Longest Valid Parentheses | LeetCode 32 | Coders Camp

Code: coderscamp.wixsite.com/codeit...
Link to Problem: leetcode.com/problems/longest...
This video solves the problem "Longest Valid Parentheses" in O(n) Time and space complexity
#Leetcode #coderscamp #whiteboardexplanation #codinginterview #stack #validparentheses

[@samridhshubham8109]

Excellent, clean cut explanation

[@RahulKumar-oq5oo]

good clean, short explanation.

[@hakoHiyo271]

Your explanation of algorithm is nice and straight forward. There are so many other videos making this more complicated that it needs to be. Thanks!

[@CodersCamp]

Thanks a lot!

[@Its_________bansal4821]

really great💯💯💯

[@malayawasthi]

Crisp and to the point. Great !!!

[@manthansrivastav9640]

nice explanation really liked it

[@CodersCamp]

Thank you :)

[@riyankadey102]

Easiest one❤️..thank you

[@ogundimuhezekiah845]

How was this solution accepted? I don't understand the magic going on here😂

[@MEGANE34]

Thank you for good explaining but i want to ask why index is starting -1?

[@indra16]

you cant iterate Stack like that , that is not allowed in stack
instead of that you can pop elements
class Solution {
public int longestValidParentheses(String s) {

if(s == null || s.length()==0){
return 0;
}
int max = 0;

Stack stack = new Stack();

for(int i=0; i

[@prakashsingh1107]

thanks

[@SANJAYSINGH-jt4hf]

Solution wont' work as iteration over stack is wrong. Instead try this same thing just from right to left iteration
class Solution {
public int longestValidParentheses(String s) {
if(s==null || s.length()==1) return 0;
ArrayDeque stk = new ArrayDeque();
for(int i=s.length()-1;i>=0;i--){
if(!stk.isEmpty() && s.charAt(i)=='(' && s.charAt(stk.peek())==')'){
stk.pop();
}
else{stk.push(i);}
}
int index=-1,res=0;
for(int n:stk){
res=Math.max(res,n-index-1);
index=n;
}
res= Math.max(res,s.length()-index-1); //ifafter ) there is prefect brack eg:)()()
return res;
}
}