Vinit Gupta

Let's discuss the 4th problem of today's LeetCode Weekly Contest 414 ... It personally took me a lot of time to optimize it before getting AC but the problem was really good... Concepts => BFS + BitMaskDP Only reason I could think of BitMask was because the size of positions vector was atmax 15 so each mask can ideally represent the state of the board. Now we also need a state to show whose turn it is which can be 0/1 where 0 represents it's Alice's turn and 1 represents it's Bob's turn. We also need information of current position of Knight. Now your Knight can move across different boxes but we just need the information when the Knight is at the position where it kills any pawn because that's when the other player will take it's turn so it will be at max 16 boxes where 15 can be those from positions vector and 1 of the initial position of Knight. We will precompute the minimum distances between any two positions in grid (here min moves) using BFS or Dijkstra. Knight can move at max 8 adjacent positions . Also we don't need to find this for each box of grid , just find it for those 16 boxes which are of our concern. Now we can define the state of DP : dp[ind][mask][0] => minimum number of moves needed to kill all pawns which are remaining (the ones whose bit is set in mask) if we are standing at {x,y} => {position[ind][0], position[ind][1]} place in grid. dp[ind][mask][1] => maximum number of moves needed to kill all pawns which are remaining (the ones whose bit is set in mask) if we are standing at {x,y} => {position[ind][0], position[ind][1]} place in grid. State Transitions : We can kill any pawn and check for further states. Let's say our current Mask is : "mask", current index is : "ind" current turn = "turn" // turn = 1 for alice and turn = 0 for bob. if(mask==0) return 0; // all pawns are killed if(dp[ind][mask][turn]!=-1) return dp[ind][mask][turn]; int ans = (turn ? -1 : 1e10); for(int i = 0;i<15;i++) { if(mask&(1<<i)) => this means the pawn at {positions[i][0],positions[i][1]} is alive so we can kill this one and get answer for further state which is : int newMask = mask& (~(1<<i)); // unset i'th bit int nextStateAns = solve(i,newMask, !turn); int currReqMoves = dist[ind][positions[i][0]][positions[i][1]]; if(turn==1) ans = max(ans, currReqMoves + nextStateAns); else ans = min(ans, currReqMoves + nextStateAns); } return dp[ind][mask][turn] = ans; You can set index of {kx,ky} as positions.size() ; Answer => dp[index of {kx,ky} ][all bits set][1]; Try to figure out the time and space complexity for this approach and mention it in comments below. Optimizations were required here just to be able to fit into ideal number of operations which are 1e7 or 1e8 at max in one second. My submission : https://lnkd.in/dAM9cpb4 #leetcode #leetcodecontest #codingcommunity #weeklycontest #competitiveprogramming #codingchallenge #cppprogramming #cpp #contest #approach #solutions #datastructures #algorithms

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If you're a frontend developer and looking for a job change then don't learn serious DSA. .. .. If, 1. You're target company is not premium product based company like Google/Microsoft etc. 2. Many startups and mid range companies also ask DSA but you don't have to solve 400+ leetcode problems for that. 3. Just by studying array, string, linked list questions of easy and medium difficulty you can crack them. 4. Along with basic DSA focus more on machine coding, web optimization etc. Which will have more weightage for you in the interview. open to hear your thoughts on how much DSA is important for frontend developer. Follow me and I will help you to find your dream job: Vasanth Bhat

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Google offers a best CTC + Good WLB to freshers and working professionals for their SDE Role! In October , multiple working professionals and 2025-2026 grads under-going Google SWE Interviews. Google is one of those companies which asks interesting and challenging 𝐃𝐒𝐀 questions to their candidates in interview rounds; this year they asked total of many different and all of them are a super hard nut to crack - mostly revolving around high level Dynamic Programming + Graphs + Binary Search + Greedy :) Generally students face issue to crack OA/Interview rounds of top-tech companies because of their relying on standard DSA sheets and only doing 200-500 easy standard DSA problems - to tackle OA/interview - you need to develop critical thinking and problem solving skills and be ready to tackle new hard and unseen DSA questions! Google SDE-3 Interview DSA Problem asked in October - Name of the Problem - 𝐏𝐚𝐫𝐭𝐢𝐭𝐢𝐨𝐧𝐬. Crux of the problem :- In the image below. Solution Idea - DP + Greedy We will solve this problem in our upcoming FAANG OA Training - Master-class on Sunday 𝐉𝐨𝐢𝐧 𝐨𝐮𝐫 850 𝐡𝐨𝐮𝐫 𝐜𝐨𝐮𝐫𝐬𝐞 𝐨𝐧 𝐃𝐒𝐀 + 𝐎𝐀 + 𝐂𝐏 𝐰𝐢𝐭𝐡 8 𝐦𝐨𝐧𝐭𝐡 𝐥𝐢𝐯𝐞 𝐭𝐫𝐚𝐢𝐧𝐢𝐧𝐠 + 𝐩𝐞𝐫𝐬𝐨𝐧𝐚𝐥 𝐦𝐞𝐧𝐭𝐨𝐫𝐢𝐧𝐠 + Referral support - https://lnkd.in/dgJtAPBG (for working professionals /college students.) Link to our FAANG OA + DSA Training group in the comments.

