String Stack | Interesting Problem | Complete Intuition Explained

Connect with me: https://www.linkedin.com/in/sanyam-jain-229052250/ OA Series playlist: https://www.youtube.com/playlist?list=PLeh6VbPaYLX-zusD0HrwH9a3Nm9YdAOVA Hello Coders! In today's video, we're tackling a fascinating string problem that challenges us to build a target string ('tar') from a pattern string ('pat') using special append and delete operations. It seems complex at first, but we'll uncover a surprisingly elegant and efficient solution using a clever two-pointer trick. The Problem You are given two strings, 'pat' and 'tar'. For each character in 'pat', you can either append it to a new string 's' or delete the last character of 's'. Can you make 's' equal to 'tar' after using every character in 'pat' exactly once? The Intuition: Think Backward! A forward approach can get confusing quickly. The magic happens when you start from the END of both strings! Here's the core logic: 🔹 An 'append' operation uses one character from 'pat' to form one character of 'tar'. 🔹 A 'delete' operation effectively cancels out a previous append. This means a 'delete' consumes TWO characters from 'pat'—the character causing the delete, and the character that was previously appended. By iterating backward, the logic becomes simple: If `pat[i]` matches `tar[j]`, great! We've found the character that forms the end of our target. We move on to the next characters (`i--`, `j--`). If `pat[i]` DOES NOT match `tar[j]`, this `pat` character must have been part of a cancelled operation. So, we skip it and the character before it (`i -= 2`). We'll walk through the C++ code line-by-line to see how this powerful idea translates into a clean and optimal solution. #GreedyAlgorithm #ProblemSolving #Optimization #stack #CodingInterview #DataStructuresAndAlgorithms #gfgstreek #Cplusplus #Programming #techinterview #gfg #gfgpotd #gfgpotdtoday #stringstack