Java Solution: Minimum Depth of Binary Tree

Day 34 of #100DaysOfLeetCode 💻✅ Solved #110. Balanced Binary Tree on LeetCode using Java. Approach: • Used a bottom-up recursive approach to calculate height • Returned -1 immediately if any subtree is unbalanced • Compared left and right subtree heights at each node • Checked if the height difference is greater than 1 • Stopped early to optimize unnecessary computations Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) ✓ Memory: 45.33 MB (Beats 95.41% submissions) Key Learning: ✓ Understood how to combine height calculation with balance checking ✓ Learned early termination technique in recursion ✓ Improved problem-solving for tree-based recursive problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTree #Recursion #TreeProblems #ProblemSolving #CodingJourney #100DaysOfCode

Day 31 of #100DaysOfLeetCode 💻✅ Solved #94. Binary Tree Inorder Traversal on LeetCode using Java. Approach: • Used recursion to perform inorder traversal • Followed Left → Root → Right order strictly • Traversed left subtree first before processing current node • Stored node values in a list during traversal • Returned the final list after complete traversal Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) ✓ Memory: 42.76 MB (Beats 98.30% submissions) Key Learning: ✓ Strengthened understanding of tree traversal patterns ✓ Improved clarity on recursion stack behavior ✓ Reinforced difference between preorder, inorder, and postorder traversal Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTree #Recursion #TreeTraversal #ProblemSolving #CodingJourney #100DaysOfCode

Day 24 of #100DaysOfLeetCode 💻✅ Solved #35. Search Insert Position on LeetCode using Java. Approach: • Applied Binary Search to achieve O(log n) time complexity • Initialized two pointers: left and right • Calculated mid using left + (right - left) / 2 to avoid overflow • Compared mid element with target to adjust search space • If target not found, returned left as the correct insert position Performance: ✓ Runtime: 0 ms (Beats 100% submissions) ✓ Memory: 43.2 MB Key Learning: ✓ Strengthened understanding of Binary Search pattern ✓ Learned how to determine insert position when element is absent ✓ Improved confidence in handling sorted array problems efficiently Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #ProblemSolving #CodingJourney #100DaysOfCode

Day 30 of #100DaysOfLeetCode 💻✅ Solved #108. Convert Sorted Array to Binary Search Tree on LeetCode using Java. Approach: • Used Divide and Conquer strategy • Selected the middle element as the root to maintain balance • Recursively built the left subtree using left half of array • Recursively built the right subtree using right half of array • Ensured the tree remains height-balanced at every step Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) ✓ Memory: 45.18 MB (Beats 44.10% submissions) Key Learning: ✓ Understood how sorted arrays can directly map to balanced BST ✓ Strengthened recursion fundamentals in tree construction ✓ Improved understanding of height-balanced binary trees Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearchTree #Recursion #DivideAndConquer #ProblemSolving #CodingJourney #100DaysOfCode

Day 32 of #100DaysOfLeetCode 💻✅ Solved #144. Binary Tree Preorder Traversal on LeetCode using Java. Approach: • Used recursion to perform preorder traversal • Followed Root → Left → Right order strictly • Added node value before traversing subtrees • Traversed left subtree first, then right subtree • Stored values in a list during traversal Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) ✓ Memory: 42.83 MB (Beats 96.77% submissions) Key Learning: ✓ Strengthened understanding of preorder traversal pattern ✓ Clearly differentiated preorder from inorder traversal ✓ Improved recursive tree traversal implementation skills Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTree #Recursion #TreeTraversal #ProblemSolving #CodingJourney #100DaysOfCode

Day 33 of #100DaysOfLeetCode 💻✅ Solved #145. Binary Tree Postorder Traversal on LeetCode using Java. Approach: • Used recursion to perform postorder traversal • Followed Left → Right → Root order strictly • Traversed left subtree first • Then traversed right subtree • Added node value after visiting both subtrees Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) ✓ Memory: 43.12 MB (Beats 71.37% submissions) Key Learning: ✓ Clearly understood difference between preorder, inorder, and postorder ✓ Strengthened recursive tree traversal concepts ✓ Improved confidence in handling binary tree problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinaryTree #Recursion #TreeTraversal #ProblemSolving #CodingJourney #100DaysOfCode

Day 25 of #100DaysOfLeetCode 💻✅ Solved #167. Two Sum II – Input Array Is Sorted on LeetCode using Java. Approach: • Used Two Pointer technique since the array is already sorted • Initialized one pointer at the beginning and one at the end • Calculated the sum of both elements • If sum < target, moved left pointer forward • If sum > target, moved right pointer backward • Returned 1-based indices as required in the problem Performance: ✓ Runtime: 1 ms (Efficient with O(n) time complexity) ✓ Memory: Constant extra space (O(1)) Key Learning: ✓ Understood when to apply Two Pointer technique instead of HashMap ✓ Improved ability to optimize space complexity ✓ Strengthened pattern recognition for sorted array problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #TwoPointers #ProblemSolving #CodingJourney #100DaysOfCode

LeetCode Problem || Find Unique Binary String (1980)🚀 Today I solved the problem "Find Unique Binary String" using Java. 🔹 Problem: We are given an array of n binary strings, each of length n. The goal is to return a binary string of length n that does not exist in the array. 🔹 Approach (Diagonal Flip Technique): The idea is simple but powerful: Traverse the array using index i. Look at the i-th character of the i-th string (nums[i][i]). Flip the bit (0 → 1, 1 → 0). Append it to a result string. 💡 Time Complexity: O(n) Practicing problems like this strengthens logical thinking and problem-solving skills. #LeetCode #Java #CodingPractice #ProblemSolving #DSA

🚀 Day 18/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1528. Shuffle String Used a direct indexing approach by placing each character at its correct position using the given indices array. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) A simple yet important problem to strengthen understanding of arrays and index mapping. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

Day 6/100 – LeetCode Challenge 🚀 Problem: #189 Rotate Array   Difficulty: Medium   Language: Java   Approach: Array Reversal Technique   Time Complexity: O(n)   Space Complexity: O(1) 🔍 Key Insight: Instead of shifting elements one by one, the array can be rotated efficiently using a three-step reversal strategy. Steps: 1️⃣ Reverse the entire array   2️⃣ Reverse the first k elements   3️⃣ Reverse the remaining elements  This achieves the required rotation in-place with constant extra space. 🧠 Solution Brief: Calculated k % n to handle cases where k is greater than array length.   Reversed the entire array first.   Then reversed the first k elements and the remaining n-k elements.   This sequence correctly rotates the array to the right. 📌 What I Learned: Understanding patterns like array reversal can simplify problems that initially seem complex.   Optimizing from brute force shifting to an in-place O(n) solution improves efficiency. #LeetCode #Day6 #100DaysOfCode #Java #DSA #Arrays #RotateArray #ProblemSolving #CodingJourney