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I stopped learning MongoDB . It's not working and won't. I have been learning MERN stack from Shradha Khapra mam and about to complete the Airbnb clone, suddenly to my surprise MongoDB stopped supporting on my windows. and guess What ? it became a reason to quit the journey. for a change I did not accept it and got into learning React that was doing good on my Laptop. Why MongoDB stopped? The reason is my laptop. I have been using a laptop that is given by government of Tamilnadu Nadu seven years back. But yes the passion for tech I have never gets down for such reasons. ps:Which laptop do you own? #MongoDb #tech #React #learning #srinivastechjourney

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Learnings from Budget 2024 : - have less/no hopes from the govt - keep working harder to beat inflation - hustle & earn more money to afford a good lifestyle Incase you're a Frontend Engineer and want to learn JavaScript, React or Frontend System Design, then checkout NamasteDev.com and upskill yourself today! Upskill Yourself --> Keep Hustling --> Make more money YoY 🚀

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DSA DSA DSA That's what the top 3 key points were, an upcoming Google SWE told me yesterday, on how to crack a FAANG company. DSA may be the optimal choice for many recruiters to test the interviewee's logical and programming skills. Well of course, if you have strong connections and a referral from the existing Google employee, your chances increase by a milestone. I have never touched DSA. That came as a shock to me, when he enlightened upon the fact that it is the most important. Just intermediate level DSA is enough. One who has solved 200-300 LeetCode questions would be able to crack the Google tech interview. Well, I gained a bunch of insights from talking with Sai Sree Ram Putta last night. He gave some really positive thoughts on the way to grabbing a FAANG internship, or a full time offer. Do check Sai's profile and schedule a meeting with him. #google #faang #maang #linkedin #connections #networks #networking #connecting #interview #internships #insights #linkedininsights #microsoft #referrals #growth #career #careertips #leadership #LearnandConnect #TechInsights #LearningJourney #TechTalks

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After Solving 300 Questions on LeetCode, Here's What I Have Learned: 1. Limit Focus on Easy Problems:   - Use easy problems primarily when learning a new data structure or algorithm.   - Move on to more challenging problems once you grasp the basics. 2. Prioritize Medium Problems:   - Concentrate more on medium problems initially.   - Don't worry too much about hard problems when you're starting out. 3. Time Management and Solution Review:   - Spend 15-25 minutes on a problem. If you can't find a solution, check the provided solutions or watch YouTube explanations.   - Dry run the solutions to understand them better. 4. Consistency:   - Solve problems daily to avoid procrastination and build a regular habit of problem-solving. 5. Efficient Practice:   - After learning a new topic, solve no more than five problems related to it. You'll naturally encounter similar problems later, which will test your understanding effectively. 6. Review and Refine Solutions:   - Mark the problems you solved with help. Revisit them after 3 to 7 days to rewrite the solutions more efficiently and optimally. ------------------------- if you you follow these steps you will definitively become a good problem solver . All the best coders #leetcode #problemSolving

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Let's talk about today's LeetCode Weekly Contest 414. Questions 1 - Just convert into binary. Submission Link: https://lnkd.in/g_dw-wkG Question 2 - Whenever we have to find the maximum of minimum or minimum of maximum, then it can be easily solved using a binary search on the answer. We will be using the same approach to get the maximum possible score. Submission Link: https://lnkd.in/g9EaYNpn Question 3 - We will be using the next greater element concept here and we will be finding the next greater element index. Submission Link: https://lnkd.in/gmpnweFg Question 4 - As the constraints are smaller, we will try the brute force approach. We will be generating the minimum steps needed from a position to all other cells. Then we can apply the DP approach to get the maximum total number of moves. Submission Link: https://lnkd.in/g3R38TB6 Follow 🤝 : Prateek Goel #leetcodechallenge #rank #linkedin #coding #learning #contest #biweekly #cp #dsa #faang #sde

