Move Pieces to Obtain a String | Brute Force | Wrong | Optimal | Leetcode 2337 | codestorywithMIK
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This is the 52nd Video of our Playlist "Strings" by codestorywithMIK
In this video we will try to solve a good string problem : Move Pieces to Obtain a String | Brute Force | Wrong | Optimal | Leetcode 2337 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Move Pieces to Obtain a String | Brute Force | Wrong | Optimal | Leetcode 2337 | codestorywithMIK
Company Tags : will update later
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/strings/Move Pieces to Obtain a String.cpp
Leetcode Link : https://leetcode.com/problems/move-pieces-to-obtain-a-string/description/
My DP Concepts Playlist : • Roadmap for DP | How to Start DP ? | ...
My Graph Concepts Playlist : • Graph Concepts & Qns - 1 : Graph will...
My Recursion Concepts Playlist : • Introduction | Recursion Concepts And...
My GitHub Repo for interview preparation : https://github.com/MAZHARMIK/Interview_DS_Algo
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Summary :
Approach 1: Recursive Backtracking with Memoization
Idea: Uses recursive backtracking to explore all possible valid transformations from the start string to the target string by swapping adjacent characters 'L' and 'R' with underscores (_), with memoization to avoid recomputation.
Time Complexity: Exponential in the worst case due to exploring all possible swaps, though memoization reduces redundant computations.
Space Complexity: High, due to the recursive call stack and memoization map.
Approach 2: Two-Pointer Greedy Approach
Idea: Uses two pointers to simultaneously traverse the start and target strings, ensuring that characters L and R appear in valid positions relative to their movement constraints (L can only move left, and R can only move right).
Time Complexity: Linear, i.e., O(n), since each character is processed once.
Space Complexity: Constant, i.e., O(1), since it uses only two pointers and no additional data structures.
✨ Timelines✨
00:00 - Introduction
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