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

Chapters:
00:00 - Intro
00:52 - Problem Statement
03:08 - Understand valid parentheses
06:43 - Using stacks for efficiency
11:21 - Dry-run of Code
13:45 - Final Thoughts

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#leetcode #programming #interview

[arslanmuhammad]

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

[toheebalawode]

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

[arslanmuhammad]

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

[priyabratapadhi]

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

[rawat]

Keep up the Great work Sir, Thank you

[SanthoshaK-wb]

Thank you sir. im confused from 4 days

[mahalasakini]

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

[Nexgenstory]

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

[tasniatahsin]

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??

[נויבןדוד-גר]

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

[pinnapureddynithinreddy]

sir can you do videos on solving contest problems every week

[arslanmuhammad]

sir When You are going to work on Graphs.

[arslanmuhammad]

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

[kurapatisabitha]

Can you please provide solution for Human Readable Time

[SanthoshaK-wb]

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<Character> 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();


}
}

[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;
}