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Very Important Interview Question For Every SDE What is the difference between ApiKeys and Token? Tokens (such as JWT - JSON Web Tokens): 1. Include user information and entitlement forms for the purpose of performing authorizations or authentications. 2. Often contains user identification, the rights of the user, and the time limit after which the user cannot be identified as a right-holder, usually in JWT. 3. Contains specific details of the user and is issued by the authentication server, after they have logged in. API Keys: 1. They are mainly used to authenticate the application or the consumer making the API call. 2. Usually a long string and part of the request header or query parameters of the API call. 3. Operated in the absence of the user context such as making public API calls or in case of service-to-service calls. 4. Long-term in nature and are generated through the console of the API vendor or administrator. In a more simple case: API keys – tell who is using the application. Tokens - are in charge of active sessions of users, their rights and surrounding information. So do you think a marriage card is a Token or just a key? 😅 For more such content, Please follow Kunal Arora #systemdesign #design #lowleveldesign #highleveldesign #hld #lld #interview #faang #jobs #dsa

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Let's talk about today's LeetCode Weekly Contest 400. Question 1 - Just take a counter, increase for E and decrease for L, and take maximum. Submission Link: https://lnkd.in/geRSp2SJ Question 2 - I used merge intervals for merging and after calculating the difference. Submission Link: https://lnkd.in/gKXcryCp Question 3 - It is a simple heap, that stores characters and its index. Submission Link: https://lnkd.in/gdNDuCnf Question 4 - I used a sliding window for this, and for calculating the AND of a range(i,j), I used segment tree. Submission Link: https://lnkd.in/g_MV9JyH Follow 🤝 : Prateek Goel #leetcodechallenge #globralRank #weekly #rank #learning #linkedin #coding #learning #contest #biweekly

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👩🏼‍💻Progressing further to #Day8 of the 160-day #DSA challenge by GeeksforGeeks Today's task was "Stock Buy and Sell – Max One Transaction Allowed", a problem that required strategic thinking and careful array analysis. Tackling such scenarios not only sharpens problem-solving skills but also brings a sense of accomplishment. 📊 Every step forward is a step toward #growth. Let's keep the #momentum going! 🚀 GeeksforGeeks #gfg160 #womenintech #geekstreak2024

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Ask yourself this everyday... Who checked on you? - When you were grinding DSA. - When you were stuck in bugs/errors. - When you were constantly visiting StackOverflow for your development optimizations. - When you were frustrated & regretted over your life by looking at current scenario of layoffs. - When you were in doubts "ki yaar mera kuch hoga bhi ya nhi 🥺." Who was there? Other than you & your family tell me? Remember, Hard times is what teaches you the reality of this world. Whole world will be there with you once you will be successful. Nobody f****ing cares where are you right now? So why would you stop? #keepgrinding #makeithappen Follow Deepak Manglani for more! #dsa #development #motivation #recession #hustle #success

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#Day 3: FrontEnd Interview QNA Hooks Today, I learned Interview Quation of React: (Repeate the Counter Inc Task). Scenario : Counter ke bad pucha jata he ye qna : It repeate the setCounter function 4 times In the useState Hook (And ask what is value after click) Answer is : We learn the concept Fiber so it is diffing algorithem it send the updates in batches so it is update and create only one batche of reapeted task ( In this case setCounter is Repeated Task ) Solution ( To work ): Directly call the new callback so It is execute after compleating previous Code : setCounter((prevCounter) => prevCounter + 1) setCounter((prevCounter) => prevCounter + 1) setCounter((prevCounter) => prevCounter + 1) Thanks Hitesh Choudhary sir For teaching Learn by doing method input : step 1: click on the increament counter button step 2 : See the result like counter = 15 (Initialy) After clicking btn useState propogate changes 3 times So, counter = 18

