Freedom Trail | Recursion | Memoization | Bottom Up | Detailed | Leetcode 514 | codestorywithMIK
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This is the 94th Video of our Playlist "Dynamic Programming : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a very good DP problem based on Strings : Freedom Trail | Recursion | Memoization | Bottom Up | Detailed | Leetcode 514 | 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 : Freedom Trail | Recursion | Memoization | Bottom Up | Detailed | Leetcode 514 | codestorywithMIK
Company Tags : Google
My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/DP/DP%20on%20Strings/Freedom%20Trail.cpp
Leetcode Link : https://leetcode.com/problems/freedom-trail/
My DP Concepts Playlist : https://youtu.be/7eLMOE1jnls
My Graph Concepts Playlist : https://youtu.be/5JGiZnr6B5w
My Recursion Concepts Playlist : https://www.youtube.com/watch?v=pfb1Zduesi8&list=PLpIkg8OmuX-IBcXsfITH5ql0Lqci1MYPM
My GitHub Repo for interview preparation : https://github.com/MAZHARMIK/Interview_DS_Algo
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Approach Summary :
Approach-1:
This approach employs recursion with memoization to find the minimum number of steps required to spell out a given key string from a ring of characters. Without memoization, the time complexity is \( O(n^m) \), where \( n \) is the length of the ring string and \( m \) is the length of the key string. However, with memoization, the time complexity improves to \( O(n^2 * m) \), where \( n \) represents the length of the ring and \( m \) represents the length of the key. The space complexity remains constant at \( O(101*101) \) or simply \( O(1) \).
In summary, this approach leverages recursion with memoization to efficiently compute the minimum steps needed.
Approach-2:
This approach utilizes a bottom-up dynamic programming approach to solve the same problem. It also computes the minimum number of steps required to spell out the key string from the ring. The time complexity is \( O(n^2 * m) \), where \( n \) is the length of the ring and \( m \) is the length of the key. The space complexity is \( O(n*m) \), which is proportional to the product of the lengths of the ring and the key.
In summary, this approach builds a dynamic programming table to store the minimum steps required and iterates through the table to compute the final result.
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✨ Timelines✨
00:00 - Introduction
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