GFG POTD | Longest Subsequence with Adjacent Diff as 1 | Brute Force to Optimal | C++

Solve *GFG POTD – Longest Subsequence with Adjacent Difference as 1* in C++. Problem Link : https://www.geeksforgeeks.org/problem... Source Link : https://github.com/Krishnkantm/DSA-Co... In this video, we discuss multiple approaches: Brute Force / Recursion Memoization (Top-Down DP) Iterative DP Optimized approach using a Hash Map The goal is to find the longest subsequence where the difference between adjacent elements is exactly 1. *Topics:* Dynamic Programming, Recursion, Memoization, Iterative DP, Hash Map, Subsequence *Complexity of Optimized Approach:* Time: O(n) Space: O(n) If you find the video helpful, Like, Share & Subscribe to *Logic Mode* for more GFG POTD and DSA solutions. #GFG #GFGPOTD #DynamicProgramming #DSA #CPlusPlus