988. Smallest String Starting From Leaf | Tree | DFS | Recursion & Backtracking | Divide & Conquer

In this video, I'll talk about how to solve Leetcode 988. Smallest String Starting From Leaf | Tree | DFS | Recursion & Backtracking | Divide & Conquer | O(n) time ?
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I am Aryan Mittal - A Software Engineer in Goldman Sachs, Speaker, Creator & Educator. During my free time, I create programming education content on this channel & also how to use that to grow :)
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[@ARYANMITTAL]

Please remove your blanket trust from youtubers, always apply your brain too & make sure you learn right things to get your basics strong, else you'll get a dopamine hit of solving a problem which ideally you didn't learn anything from. Adding, whenever i miss any optimal approach, do comment also, that other people watching realise that they are missing something. 🤓
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[@priyanshuawasthi9479]

Bro, You're such a great guy! Now it may seem like you're not getting reach but trust the process. For me you're by far the best DSA guide.

[@Ashutosh07535]

Thanks Aryan i was very Confused About TC but you cleared it very well. Probably the best explanation out there on internet

[@Authenticbharatiya]

Sir hat's off to your dedication for making such good videos. Even though i have been doing dsa for 4 months i still get to know the very basic things from you which i can't find anywhere over the internet. Please sir keep on uploading regularly it really helps me a lot and i don't know about others but i am a regular viewer of your channel. Really grateful to you sir for the efforts you put.

[@-lyxics]

Amazing You deserve my like and Sub

[@zebra-er6xc]

learnings:
// s = s+ ch // takes O(n) time
// s += ch // takes O(1) time
// s.push_back(ch) // also takes O(1) time
// by passing the string as reference we have prevented the copying of string at every stage
// which will save us space and also time used in copying the string ie O(n)
// we will have maximum number of leaf nodes in a complete binary tree, 2 nodes at every node
// maximum number of leaf nodes = N/2 if total number of nodes = N
// Height of balanced binary tree = log2(N) = length of each string
// which gives us maximum time complexity
// worst case time taken to traverse till leaf nodes*time taken to reverse all the strings
// as O(N/2 * logN + N/2(visiting all nodes) ) = O(N*logN)
// space = O(logN) // height of binary tree in a balanced tree
// for worst case space complexity, we will have a skewed tree
// space complexity = O(N)
// and for that time complexity will be O(N) since there is only one leaf node
// and we will reverse the string only once that will be of length N

[@ok-google-run]

thanks for the summary.. And kudos to Aryan for such consistency with great explanation

[@gilfoyle2211]

kaisi jeeb laplapai

[@rajrajesh1669]

This problem is a decent problem but I messed up a lot handling edge cases😢💯.
I tried to solve without using global variable and that was the worst decision I made😢.

[@RohanSharma-tl9wh]

not my white thing 🌚🌚

[@codingkart245]

Well Explained brother! Hats off to your effort!

[@mohdsahil3191]

bro u explained very well

[@the_ved11]

Thanks, Aryan

[@zebra-er6xc]

thanks for the worst case time and space complexity thing

[@dhruvrawatt9]

05:34 🤣

[@atul9203]

apart from your english loved the video and in dept explanation of solution

[@abhinay.k]

5:36 💀

[@-lyxics]

BTW don't eat center fruit gum, cause well you know.

[@Owais486]

bruh, reversing also taking o(n) right ?

[@Owais486]

sorry u later took that part, my bad😅