Maximize Subarrays After Removing One Conflicting Pair | Detailed Explanation | Leetcode 3480 | MIK
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Hi Everyone, this is the 46th video of our Playlist "Greedy : Popular Interview Problems".
Now we will be solving a very good Greedy based Problem - Maximize Subarrays After Removing One Conflicting Pair | Detailed Explanation | Leetcode 3480 | codestorywithMIK
I will explain it in full detail so that it becomes easy to understand. Each line will be explained and you will know the WHY behind everything.
Problem Name : Maximize Subarrays After Removing One Conflicting Pair | Detailed Explanation | Leetcode 3480 | codestorywithMIK
Company Tags : will update later
Code Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Greedy/Maximize Subarrays After Removing One Conflicting Pair.cpp
Leetcode Link - https://leetcode.com/problems/maximize-subarrays-after-removing-one-conflicting-pair
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Video Summary :
We want to count all valid subarrays [l, r] where no conflicting pair (a, b) exists such that l less than a less than b less than equal to r. For each r, we track the maximum conflicting a values (maxConflictStart, secondMaxConflictStart) that limit how far back l can go. We compute the number of valid subarrays ending at each r as r - maxConflictStart. Separately, we calculate the potential gain if we remove the most restrictive conflict, and add the best such gain to our total. This gives the maximum possible count with one conflict removed.
✨ Timelines✨
00:00 - Introduction
0:15 - Motivation
0:52 - Problem Explanation
5:00 - Thought Process
17:06 - Good Example Dry Run
22:58 - Important Observations
27:05 - Complete Dry Run
45:09 - Coding it up
53:29 - Bonus Tip
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