Binary Search Insert Position Challenge

𝐃𝐚𝐲 87/100 – 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 🚀 Problem: 228. 𝐒𝐮𝐦𝐦𝐚𝐫𝐲 𝐑𝐚𝐧𝐠𝐞𝐬  Today I solved a problem where we need to summarize consecutive numbers in a sorted unique array into ranges. 🔑 𝐈𝐝𝐞𝐚: Traverse the array and keep extending the range while consecutive numbers continue. Once the sequence breaks, close the range and store it. 💡 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Start with the first element as start Move forward while nums[i] + 1 == nums[i+1] If range exists → "start->end" Else → single number "start" 𝐊𝐞𝐲 𝐈𝐧𝐬𝐢𝐠𝐡𝐭: Efficient single pass solution (O(n)) by grouping consecutive elements on the fly. #LeetCode #Java #ProblemSolving #DSA #100DaysOfCode #CodingJourney

🔥 Day 58 of #100DaysOfCode Solved – Check Balanced String 🔍 What I did: Traversed the string and calculated the sum of digits at even and odd indices separately, then compared both sums to determine if the string is balanced. 💡 Key Learning: Practiced string traversal and indexing Improved understanding of even vs odd index logic Learned how to convert char to integer (c - '0') 🎯 Takeaway: Even simple index-based problems can strengthen your fundamentals and attention to detail. #Day58 #LeetCode #Java #ProblemSolving #CodingJourney #100DaysOfCode #DSA

Day 74 - Univalued Binary Tree Checking whether all nodes in a binary tree share the same value using recursion. Approach: • Store root value • Traverse the tree recursively • Compare each node with root value • Return false if any mismatch is found Key Insight: Recursion simplifies tree traversal and validation Time Complexity: O(n) #Day74 #LeetCode #Java #CodingPractice #TechJourney #DSA #BinaryTree

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 61/100 Today’s problem: Check Balanced String 📌 Problem Summary: A string is called balanced if the sum of digits at even indices is equal to the sum at odd indices. 💡 What I learned: - How to handle index-based logic - Difference between even & odd indexing - Clean iteration and condition checking 🧠 Approach: - Traverse the string - Maintain two sums → even index & odd index - Compare both sums ⚡ Time Complexity: O(n) ⚡ Space Complexity: O(1) Consistency is the key — small steps every day lead to big results 💪 #Day61 #CodingJourney #Java #DSA #ProblemSolving #Consistency #KeepLearning #100DaysOfCode

Day 76 - Symmetric Tree Applied mirror comparison technique to validate symmetric binary trees. Approach: • Base cases for null nodes • Compare node values • Cross-check children recursively Time Complexity: O(n) Space Complexity: O(h) #Day76 #LeetCode #CodingJourney #DSA #Java #BinaryTree #LearningDaily

💡 Day 41 of LeetCode Problem Solved! 🔧 🌟28. Find the Index of the First Occurrence in a String🌟 Task : • Given two strings needle and haystack, return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Example 1: Input: haystack = "sadbutsad", needle = "sad" Output: 0 Explanation: "sad" occurs at index 0 and 6. The first occurrence is at index 0, so we return 0. Example 2: Input: haystack = "leetcode", needle = "leeto" Output: -1 Explanation: "leeto" did not occur in "leetcode", so we return -1. #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfChallenge #CodingJourney #KeepGrowing

Day 81 - Balanced Binary Tree Checked whether a binary tree is height-balanced using a bottom-up approach. Approach: • Recursively compute height of left and right subtrees • If height difference > 1 → not balanced • Use -1 as a signal to stop early Time Complexity: O(n) Space Complexity: O(h) #Day81 #LeetCode #BinaryTree #DSA #Java #CodingJourney #ProblemSolving

Some of the hardest problems become manageable once you recognize a repeating pattern. 🚀 Day 105/365 — DSA Challenge Solved: Subarrays with K Different Integers Problem idea: We need to count subarrays that contain exactly k distinct integers. Efficient approach: Use the powerful trick: subarrays with exactly k distinct = subarrays with ≤ k distinct − subarrays with ≤ (k − 1) distinct Steps: 1. Use a sliding window with a hashmap to track frequency of elements 2. Expand window by moving right pointer 3. If distinct count exceeds k, shrink window from the left 4. Count valid subarrays ending at each index 5. Subtract results to get exact count This pattern converts a hard problem into a manageable one. ⏱ Time: O(n) 📦 Space: O(n) Day 105/365 complete. 💻 260 days to go. Code: https://lnkd.in/dad5sZfu #DSA #Java #SlidingWindow #HashMap #LeetCode #LearningInPublic

🚀 Day 88 / 100 — LeetCode Challenge Solved: Find Minimum in Rotated Sorted Array 💡 Approach: Used Binary Search (O(log n)) to efficiently locate the minimum element in a rotated sorted array. Instead of scanning linearly, compared mid with high to decide the search direction. 🔍 Key Insight: If nums[mid] > nums[high] → minimum is in the right half Else → minimum is in the left half (including mid) ⚡ Performance: Runtime: 0 ms (Beats 100%) Memory: 43.92 MB 🧠 Learning: Understanding how sorted + rotated arrays behave helps reduce time complexity drastically from O(n) → O(log n) Consistency is key. 12 days to go 💯🔥 #Day88 #LeetCode #BinarySearch #Java #DSA #100DaysOfCode