Insertion Sort on Linked List LeetCode Solution

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 64/100 Today’s problem: Find all strings that are substrings of another word 🧠 What I learned: - How to compare strings using nested loops - Using ".contains()" to check substrings efficiently - Importance of breaking early to optimize performance - Strengthening problem-solving with brute-force approach 💡 Key Insight: Sometimes simple solutions (O(n²)) are enough when constraints are small. No need to overcomplicate! 🔁 Consistency > Perfection #Day64 #DSA #Java #CodingJourney #Consistency #KeepLearning #100DaysOfCode

🚀 Day 58 of #100DaysOfCode Solved 165. Compare Version Numbers on LeetCode 🔗 🧠 Key Insight: Version strings are split by "." into multiple revisions. We compare each revision numerically (not lexicographically). Example: "1.01" = "1.1" (leading zeros don’t matter) ⚙️ Approach (Split + Compare): 1️⃣ Split both versions using "." 🔹 version1 → s1[] 🔹 version2 → s2[] 2️⃣ Traverse till max length of both arrays 3️⃣ For each index i: 🔹 num1 = i < s1.length ? parseInt(s1[i]) : 0 🔹 num2 = i < s2.length ? parseInt(s2[i]) : 0 4️⃣ Compare: 🔹 if num1 < num2 → return -1 🔹 if num1 > num2 → return 1 5️⃣ If all equal → return 0 ⏱️ Time Complexity: O(n + m) 📦 Space Complexity: O(n + m) (for split arrays) #100DaysOfCode #LeetCode #DSA #Strings #TwoPointers #Java #InterviewPrep #CodingJourney

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 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

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

LeetCode — Problem 189 | Day 3 💡 Problem: Rotate Array Given an array, rotate it to the right by k steps. 🧠 My Approach: - Used reverse technique for in-place rotation - First reversed the entire array - Then reversed first k elements - Finally reversed remaining elements - Handled k using k = k % n This problem gave a good understanding of: ✔️ Array manipulation ✔️ In-place optimization (O(1) space) ✔️ Reverse logic #LeetCode #DSA #Java #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 84/100 – 𝐒𝐞𝐚𝐫𝐜𝐡 𝐢𝐧 𝐑𝐨𝐭𝐚𝐭𝐞𝐝 𝐒𝐨𝐫𝐭𝐞𝐝 𝐀𝐫𝐫𝐚𝐲 𝐈𝐈  🔍  𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠: Binary search works great on sorted arrays, but duplicates introduce ambiguity — making it harder to decide which half is sorted. 💡 𝐂𝐨𝐫𝐞 𝐈𝐝𝐞𝐚: Use modified binary search Identify the sorted half Handle duplicates by shrinking the search space ⚡ 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Find 𝐦𝐢𝐝 If 𝐭𝐚𝐫𝐠𝐞𝐭 𝐟𝐨𝐮𝐧𝐝 → 𝐫𝐞𝐭𝐮𝐫𝐧 𝐭𝐫𝐮𝐞 𝐇𝐚𝐧𝐝𝐥𝐞 𝐝𝐮𝐩𝐥𝐢𝐜𝐚𝐭𝐞𝐬 (𝐥𝐨𝐰++) Check which half is sorted Narrow down search accordingly #Day84 #100DaysOfCode #Java #DSA #LeetCode #BinarySearch #CodingJourney

🚀 Day 60/100 📌 Problem: String to Integer Given a string, convert it into a 32-bit signed integer while: • Ignoring leading whitespaces • Handling '+' and '-' signs • Reading digits until a non-digit appears • Returning INT_MAX or INT_MIN in case of overflow 💡 What I Learned: • Importance of handling edge cases • How to safely manage overflow • Writing clean and efficient parsing logic ⚡ Result: Runtime 1 ms (Beats 100%) Consistency + Practice = Improvement 📈 #Day60 #Java #DSA #LeetCode #CodingJourney #ProblemSolving #100DaysOfCode #LearnToCode