Minimum Remove to Make Valid Parentheses - Leetcode 1249 - Python 

Even though I solve LeetCode problems daily on my own, I still find immense value in watching your NeetCode videos. Your insights and perspectives add depth to my understanding and help me approach problems from different angles. ❤

a great explanation, I solved it today using a stack. But this way is giving another way of thinking

it would be also good to see how to optimize this code more

Thanks for sharing your explanation, really appreciate that!

Thanks for sharing! One small note for future videos is that the singular of "parentheses" is "parenthesis"

can u create playlist of leetcode on the basis of topics like arrays , list , binary tree u are great by the way

Death taxes and neetcode making a video on the leetcode daily ❤

awesome explanation thanks bro

Solved it !

Why not do `for c in reversed(res)`? I think it's just as Pythonic and might be easier to understand for beginners

Please make a video on leetcode 31: next permutaion

Please make a video on 443. String Compression

awesome thanks

Can you please show the most concise solution you can come up with?

Why didn't we just do del res[index] instead of making another array called filtered?

Glorious

Do you think using Java instead makes it harder to do these problems?

Hi...sir , i am from india. I have complete more than 250 problems and pratice daily basis , POTD and some extra questions but when i participate on weekly and biweekly contest i can't solve more than one question . Can you make any video about this..or any solution. Thank you sir.

omg, how would you come up with the solution in an interview if you have never seen this question before.

You made it hard unnecessarily, just solve in following easy way it has same TC + SC HAPPY TO BE DISCUSS IF YOU WANT TO :) string minRemoveToMakeValid(string s) { // get all invalids and its indexes in the stack stack st; for (int i = 0; i < s.size(); i++) { if (s[i] == '(') st.push({s[i], i}); else if (s[i] == ')') { if (!st.empty() and st.top().first == '(') st.pop(); else st.push({s[i], i}); } } string res; while (!st.empty()) { int idx = st.top().second; st.pop(); s[idx] = '#'; // marking the indexes which we have to ignore in // final result string } for (auto& c : s) { if (c != '#') res.push_back(c); } return res; }

sir what if there's a surplus of opening brackets "("

class Solution: def minRemoveToMakeValid(self, s: str) -> str: stack = list() spare = list() for i in range(len(s)): if s[i] == '(': stack.append(i) elif s[i] == ')': if len(stack) > 0: stack.pop() else: spare.append(i) res = '' badChars = set(stack) | set(spare) for i in range(len(s)): if i in badChars: continue else: res += s[i] return res my solution... It looks like O(n) but it was only better than 25%, what could be the reason?

how's the tooth pain?

I just kept track of the elements to be deleted along with the queue: class Solution: def minRemoveToMakeValid(self, s: str) -> str: q = [] ind = [] for i in range(len(s)): if s[i] == '(': q.append(s[i]) ind.append(i) elif s[i] == ')': if q and q[-1] == '(': q.pop() ind.pop() else: q.append(s[i]) ind.append(i) ind.sort(reverse=True) for j in ind: s = s[:j] + s[j+1:] return s

as a terrible beginner myself, id say that res[::-1]: is something i had to look into because you kinda skipped over how it worked, idk if youd wanna bother wasting time explaining something like that, especially on a medium problem video, but i just mention it because of what you said towards the end of the video.

string minRemoveToMakeValid(string s) { stack index; int i=0; for(int j=0;j this is could potenially be the best-code for this sum

Is this solution is it efficient ? : class Solution: def minRemoveToMakeValid(self, s: str) -> str: res = [""] * len(s) st = 0 idx = 0 for ch in s: if ch == "(": res[idx] = ch idx += 1 st += 1 elif ch == ")": if st > 0: res[idx] = ch idx += 1 st -= 1 else: res[idx] = ch idx += 1 if st == 0: return "".join(res) cnt = st i = idx - 1 while i >= 0 and cnt > 0: if res[i] == "(": res[i] = "" cnt -= 1 i -= 1 return "".join(res)

Python more efficient solution - class Solution: def minRemoveToMakeValid(self, s: str) -> str: count = 0 res = [] for char in s: if char != ')': res.append(char) if char == '(': count += 1 else: if count > 0: count -= 1 res.append(char) if count != 0: for i in range(len(res) - 1, -1, -1): if res[i] == '(' and count > 0: del res[i] count -= 1 elif count == 0: break return ''.join(res)

in CPP ``` class Solution { public: string minRemoveToMakeValid(string s) { string ans; stack tmp; for(int i=0;i=0;--i) { if(!tmp.empty()&&tmp.top()==i) tmp.pop(); else ans+=s[i]; } reverse(ans.begin(),ans.end()); return ans; } }; ```