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LeetCode Knight Badge Progress – Post #3 🌟 Hi guys, In Weekly Contest 422, I managed to solve three questions getting a rank of 2759, and my rating finally crossed the 1700 mark 🎉. Here’s a summary of each question and how I approached them: Q1. Easy – This was straightforward, with basic logic that most could handle quickly. Q2. Medium – This problem involved finding the shortest path from the cell (0, 0) to (n-1, m-1) in a grid. Recognizing it as a shortest-path problem, I implemented Dijkstra's algorithm using a priority queue. The priority queue stored i, j, and time values to track each cell's position and time cost. The traversal followed these steps: - Starting from cell (0,0), I moved in all four directions: up, down, left, and right. - For each move, I calculated newTime as Math.max(currTime, moveTime[nRow][nCol]) + 1, where nRow and nCol represented the next cell to move to. - This approach ensured that each move's time was incremented by 1, reflecting the minimum time required to reach that cell. - The minimum time to reach the destination was recorded in vis[n-1][m-1]. Q3. Medium – This was a variation of Q2, with a slight twist. Instead of adding 1 second for each move, the time alternated between 1 and 2 seconds with each step. For example, moving from (0,0) to (0,1) added 1 second, then moving from (0,1) to (1,1) added 2 seconds, and so on. To handle this, I: - Created a 3D matrix for distances (vis) with the third dimension of size 3 (can be done with size 2), to account for alternating times. - Initialized the priority queue to store alternating time costs as I traversed the grid. - Modified the traversal logic from Q2 to use alternating time increments (1, then 2, and so on). - At the end, I calculated the answer by taking the minimum across dimensions: Math.min(vis[n-1][m-1][0], vis[n-1][m-1][1], vis[n-1][m-1][2]) [Can be done finding the min of just index 0, 1 of the alternating time]. This contest was a great challenge and crossing 1700 is a huge motivation boost! #leetcodecontest #leetcode #codingjourney #codingcommunity #weeklycontest #competitiveprogramming #codingchallenge #java #contest #intuition #solutions #leetcodeweeklycontest #competitivecoding #dsa #datastructuresandalgorithms #algorithms

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I know it's 2024 end almost and thse days people are solving leetcode problems massively, but, if you are starting with #DSA as a newbie, you still need to learn basic pattern printing, swap 2 numbers without 3rd variable such kind of easy question. Similarly, no matter how fancy Spring framework sounds, you still need to learn core java i.e. #J2SE, print odd even number using 2 threads in #Java like basic topics or questions before jumping on to frameworks. Remember, in learning path, if you skip initial steps, you will keep on struggling, so, decide properly and execute properly, game of patience, no shortcut 🙂 Learning from scratch requires path, not a direct jump, things don't work well in the long run. #dsa #java #learningcurve #softwareengineering

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🚀 **Cracking the Code: Solving LeetCode Problem 2181 - Merge Nodes in Between Zeros** Hey LinkedIn Network, I wanted to share an interesting problem I tackled recently on LeetCode: **"2181. Merge Nodes in Between Zeros"**. This problem provided a fantastic exercise in linked list manipulation and problem-solving strategy. ### Problem Overview The challenge is to merge nodes in a linked list that are between zero-value nodes. Each segment of nodes between zeros should be summed up, and the zeros themselves act as delimiters. The goal is to transform the list such that only the sum of each segment remains, with the rest of the nodes removed. ### Key Learnings and Approach - **Understanding Linked Lists**: This problem required a solid grasp of linked list operations—especially traversing and manipulating node pointers. - **Pointer Manipulation**: Efficiently using two pointers (slow and fast) allowed for seamless merging of nodes and accurate summation of values between zeros. - **Memory Management**: Ensuring proper deletion of unused nodes is crucial to prevent memory leaks, a fundamental aspect when working with dynamic data structures in C++. ### Approach Breakdown 1. **Initialization**: Start with two pointers: `slow` for placing the summed values and `fast` for traversing through the list. 2. **Summation Loop**: Traverse the list, summing up values between zeros. 3. **Node Adjustment**: Once a zero is encountered, place the sum in the `slow` pointer's node, move the `slow` pointer forward, and reset the sum. 4. **Cleanup**: Properly handle the end of the list by deleting unused nodes to maintain efficient memory usage. This problem was a great reminder of the importance of understanding underlying data structures and efficient memory management in C++. Tackling such problems not only sharpens coding skills but also enhances logical thinking and problem-solving abilities. Feel free to reach out if you're interested in discussing the solution further or have any insights on optimizing linked list operations. Happy coding! 💻✨ #LeetCode #CodingChallenge #LinkedLists #ProblemSolving #SoftwareEngineering #TechSkills #MemoryManagement #DataStructures

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Top-7 creators I follow for tech insights. (& interview preparation) • Akshay Saini 🚀 - For his most practical interview tips. • Anuj Saklani - His inspiring leetcode journey, and motivation. • Aanshul Sadaria - For in depth analysis of the greatest technologies. • Gunjan Madaan - Explaining everything with 3 pointers. Short, crisp and precise. • Hina Arora - For the perfect personal branding tips for developers. • Revanth Murigipudi - For practical advice about switching to a PBC. • raj vikramaditya (Striver) - Not just for content, but for motivation of how much a single person can achieve. Who's your favourite?

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