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2791. Count Paths That Can Form a Palindrome in a Tree | O(N^3) - O(N*N) - O(N) |Leetcode Weekly 355

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Описание к видео 2791. Count Paths That Can Form a Palindrome in a Tree | O(N^3) - O(N*N) - O(N) |Leetcode Weekly 355

Segment Trees Series - https://bit.ly/segment-trees
Part 1 -    • Segment Trees Series | Basics of Segment T...
Part 2 -    • Segment Trees Series | Queries on prefix o...
Part 3 -    • Segment Trees Series | Part 3 | Solving Co...
Part 4 -    • Segment Trees Series | Part 4 | Solving Co...
Part 5 -    • Segment Trees Series | Part 5 | Longest In...
Part 6 -    • Segment Trees Series | Part 6 | Solving Sp...
Part 7 -    • Segment Trees Series | Part 7 | Binary Sea...
Part 8 -    • Segment Trees Series | Part 8 | Binary Sea...
Part 9 -    • Segment Trees Series | Part 9 | Merge Sort...

Segment Trees Problems Playlist -    • Segment Trees

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Contest Link - https://leetcode.com/contest/weekly-c...
Problem Link - https://leetcode.com/contest/weekly-c...

Solution - https://leetcode.com/problems/count-p...

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Timestamps -

00:00 - Agenda
00:30 - Problem Description
03:40 - Brute force solution (N*N*N)
08:25 - Intuition to optimize the solution (N*N*logN)
14:20 - Pseudo code
15:00 - Optimizing the solution further by removing LCA in O(N*N)
17:30 - Intuition to optimize the solution
18:37 - Finding # of even palindromes in O(N*C)
24:15 - Pseudo code (Even length palindromes)
26:20 - Finding # of odd length palindromes in O(N*C)
33:00 - Dry run of the Algorithm
35:30 - Every odd length palindrome pair are counted twice
36:20 - How to implement it cleanly (using decimal instead of binary array)
38:38 - Pseudo code & Time Complexity
40:46 - Code Walkthrough

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Interview Experiences Playlists -

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Amazon -    • Amazon Interview Questions
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