Valid Parentheses (LeetCode 20) | Full solution with visuals and animations | Stack Data Structure

[@arslanmuhammad4190]

Love You Sir, Got A Internship. 6 months of struggle. Got Pass my DSA test. Thanks.

[@nikoo28]

all the very best...congrats 😄

[@adityasingh2883]

Bhai mujhe guide kardo yaar konsi company main internship mili or kaise mili kya criteria tha?

[@ashishkarthik2817]

Hey bro can u tell how and what all did u practice

[@ForWork-mj9fv]

Einstein: if you can't explain it simply, you don't know it well enough 🐐, I use to kill myself crying i was dull, not until i found it maybe I was getting a proper explanation, and you did prove me right, imagine after your explanation, i didn't even wait to see you write the code, i went straight ahead to try it out myself, cause i now understand the problem 100% 🔥❤, Thank you sir 🙏

[@toheebalawode1634]

your explanation sticks to the brain, please continue the good work

[@arslanmuhammad4190]

By Watching Your Videos now. I solved problems in efficient complexity. Thanks.

[@nikoo28]

Excellent!

[@priyabratapadhi2421]

thanks,sir, really loved your CLEAR STEP by step explanations.

[@ashmita_shrivastava]

best solution till date

[@rawat7203]

Keep up the Great work Sir, Thank you

[@sarthakmahadik9760]

Very informative . Love you Man

[@SanthoshaK-w5b]

Thank you sir. im confused from 4 days

[@ishimwezachee2179]

You are the best for sure

[@contentorwhat6452]

Best Explanation Sir!!!!

[@tasniatahsin8637]

hi nikhil, there is a line else which made me confused, if(is.empty || stack.pop()!=c) why did you write the condition is.empty?? wont it be !is.empty() since if its empty, we are gonna be returning true??

[@nikoo28]

We are doing this to check if the stack becomes empty without matching all the brackets.

[@rohithbhandari7836]

​@@nikoo28 That is only possible when input contains numbers or letters Is not input always contains parenthesis or brackets ?

[@nikoo28]

Check the problem constraints

[@ChppuKaPaati]

What if string starts with close bracket like "]()" it will try to pop empty stack and will throw error so that stack.isEmpty() Condition is there

[@Nexgenstory]

Very informative video… could you please make a DSA playlist from a scratch

[@nikoo28]

what kind of videos are you looking for? I have videos on stacks/queue, other data structures...

[@mahalasakini5045]

Sir please make some video on design patterns …it will be helpful… and thanks for wonderful videos on DSA

[@nikoo28]

design patterns, system design, graphs...these are the next series of videos in my pipeline..stay tuned!!

[@pinnapureddynithinreddy4805]

sir can you do videos on solving contest problems every week

[@nikoo28]

I post a new video every week :)

[@jashanjotbedi1796]

Explaination is at par level .

[@FR0ZING]

thank you so much 🙏🙏🙏

[@appikeeru5785]

Sir very clear explanation 👌

[@arslanmuhammad4190]

sir When You are going to work on Graphs.

[@nikoo28]

very soon. My next week video will be on introduction to graphs

[@arslanmuhammad4190]

@@nikoo28 Thanks Sir

[@nikoo28]

The complete playlist on graphs is now available: kzbin.info/aero/PLFdAYMIVJQHNFJQt2eWA9Sx3R5eF32WPn

[@MdShamuel]

Amazing Bro . keep it up💘.

[@nikoo28]

Thanks 🔥

[@נויבןדוד-ג5ר]

Thank you! You’re awesome! Can you please also upload a solution of Leetcode 200?

[@nikoo28]

yes, that is a very popular problem indeed. Will make a video soon

[@SanthoshaK-w5b]

class Solution { public boolean isValid(String s) { //valid parentheses would mean that all types of opening parentheses are correctly matched with their corresponding closing parentheses, and they are properly nested within each other. Stack stack = new Stack(); for(char par : s.toCharArray()){ if(par == '('){ stack.push(')'); }else if(par == '{'){ stack.push('}'); }else if(par == '['){ stack.push(']'); }else if(stack.pop() != par){ this code is also working , but i dont know why at last else if condition we want 2 condition to check. can any one explain return false; } } return stack.isEmpty(); } }

[@ramanarao4646]

@user-ei3ng5kt4n check for this input String s =")";

[@ChppuKaPaati]

Here problem is what if string stars with close bracket? Like this "]()[]" it will not push any charecter to stack but will try to pop the empty stack which will throw error so you must add isEmpty() Condition to avoid error

[@subee128]

Thanks

[@nandanihada4461]

great explanation😄

[@kurapatisabitha]

Can you please provide solution for Human Readable Time

[@nikoo28]

Can you provide me a link to the problem?

[@ssss22144]

❤

[@arslanmuhammad4190]

Sir, Which Tree can I learn. I Know about BST and AVL. Can You guide me for More Trees.

[@abhi_coder6]

Bhai kafi h yaar bas rulayega kya itna hi karle aur question practice kar bas aur kuch nhi mostly tujhe bst pe bt pe bhi sawal milega avl rare h

[@nikoo28]

exactly, you don't need so much. BST and binary trees are usually enough.

[@arslanmuhammad4190]

sir Please I know about BST and AVL but What are other Binary trees.

[@arslanmuhammad4190]

@@abhi_coder6 yaar to sai khta ha lekin smj nai a rai or konsay binary trees.

[@nikoo28]

@@arslanmuhammad4190 solve more problems on binary trees, just understanding the concept is not enough. Try to find the lowest common ancestor, least paths, backtracking in trees etc.

[@jjoelthomas]

Hi Nikhil, Your solutions are amazing. But for this particular problem, I feel this can be done in a better way instead of complicating it with a stack data structure. Please find my solution which was advised to me in an interview at Apexon UK. I find the solution to be intuitive and very simple. private static boolean isBalancedParanthesis(String s){ do { if(s.contains("()")) s = s.replace("()", ""); if(s.contains("[]")) s = s.replace("[]",""); if(s.contains("{}")) s = s.replace("{}", ""); }while(s.contains("[\\W]+")); if(s.length()>0) return false; else return true; }

[@nikoo28]

that works as well, but I was also trying to give an idea about how to solve mathematical expressions also. :) You are also relying on "contains" functionality.

[@jjoelthomas]

@nikoo28 I don't mean to discredit you. Because I learn a lot from you. And this solution I posted just in case someone's looking for an easy solution. Anyway your videos are all amazing and you are really doing a great work

[@vishalrai6027]

@@jjoelthomas you have to tell the interviewer that contains and replace function are costly. makes whole program as O(n^2)

[@parthmodi2028]

Greak Work Sir

[@nikoo28]

Thanks

[@MrShrane]

@@nikoo28 Sir please answer my one doubt , are we not required to return true?? why was there no returning of true? else how will we get it

[@MrShrane]

ohh got it ,thanks sir understood the last part

[@kinderhero8897]

классная кофта со спайдером