🔥 𝗗𝗮𝘆 𝟵𝟴/𝟭𝟬𝟬 — 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝟳𝟬. 𝗖𝗹𝗶𝗺𝗯𝗶𝗻𝗴 𝗦𝘁𝗮𝗶𝗿𝘀 | 🟢 Easy | Java The gateway problem to Dynamic Programming — and one of the most iconic problems in DSA. 🪜 🔍 𝗧𝗵𝗲 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 You can climb 1 or 2 steps at a time. How many distinct ways can you reach the top of n stairs? 💡 𝗧𝗵𝗲 𝗜𝗻𝘀𝗶𝗴𝗵𝘁 To reach step i, you either came from step i-1 (1 step) or step i-2 (2 steps). So ways(i) = ways(i-1) + ways(i-2) Sound familiar? It's literally the Fibonacci sequence. 🐇 ⚡ 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵 — 𝗕𝗼𝘁𝘁𝗼𝗺-𝗨𝗽 𝗗𝗣 ✅ Base cases: dp[1] = 1, dp[2] = 2 ✅ Fill dp[i] = dp[i-1] + dp[i-2] for i from 3 to n ✅ Return dp[n] No recursion. No stack overflow. No repeated subproblems. Just a clean bottom-up table. 🧠 📊 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆 ⏱ Time: O(n) — single pass 📦 Space: O(n) — can be further optimised to O(1) with just two variables This problem teaches the most important DP lesson: identify overlapping subproblems, store results, build up from the base. Every hard DP problem starts with this same thinking. 𝗔𝗻𝗱 𝘆𝗲𝘀 — 𝗷𝘂𝘀𝘁 𝗹𝗶𝗸𝗲 𝘁𝗵𝗲𝘀𝗲 𝘀𝘁𝗮𝗶𝗿𝘀, 𝘁𝗵𝗶𝘀 𝗰𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝘄𝗮𝘀 𝗰𝗹𝗶𝗺𝗯𝗲𝗱 𝗼𝗻𝗲 𝗱𝗮𝘆 𝗮𝘁 𝗮 𝘁𝗶𝗺𝗲. 🏔️ 📂 𝗙𝘂𝗹𝗹 𝘀𝗼𝗹𝘂𝘁𝗶𝗼𝗻 𝗼𝗻 𝗚𝗶𝘁𝗛𝘂𝗯: https://lnkd.in/gS3DX_YW 𝗝𝘂𝘀𝘁 𝟮 𝗺𝗼𝗿𝗲 𝗱𝗮𝘆𝘀. 𝗧𝗵𝗲 𝘀𝘂𝗺𝗺𝗶𝘁 𝗮𝘄𝗮𝗶𝘁𝘀! 🏁 #LeetCode #Day98of100 #100DaysOfCode #Java #DSA #DynamicProgramming #Fibonacci #CodingChallenge #Programming

🔥 𝗗𝗮𝘆 𝟴𝟵/𝟭𝟬𝟬 — 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝟳𝟰𝟰. 𝗙𝗶𝗻𝗱 𝗦𝗺𝗮𝗹𝗹𝗲𝘀𝘁 𝗟𝗲𝘁𝘁𝗲𝗿 𝗚𝗿𝗲𝗮𝘁𝗲𝗿 𝗧𝗵𝗮𝗻 𝗧𝗮𝗿𝗴𝗲𝘁 | 🟢 𝗘𝗮𝘀𝘆 | 𝗝𝗮𝘃𝗮 Looks easy — but the circular wrap-around is the trap most people miss. 𝗧𝗵𝗲 𝗽𝗿𝗼𝗯𝗹𝗲𝗺: Given a sorted circular array of letters, find the smallest letter strictly greater than the target. If none exists, wrap around to the first letter. 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵 — 𝗕𝗶𝗻𝗮𝗿𝘆 𝗦𝗲𝗮𝗿𝗰𝗵: ✅ Standard binary search for the first letter > target ✅ If letters[mid] > target → go left (end = mid - 1) ✅ Else → go right (start = mid + 1) ✅ After the loop, start points to the answer ✅ Use start % letters.length to handle the circular wrap-around 𝗧𝗵𝗲 𝗰𝗹𝗲𝘃𝗲𝗿 𝗯𝗶𝘁: If target is greater than or equal to all letters, start lands at letters.length — modulo wraps it back to index 0 automatically. No special case needed. 🔄 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆: ⏱ Time: O(log n) 📦 Space: O(1) One modulo operation handles the entire circular edge case cleanly. This is binary search with a twist — and the wrap-around trick is worth remembering! 🧠 📂 𝗙𝘂𝗹𝗹 𝘀𝗼𝗹𝘂𝘁𝗶𝗼𝗻 𝗼𝗻 𝗚𝗶𝘁𝗛𝘂𝗯: https://lnkd.in/gvK_p44b 11 more days. So close! 💪 #LeetCode #Day89of100 #100DaysOfCode #Java #DSA #BinarySearch #Arrays #CodingChallenge #Programming

