Task Scheduler | Using Greedy Only | No Heap | Leetcode 621 | codestorywithMIK
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This is the 30th Video of our Playlist "Greedy : Popular Interview Problems".
In this video we will try to solve a very good and famous heap problem : Task Scheduler | Using Greedy Only | No Heap | Leetcode 621 | codestorywithMIK
NOTE - I already made a video on this using Heap (See my Heap Playlist)
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 : Task Scheduler | Using Greedy Only | No Heap | Leetcode 621 | codestorywithMIK
Company Tags : META
My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Greedy/Task%20Scheduler.cpp
GfG Problem Link : https://leetcode.com/problems/task-scheduler/
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 :
The provided solution aims to find the minimum time required to execute a list of tasks, given a cooling period `p` between two same tasks. It does so by first calculating the frequency of each task, then determining the number of chunks and idol spots needed to accommodate the tasks. Chunks represent intervals between identical tasks, and idol spots represent spaces that need to be filled within these chunks. If the idol spots are not sufficient to accommodate all tasks, additional spots are added, resulting in a longer execution time. If there are enough idol spots, the execution time equals the number of tasks. Finally, the solution returns the minimum time required for task execution.
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✨ Timelines✨
00:00 - Introduction
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