Extra Characters in a String | Recursion | Bottom Up | Easy | Leetcode 2707 | codestorywithMIK
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This is the 101st Video of our Playlist "Dynamic Programming : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a classic DP problem : Extra Characters in a String | Recursion | Bottom Up | Easy | Leetcode 2707 | 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 : Extra Characters in a String | Recursion | Bottom Up | Easy | Leetcode 2707 | codestorywithMIK
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My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Intervie...
Leetcode Link : https://leetcode.com/problems/extra-c...
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My Recursion Concepts Playlist : • Introduction | Recursion Concepts And...
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Summary :
The two approaches to solving the problem of minimizing extra characters in a string when compared with a dictionary of words are:
Recursion + Memoization:
This approach uses recursion to explore every substring starting from each index and checks if it matches any word in the dictionary.
Memoization is applied to store the results of subproblems in an array to avoid redundant calculations, reducing the overall time complexity.
Time complexity is O(n3) due to substring generation, and space complexity is O(n+k), where k is the total number of characters in the dictionary.
Bottom-Up Dynamic Programming:
This approach builds the solution iteratively from the end of the string to the beginning. A DP array stores the minimum extra characters needed for substrings from each index.
By considering every possible substring from each index, the minimum extra characters are computed in a bottom-up manner.
It has the same time complexity O(n3), but the space complexity is O(n) since it only uses an array of size n+1.
✨ Timelines✨
00:00 - Introduction
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