🔥 𝗗𝗮𝘆 𝟴𝟲/𝟭𝟬𝟬 — 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝟳𝟮𝟰. 𝗙𝗶𝗻𝗱 𝗣𝗶𝘃𝗼𝘁 𝗜𝗻𝗱𝗲𝘅 | 🟢 𝗘𝗮𝘀𝘆 | 𝗝𝗮𝘃𝗮 Deceptively simple — and a perfect introduction to the prefix sum pattern. 𝗧𝗵𝗲 𝗽𝗿𝗼𝗯𝗹𝗲𝗺: Find the index where the sum of all elements to the left equals the sum of all elements to the right. 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵 — 𝗣𝗿𝗲𝗳𝗶𝘅 𝗦𝘂𝗺: ✅ Compute the total sum in one pass ✅ Track leftTotal as we iterate ✅ rightTotal = total - leftTotal - nums[i] ✅ If leftTotal == rightTotal → pivot found! No extra arrays. No nested loops. Just one pre-computation and one clean scan. 𝗧𝗵𝗲 𝗸𝗲𝘆 𝗶𝗻𝘀𝗶𝗴𝗵𝘁: Instead of recalculating both sides at every index, derive the right sum from what you already know — total and left. That drops it from O(n²) to O(n) instantly. 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆: ⏱ Time: O(n) — two passes 📦 Space: O(1) This pattern — precompute total, derive the other side on the fly — shows up everywhere: product arrays, equilibrium points, range queries. A must-know! 🧠 📂 𝗙𝘂𝗹𝗹 𝘀𝗼𝗹𝘂𝘁𝗶𝗼𝗻 𝗼𝗻 𝗚𝗶𝘁𝗛𝘂𝗯: https://lnkd.in/g_E5Ahfe 14 more days. The finish line is in sight! 💪 #LeetCode #Day86of100 #100DaysOfCode #Java #DSA #PrefixSum #Arrays #CodingChallenge #Programming

“Most people try to overcomplicate this one… but the simplest approach wins.” Day 69 — LeetCode Progress Problem: Height Checker Required: Given an array of student heights, return the number of indices where the heights are not in the expected non-decreasing order. Idea: If we sort the array, we get the expected order. Now just compare the original array with the sorted version — mismatches are the answer. Approach: Create a copy of the original array Sort the copied array Traverse both arrays: Compare elements at each index If they differ → increment count Return the count Time Complexity: O(n log n) Space Complexity: O(n) #LeetCode #DSA #Java #Arrays #Sorting #ProblemSolving #CodingJourney

Day 79 - Binary Tree Paths Solved a problem on generating all root-to-leaf paths in a binary tree using DFS recursion. This improves understanding of tree traversal and path construction techniques. Time Complexity: O(N) Space Complexity: O(N) #Day79 #CodingChallenge #LeetCode #BinaryTree #DSA #Java #Recursion #CodingJourney

