linked list problem that strengthens pointer manipulation skills. Time Complexity: O(n + m) Space Complexity: O(1) #Day59 #LeetCode #Java #LinkedList #CodingPractice #DSA #TechJourney

🚀 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 41 - Find Pivot Index Solved using a prefix sum approach. First computed the total sum of the array, then traversed once while maintaining a running left sum. For each index, the right sum is calculated using total - left - nums[i]. If both sums match, that index is the pivot. Time Complexity: O(n) #Day41 #LeetCode #Java #PrefixSum #DSA #ProblemSolving #CodingJourney

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

Day 25/100: Finding the "Gap" 🎯 Today's challenge: Search Insert Position. We all know Binary Search finds an element in O(log n), but what if the element isn't there? I learned that by the end of the search, the `left` pointer doesn't just give up—it points exactly to where that missing number *should* be inserted to keep the array sorted. It’s a powerful way to handle dynamic data without breaking the order. Quarter of the way through the challenge! 🚀 #100DaysOfCode #Java #DSA #BinarySearch #ProblemSolving #Unit3 #Day25 #LearnInPublic

Day 60 - Remove Duplicates from Sorted List Worked on removing duplicate elements from a sorted linked list. Approach: • Traverse the list once • Compare current node with next node • Skip duplicate nodes by adjusting pointers Time Complexity: O(n) Space Complexity: O(1) #Day60 #LeetCode #Java #LinkedList #CodingPractice #DSA #TechJourney