Maximize the Number of Partitions After Operations | Leetcode Weekly Contest 379 | Leetcode 10038
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This is the 81st Video on our playlist "Dynamic Programming : Popular Interview Problems".
In this video we will try to solve a very good Digit DP problem - Maximize the Number of Partitions After Operations (Leetcode 10038).
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 : Maximize the Number of Partitions After Operations
Company Tags : Will update soon
My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/DP/Digit%20DP/Maximize%20the%20Number%20of%20Partitions%20After%20Operations.cpp
Leetcode Link : https://leetcode.com/problems/maximize-the-number-of-partitions-after-operations
My DP Concepts Playlist : https://youtu.be/7eLMOE1jnls
My Graph Concepts Playlist : https://youtu.be/5JGiZnr6B5w
My GitHub Repo for interview preparation : https://github.com/MAZHARMIK/Interview_DS_Algo
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Approach Summary : The solve function recursively calculates the maximum number of partitions based on the current index i, a bitmask uniqueChars representing unique characters encountered so far, and a boolean flag canChange indicating whether a character change is allowed. The function employs a memoization technique using an unordered map mp to store and retrieve previously computed results. The main logic involves iterating through the characters of the string and determining the number of unique characters encountered. If the count exceeds the given limit k, a character change is allowed, and the function is called recursively with a new set of unique characters. The result is updated based on the maximum value obtained from different recursive calls. The maxPartitionsAfterOperations function initializes the required parameters and starts the recursion from the beginning of the string, ultimately returning the maximum number of partitions possible plus one.
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✨ Timelines✨
00:00 - Introduction
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