Max Distance Between Houses of Different Colors LeetCode Solution

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/g9SGp2Qa 💡 My thought process: First, create a mapping from each value to a sorted list of its indices using a map<int, vector<int>>. This allows for easy lookup of all positions where a value occurs. For each query index idx, it retrieves the corresponding value num. If that value appears only once, the result is -1 because no valid pair exists. In case of multiple occurrences, we use binary search on the index list of num: * upper_bound finds the next occurrence that is strictly greater than idx. * lower_bound locates the current position and checks the previous occurrence. It calculates distances in both directions: * Forward distance: either direct (next - idx) or circular wrap (n - (idx - first)). * Backward distance: either direct (idx - prev) or circular wrap (n - (last - idx)). The minimum of these distances is stored as the answer. 👉 My Solution: https://lnkd.in/gjmKVa3y If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

I just returned false on a LeetCode problem… and it passed ALL 67 test cases. No loops. No base conversions. No palindrome checks. Just one line: return false; And it got accepted instantly. The problem? 2396.  Strictly Palindromic Number You had to check if a number n is a palindrome in every base from 2 to n-2. Sounds brutal, right? Then I actually read the problem properly. Here's the mind-blowing part: For any n ≥ 4, when you convert it to base (n-2), it always becomes "12". n=5 → base 3 → "12" n=6 → base 4 → "12" n=10 → base 8 → "12" "12" is never a palindrome. So it's mathematically impossible for any number ≥ 4 to be strictly palindromic. For n=1,2,3 the base range is invalid anyway. The answer is always false. The entire "hard" problem collapses into a single line of code. Lesson for every developer: Before you write clever code, ask yourself: Is this problem even possible? Sometimes the best solution isn’t optimized algorithms. It’s pure elimination.  O(1) thinking. Read the problem. Really read it. #LeetCode #ProblemSolving #SoftwareEngineering #Coding #DSA #BackendEngineering #CodingInterview #DeveloperMindset #CleanCode

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/g8Xgd5wx The function goes through each word in the `queries` array. For every query word, it compares it with each word in the `dictionary`. Since all words are assumed to have the same length, it performs a character-by-character comparison. For each pair of words, a counter called `unequal` tracks how many positions have different characters. The code loops through each index of the word and increases this counter whenever the characters at the same position do not match. If the number of differing characters is 2 or less at any point, it means the query word can change into that dictionary word with at most two edits. In this case, the query word is added to the result list, and the inner loop stops early, using `break` to avoid unnecessary comparisons with other dictionary words. Finally, after checking all query words, the function returns the list of valid words that meet the condition. 👉 My Solution: https://lnkd.in/gMnYvSvx If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 Day 12 of My LeetCode Journey — Linked List Patterns Getting Stronger Today’s problems: 🔹 Intersection of Two Linked Lists (LeetCode 160) 🔹 Remove Linked List Elements (LeetCode 203) 💡 Problem 1: Intersection of Two Linked Lists Explored two approaches: ✅ Brute Force Traverse headA and check every node in headB ⏱️ Time: O(m × n) ✅ Using Set Store all nodes of headB in a Set Traverse headA and check for intersection ⏱️ Time: O(m + n) | 📦 Space: O(n) 👉 Helped me understand how hashing can optimize comparisons. 💡 Problem 2: Remove Linked List Elements Solved using a Sentinel (Dummy) Node 🔥 👉 Create a dummy node before head 👉 Use prev pointer to skip unwanted nodes: prev.next = prev.next.next This approach simplifies edge cases like: Removing head node Multiple consecutive deletions 🧠 What I Learned: Sentinel nodes make linked list problems cleaner Hashing can reduce time complexity significantly Thinking in terms of nodes (not values) is key 🔥 Key Takeaways: Always look for ways to reduce nested loops Dummy nodes = lifesaver in linked list problems Practice is making pointer manipulation more intuitive Grateful for the guidance from Namaste DSA and Akshay Saini 🚀 Day 13 loading… 💪 #LeetCode #DataStructures #Algorithms #CodingJourney #100DaysOfCode #SoftwareEngineering #Programming #InterviewPrep #JavaScript #CodingLife #TechGrowth #ProblemSolving #Developers #LearnToCode #LinkedList #DSA #NamasteDSA #AkshaySaini

