Nitin Yadav’s Post

#Day359 of #1001DaysOfCode 📘 LeetCode Daily Challenge Problem: Walking Robot Simulation (LeetCode 874) 💡 Approach: Simulated the robot’s movement step-by-step while tracking direction changes. Used a HashSet to store obstacle positions for O(1) lookup, ensuring the robot stops correctly when encountering obstacles. Updated the maximum distance from origin after each move. ⏱ Time Complexity: O(n + k) 🧠 Space Complexity: O(m) Staying consistent and improving problem-solving skills every day 🚀 #DSA #Java #LeetCode #ProblemSolving #Coding

#Day358 of #1001DaysOfCode 📘 LeetCode Daily Challenge Problem: Robot Return to Origin (LeetCode 657) 💡 Approach: Simulated the robot’s movement on a 2D plane by tracking vertical and horizontal positions. Each move updates the position, and if the robot returns to (0,0), it means all movements are balanced. ⏱ Time Complexity: O(n) 🧠 Space Complexity: O(1) Consistency is key — one problem every day 🚀 #DSA #Java #LeetCode #ProblemSolving #Coding

🚀 Day 5 of Daily LeetCode Challenge (06 April 2026) Staying consistent and pushing forward 💪 Today’s problem: “Robot Return to Origin” (Easy) 📌 Problem Insight: A robot moves on a 2D plane using steps (U, D, L, R). The goal is to determine whether it returns back to the origin after completing all moves. 🧠 My Approach: Used (x, y) coordinates to track movement Increment/decrement based on direction Final check: (x == 0 && y == 0) → back to origin ✅ ✨ Key Takeaway: Even easy problems strengthen logic and help build consistency — the real game is discipline, not difficulty. 💻 Time Complexity: O(n) 💾 Space Complexity: O(1) 🔥 5 days in — building momentum, one problem at a time! #LeetCode #Day5 #DSA #CodingJourney #Java #ProblemSolving #Consistency #100DaysOfCode

Solved a basic but interesting problem: Robot Return to Origin 🤖 In this problem, a robot moves based on given directions (Up, Down, Left, Right). The goal is to check whether the robot comes back to the starting point (0,0) after completing all moves. 👉 Approach: I tracked the movement using x and y coordinates. Left/Right affects x Up/Down affects y If both x and y become 0 at the end, it means the robot returned to origin. Simple logic, but a good reminder that strong basics are very important 💡 #LeetCode #DSA #ProblemSolving #CodingJourney #Java #Stack #Algorithms #Consistency #Learning #Tech

🚀 Solved LeetCode 2463 – Minimum Total Distance Traveled Today I worked on a challenging Hard-level Dynamic Programming problem that combines Greedy + Sorting + DP with Memoization — a great exercise for strengthening problem-solving skills. 🔍 Problem Insight: We are given positions of robots and factories (with limited capacity). The goal is to assign each robot to a factory such that the total distance traveled is minimized, while respecting capacity constraints. 💡 Key Learnings: Sorting plays a crucial role in optimizing assignments Greedy intuition helps in pairing closest robots with factories Dynamic Programming is essential to explore all valid assignments efficiently Memoization avoids recomputation and improves performance ⚙️ Approach Used: Sort robots and factories based on position Use DP to decide whether to skip or use a factory Try assigning multiple robots (within capacity) and track minimum cost 📈 Complexity: Efficient solution using DP reduces exponential possibilities to manageable computation. 🔥 Takeaway: This problem is a perfect example of how combining multiple concepts leads to an optimal solution. It really helped me improve my thinking around state transitions and optimization strategies. #LeetCode #DataStructures #Algorithms #DynamicProgramming #CodingJourney #ProblemSolving #Java #CompetitiveProgramming

🚀 Day 32 of #LeetCode Journey ✅ Problem: Robot Return to Origin (LeetCode 657) Today’s problem was simple yet a great way to strengthen basic logic and coordinate tracking skills. 🔍 Problem Statement: Given a string of moves (U, D, L, R), determine if the robot returns to the origin after completing all moves. 💡 Approach: * Start from position (0,0) * Track horizontal (x) and vertical (y) movements * Update position based on each move * Finally, check if we are back at (0,0) 🧠 Key Insight: Equal number of opposite moves cancel each other out: * U cancels D * L cancels R ⏱️ Complexity: Time: O(n) Space: O(1) 📌 Takeaway: Even simple problems help build strong fundamentals in problem-solving and coding logic. #Java #LeetCode #CodingJourney #100DaysOfCode #Programming #Developer

