Anurag Singh’s Post

🚀 Day 88 of #100DaysOfLeetcode 🧩 Problem: Next Smaller Element You are given an array of integers, and for each element, you need to find the next smaller element to its right. If none exists, return -1. 💡 Example: arr = [2, 3, 1] → Output: [1, 1, -1] Because for 2 → next smaller is 1, for 3 → next smaller is 1, and for 1 → none smaller → -1. 🧠 Brute Force Approach Logic: For every element, traverse the elements to its right and find the first smaller one. Code Complexity: ⏱ Time Complexity: O(n²) – two nested loops 💾 Space Complexity: O(1) – no extra space used ⚙️ Optimal Approach (Using Stack) Logic: We iterate from right to left while maintaining a stack to keep track of potential smaller elements. Pop elements from the stack while the top is greater or equal to the current element. The top of the stack after popping gives the next smaller element. Push the current element into the stack for future comparisons. Code Complexity: ⏱ Time Complexity: O(n) – each element is pushed and popped once 💾 Space Complexity: O(n) – stack used for storing elements 🧮 Key Learnings: Stack-based problems often involve processing from right to left. Understanding Monotonic Stack pattern helps solve many similar problems like “Next Greater Element” and “Stock Span”. Helps in real interview scenarios where time optimization is key. ⚠️ Edge Cases: All elements in increasing order → Output will be all -1s. All elements in decreasing order → Each next element will be the answer for previous one. Single element array → Output should be -1. 💬 Final Thought: Learning the concept of Monotonic Stack not only simplifies the solution but also builds intuition for multiple real-world stack-based interview questions. 🔥 #100DaysOfCode #Leetcode #DSA #Stack #JavaScript #CodingChallenge #ProblemSolving #TechCommunity #LearningNeverStops #ProgrammingJourney #CodeNewbie #FrontendDeveloper #DataStructuresAndAlgorithms

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