🚀 Day 6 of #100DaysOfCode | Pattern: Two Pointers Today I tackled a classic Linked List problem: 👉 Remove Nth Node From End of List At first, I thought of a brute force approach: Traverse the list to find length Then remove (length - n) node ❌ Works, but requires 2 passes 💡 Then I learned the optimal approach (Two Pointers): Use fast and slow pointers Move fast n steps ahead Move both together until fast reaches end Delete the node using slow ✅ Only 1 pass ✅ Cleaner and efficient ⚠️ Mistakes I made (important learning): Forgot to use a dummy node → failed for head deletion Got confused between moving n vs n+1 steps Faced null pointer issues in edge cases 🔥 Key Takeaway: Two pointer technique is not just for arrays, it’s super powerful in Linked Lists too 📈 Progress Update: Consistency building day by day Next week: Starting a new pattern Along with revision of Two Pointers #100DaysOfCode #DSA #Java #LinkedList #CodingJourney #ProblemSolving #LearnInPublic #PatternWise

Day 69 of #100DaysOfLeetCode 💻✅ Solved #349. Intersection of Two Arrays problem in Java. Approach: • Iterated through each element of the first array • Checked if the element exists in the second array • Used a temporary array to store intersection elements • Ensured no duplicates by checking already added elements • Copied the result into a final array of correct size Performance: ✓ Runtime: 4 ms (Beats 35.60%) ✓ Memory: 44.41 MB (Beats 96.71%) Key Learning: ✓ Practiced array traversal using nested loops ✓ Learned how to handle duplicates manually ✓ Improved understanding of set-like operations without using extra data structures Learning one problem every single day 🚀 #Java #LeetCode #DSA #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

Day 50/75 — Climbing Stairs Today’s problem was a classic dynamic programming question — counting the number of distinct ways to reach the top. Approach: • Recognize it as a Fibonacci pattern • Each step = sum of previous two steps • Optimize space using two variables instead of an array Key logic: int current = prev1 + prev2; prev2 = prev1; prev1 = current; Time Complexity: O(n) Space Complexity: O(1) A fundamental problem that builds strong intuition for DP and recurrence relations. Halfway there — consistency paying off 🔥 50/75 🚀 #Day50 #DSA #DynamicProgramming #Java #Algorithms #LeetCode

🚀 Day 13 of 180 — 3Sum ✅ Today I did the 3Sum problem. How I did it? 1. Using the two pointer technique. 2. Ran a for loop from i = 0 to n - 2, where n = array length. Checked two things: 3. If current element > 0 → not possible to get a valid answer, so break. 4. Avoid same ith number by comparing it with the previous value. If previous value is same as current value, move to next ith element. Initialized pointers: left = i + 1 right = n - 1 5. Found sum: If it is 0 → added triplet to list 6. Checked for duplicates: If left points to same second number → move left If right points to same third number → move right 7. If sum < 0 → left++ to increase sum 8. If sum > 0 → right-- to decrease sum 9. At last, returned the list. Day 13 done. 167 to go. 🔥 #180DaysDSA #Day12 #LeetCode #Java #DSA #TwoPointers #Strings #DSAJourney #CodingJourney #Programming #DataStructures #Algorithms #ProblemSolving #BuildInPublic #CodeNewbie #LearnToCode #100DaysOfCode #SoftwareDevelopment #Developer #StudentDeveloper #TechCommunity #LinkedInTech #CompetitiveProgramming

🚀 Day 44/100 Today’s problem: Reverse Vowels of a String. 🔹 Problem Summary: Given a string, reverse only the vowels while keeping other characters in the same position. 🔹 Approach I Used: I applied the Two Pointer Technique: - One pointer from the start, one from the end - Move both until vowels are found - Swap them and continue 🔹 Key Learning: ✔ Two pointers can simplify string problems a lot ✔ Always break the problem into smaller steps ✔ Practice improves pattern recognition. 🔹 Example: Input: "IceCreAm" Output: "AceCreIm" 🔹 Time Complexity: O(n) 🔹 Space Complexity: O(n) Consistency is the key 🔑 — showing up every day! #Day44 #CodingJourney #DSA #Java #Learning #100DaysOfCode