🌞 Back to LeetCode 75 – Day 43 ✅ 841. Keys and Rooms After a short break, I’m getting back into my daily problem-solving routine. Today’s problem was about checking whether we can visit all rooms starting from room 0, where each room contains keys to other rooms. Approach : This is basically a graph traversal problem in disguise. - Each room = a node - Keys = edges to other nodes So the task becomes: - Can we reach all nodes starting from node 0? I used an iterative DFS with a stack: - Start from room 0 - Use a visited array to track visited rooms - Keep exploring using available keys - Push unvisited connected rooms into the stack Complexity : - Time: O(N + K) - Space: O(N) Key Takeaway : Once you recognize this as a graph problem, the solution becomes straightforward. This is a good reminder that many array-like problems are actually graph problems in disguise. Restarting the consistency journey again — one problem at a time. 🚀 #LeetCode75 #Day43 #DSA #JavaScript #Graph #DFS #ProblemSolving #Coding #LearningInPublic #Consistency

Day 36 of My DSA Journey Today I solved LeetCode 150 – Evaluate Reverse Polish Notation (RPN) on LeetCode. 📌 Problem Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are: +, -, *, / Each operand may be an integer. Example: Input: ["2","1","+","3","*"] Output: 9 Explanation: (2 + 1) * 3 = 9 🧠 Approach – Stack This problem is a perfect application of the Stack data structure. Steps I followed: • Traverse the tokens array. • If the element is a number → push it onto the stack. • If it is an operator: Pop the top two elements (a and b) Perform the operation (a operator b) Push the result back into the stack • At the end, the stack contains the final result. ⏱ Time Complexity: O(n) — We process each token once 📦 Space Complexity: O(n) — Stack to store operands 💡 Key Learnings ✔ Understanding Reverse Polish Notation (Postfix Expression) ✔ Applying stack for expression evaluation ✔ Handling operator precedence implicitly using stack This problem connects DSA with compiler/interpreter concepts, which makes it even more interesting 🚀 Consistency continues — improving every day 💪 #100DaysOfCode #DSA #Stack #LeetCode #Java #ProblemSolving #CodingJourney #DeveloperJourney #Programming #SoftwareEngineering

🚀 Day 11 of My LeetCode Journey — Deep Dive into Linked Lists Today was all about mastering patterns in Linked Lists: 🔹 Linked List Cycle (LeetCode 141) 🔹 Palindrome Linked List (LeetCode 234) 💡 Problem 1: Linked List Cycle Tried two approaches: ✅ Using Set Store visited nodes If node already exists → cycle detected ⏱️ Time: O(n) | 📦 Space: O(n) ✅ Fast & Slow Pointer (Floyd’s Algorithm) 🔥 Slow → 1 step Fast → 2 steps If they meet → cycle exists ⏱️ Time: O(n) | 📦 Space: O(1) 👉 This approach is elegant and optimal! 💡 Problem 2: Palindrome Linked List Again explored two approaches: ✅ Convert to Array Store values in array Compare from both ends ⏱️ Time: O(n) | 📦 Space: O(n) ✅ Optimal Approach (In-place) 🔥 Find middle node Reverse second half of linked list Compare both halves ⏱️ Time: O(n) | 📦 Space: O(1) 👉 This one really tested pointer manipulation skills! 🧠 What I Learned: One problem can have multiple valid approaches Optimal solutions often reduce space complexity Linked Lists = pointer mastery + careful thinking 🔥 Key Takeaways: Always try brute force first → then optimize Fast & slow pointer is a game-changing technique In-place solutions are highly valued in interviews Big thanks to Namaste DSA and Akshay Saini 🚀 for the guidance Day 12 loading… 💪 #LeetCode #DataStructures #Algorithms #CodingJourney #100DaysOfCode #SoftwareEngineering #Programming #InterviewPrep #JavaScript #CodingLife #TechGrowth #ProblemSolving #Developers #LearnToCode #LinkedList #TwoPointers #DSA #NamasteDSA

