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🔥 Day 56 of my LeetCode Daily Challenge! Today we solve LeetCode #300 — Longest Increasing Subsequence (LIS), one of the most iconic Dynamic Programming problems in competitive programming and coding interviews — solved using Bottom-Up DP in JavaScript! ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 📌 PROBLEM: Longest Increasing Subsequence 🔗 https://leetcode.com/problems/longest... 📊 Difficulty: Medium 🗂️ Topics: Array, Dynamic Programming, Binary Search ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 💡 WHAT YOU'LL LEARN: ✅ What a subsequence is and why it differs from a subarray ✅ Why we traverse RIGHT to LEFT and look ahead at j greater than i ✅ Initializing LIS[i] = 1 (every element is a subsequence of length 1) ✅ Recurrence: LIS[i] = Math.max(LIS[i], 1 + LIS[j]) when nums[i] less than nums[j] ✅ Why we take the max of ALL LIS[i] values as the final answer ✅ Full Time & Space Complexity Breakdown → Time Complexity: O(n²) → Space Complexity: O(n) ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 🔑 KEY INSIGHTS: 💡 LIS[i] = length of the LIS starting AT index i — so we fill from RIGHT to LEFT because LIS[i] depends on future values LIS[j] (where j greater than i)! 💡 We initialize ALL cells to 1 because every single element is trivially a valid increasing subsequence by itself! 💡 We must scan the ENTIRE LIS[] array for the max answer because the longest subsequence can START at any index, not just index 0! This is the classic LIS gotcha! 🚀 💡 Follow-up challenge: Can you solve it in O(n log n) using Binary Search + Patience Sorting? Try it next! 🔥 ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 📚 MY LEETCODE JOURNEY PLAYLIST: • Leetcode 🤝 CONNECT WITH ME: • GitHub: https://github.com/ashwinikemshetty • LinkedIn: / ashwinikemshetty 📅 #100DaysOfCode #LeetCodeChallenge 🌟 If this helped you, LIKE 👍, COMMENT 💬 & SUBSCRIBE 🔔! ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ #LeetCode #JavaScript #DSA #Day56 #LongestIncreasingSubsequence #LeetCode300 #CodingChallenge #ProblemSolving #DynamicProgramming #DP #LIS #100DaysOfCode #JavaScriptDeveloper #DataStructures #Algorithms #TechInterview #CodingInterview #BottomUpDP #RightToLeft #Subsequence