𝐃𝐚𝐲 𝟖𝟔/𝟑𝟔𝟓 🚀 📌 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐏𝐎𝐓𝐃: 𝐌𝐢𝐧𝐢𝐦𝐮𝐦 𝐓𝐨𝐭𝐚𝐥 𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐓𝐫𝐚𝐯𝐞𝐥𝐞𝐝 Continuing my 𝟑𝟔𝟓 𝐃𝐚𝐲𝐬 𝐨𝐟 𝐂𝐨𝐝𝐞 journey with a focus on 𝐩𝐫𝐨𝐛𝐥𝐞𝐦-𝐬𝐨𝐥𝐯𝐢𝐧𝐠, 𝐃𝐒𝐀, 𝐚𝐧𝐝 𝐜𝐨𝐧𝐬𝐢𝐬𝐭𝐞𝐧𝐜𝐲. 💪 🔎 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Sort robots and factories by position. Use DP where dp[i][j] represents minimum distance to assign first i robots using first j factories. For each factory, either skip it or assign up to its capacity of robots. Accumulate distances incrementally to avoid recomputation. 🔍 𝐀𝐥𝐠𝐨𝐫𝐢𝐭𝐡𝐦 𝐮𝐬𝐞𝐝: Dynamic Programming + sorting + greedy assignment. ⏱ 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(𝐧 × 𝐦 × 𝐥𝐢𝐦𝐢𝐭) 🧠 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(𝐧 × 𝐦) 📈 𝐊𝐞𝐲 𝐭𝐚𝐤𝐞𝐚𝐰𝐚𝐲: Sorting simplifies matching problems, and DP helps explore optimal assignments under constraints. #LeetCode #LeetCodeDaily #365DaysOfCode #DSA #Java #DynamicProgramming #Greedy #Sorting #ProblemSolving #LearningInPublic 👨💻 🔗 Problem link in comments 👇

𝐃𝐚𝐲 𝟕𝟓/𝟑𝟔𝟓 🚀 📌 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐏𝐎𝐓𝐃: 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐖𝐚𝐥𝐥𝐬 𝐃𝐞𝐬𝐭𝐫𝐨𝐲𝐞𝐝 𝐛𝐲 𝐑𝐨𝐛𝐨𝐭𝐬 Continuing my 𝟑𝟔𝟓 𝐃𝐚𝐲𝐬 𝐨𝐟 𝐂𝐨𝐝𝐞 journey with a focus on 𝐩𝐫𝐨𝐛𝐥𝐞𝐦-𝐬𝐨𝐥𝐯𝐢𝐧𝐠, 𝐃𝐒𝐀, 𝐚𝐧𝐝 𝐜𝐨𝐧𝐬𝐢𝐬𝐭𝐞𝐧𝐜𝐲. 💪 🔎 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Sort robots and walls to process in order. For each robot, compute valid left and right intervals. Convert intervals into indices using binary search. Use DP to decide whether to go left or right, while avoiding double-counting overlapping walls. 🔍 𝐀𝐥𝐠𝐨𝐫𝐢𝐭𝐡𝐦 𝐮𝐬𝐞𝐝: Sorting + Binary Search + Dynamic Programming (interval optimization). ⏱ 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(𝐧 log 𝐧) 🧠 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(𝐧) 📈 𝐊𝐞𝐲 𝐭𝐚𝐤𝐞𝐚𝐰𝐚𝐲: When multiple choices overlap, DP helps track optimal decisions while preventing double counting. #LeetCode #LeetCodeDaily #365DaysOfCode #DSA #Java #DynamicProgramming #BinarySearch #Greedy #ProblemSolving #LearningInPublic 👨💻 🔗 Problem link in comments 👇

🚀 Day 7 of My LeetCode Journey Today’s problem: Reverse Integer At first glance, it looks simple — just reverse digits. But the real challenge was handling 32-bit integer overflow without using 64-bit storage. 🔍 What I learned: How to extract digits using modulo (%) Building the reversed number step by step Most importantly, checking overflow before updating the result ⚠️ Edge Case: If the reversed number goes beyond the 32-bit range [-2³¹, 2³¹ - 1], we must return 0 💡 Example: Input: 123 → Output: 321 Input: -123 → Output: -321 Input: 1534236469 → Output: 0 (Overflow case) Consistency over perfection 💪 #Day7 #LeetCode #DSA #CProgramming #CodingJourney #ProblemSolving

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