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gdjyFXCw First, we note that swaps aren't random; they are only allowed between specific index pairs. If index `a` can swap with `b`, and `b` can swap with `c`, then `a`, `b`, and `c` become connected. We can rearrange values freely among them. This means we need to identify groups of indices where swapping is allowed within the group. To find these groups efficiently, we use Disjoint Set Union (DSU). Each index starts in its own set. For every allowed swap pair `[x, y]`, we combine their sets. After processing all the swaps, indices that share the same parent belong to the same connected component or group. Next, we organize all indices by their DSU parent. Each group shows a set of positions where we can rearrange values without restrictions. For each group, we aim to match values from `source` to `target` as closely as possible: * First, we count how often each value appears in `source` for that group. * Then, we go through the same indices in `target`. For each value:  * If the value exists in our frequency map (meaning we have it in `source`), we use it and decrease its count.  * If it doesn't exist, it means we can't match this value in the group, contributing to the Hamming distance. By doing this for all groups, we maximize matches within each connected component, which minimizes the overall Hamming distance. 👉 My Solution: https://lnkd.in/gfx3iMut If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 Day 4 of my Coding Challenge Today's problem was simple but a great reminder of string fundamentals in JavaScript. Problem Statement Given a string s, return the last character of the string. Example Input: "learncoding" Output: "g" Any language Approach Use string indexing with length - 1 to access the last character. function getLastCharacter(s) { return s[s.length - 1]; } console.log(getLastCharacter("learncoding")); // g Key Learning:- Strings are index-based length - 1 always points to the last character Small problems help strengthen core programming logic Consistency matters more than complexity. One problem every day 📚 #CodingChallenge #Day4CodingChallenge #JavaScriptDeveloper #DSA #DataStructures #Algorithms #ProblemSolving #CodingJourney #CodeDaily #LearnToCode #DevelopersLife #SoftwareDeveloper #FrontendDeveloper #FullStackDeveloper #TechCommunity #ProgrammingLife #CodeNewbie #CodingPractice #DeveloperJourney #CodingSkills #TechLearning #JavaScriptCoding #WebDeveloper #ProgrammingChallenge #DailyCoding #CodeMotivation #CareerInTech #DeveloperMindset #CodingGrind #TechGrowth

🚀 Day 7 of My LeetCode Journey — Sorting Fundamentals Today I focused on two classic sorting algorithms: 🔹 Bubble Sort 🔹 Selection Sort 💡 Bubble Sort The idea is simple: 👉 Compare adjacent elements 👉 Swap if they are in the wrong order 👉 Repeat until the array is sorted It’s easy to understand, but not efficient for large datasets. ⏱️ Time Complexity: O(n²) 💡 Selection Sort A slightly different approach: 👉 Find the minimum element 👉 Place it at the correct position 👉 Repeat for the rest of the array Also simple, but still not optimal for big inputs. ⏱️ Time Complexity: O(n²) 🔥 Key Takeaways: These algorithms may not be optimal, but they build strong fundamentals Understanding how sorting works internally is more important than memorizing Optimization comes later — basics come first Big thanks to Namaste DSA and Akshay Saini 🚀 for guiding this journey Consistency is the real win — Day 8 loading 💪 #LeetCode #DataStructures #Algorithms #CodingJourney #100DaysOfCode #SoftwareEngineering #Programming #InterviewPrep #JavaScript #CodingLife #TechGrowth #ProblemSolving #Developers #LearnToCode #Sorting #BubbleSort #SelectionSort #NamasteDSA