1190. Reverse Substrings Between Each Pair of Parentheses | StacK | Wormhole Teleportation technique

In this video, I'll talk about how to solve Leetcode 1190. Reverse Substrings Between Each Pair of Parentheses | StacK | Wormhole Teleportation technique | 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 :)
✨ Timelines✨
0:00 - Problem Explanation
00:46 - Stack O(n^2) approach
8:12 - Wormhole Teleportation technique O(n) time
✨ Hashtags ✨
#programming #Interviews #leetcode #faang #maang #datastructures #algorithms

[@rajrajesh1669]

Been solving daily challenges for 6 months now, warmhole technique was smtg new for me!!! Thank you Aryan I learnt yet another technique.

[@sumitgupta310]

Second approach 🔥

[@think_n_code]

Did you find any lectures that explain in deep the Wormhole Teleportation technique?

[@kondekarvaishnavi2348]

After long gap aryan is back with super content. Thankyou aryan🥳🥳

[@rishabhranjan5162]

bruh he was uploading every single day.

[@aarkojha5710]

actually in the first method theres an issue, we are not removing the opening bracket from the ans. To do so just write in the loop for closing bracket, after the while statement, if(st.top()=='(') st.pop()

[@user-wq3jq7wc3s]

gajab solution!!

[@italk-gj5kk]

wonderful explanantion

[@akmarkan2490]

Thanks Bro

[@shivamgurjar8979]

The second solution is just so good

[@RatulRoy-gc8yj]

@Aryan bhaiya can u explain the outward portion again like in 11:32 when we are going from index 6 to 9 but in outward direction this part why we did ?

[@harshsharma9322]

Perfect timing bhai! Thanks

[@ARYANMITTAL]

Kyaa baat hai Harsh bhai 🫡❤️

[@parasd2802]

bhai that wormhole technique is so good.

[@naamnhibataunga5897]

didn't knew about the wormhole teleportation technique... this was my initial solution : O(n^2) class Solution { public: string reverseParentheses(string s) { stack s1; stack s2; string word = ""; int i = 0; int n = s.length(); // Process characters before the first '(' while (i < n && s[i] != '(') { word += s[i++]; } string str = ""; while (i < n) { char c = s[i]; if (c == '(') { s1.push('('); s2.push(str); str = ""; } else if (c == ')') { if (!s1.empty()) { s1.pop(); reverse(str.begin(), str.end()); if (!s2.empty()) { str = s2.top() + str; s2.pop(); } } } else { str += c; } i++; } word += str; return word; } };

[@abhayraghuwanshi4869]

Hello everyone, Can you suggest more questions where we can use warmhole techniques?

[@divyagarg4821]

Please cover 1017 th problem of leetcode

[@mdyeamentalukder1473]

can we solve it useing prefix sum?

[@darth_vader_11]

Watched twice for understanding wormhole teleportation technique, still little difficult to understand 🤒

[@nooob_coder]

dry run bro on test cases

[@rev_krakken70]

I solved using recursion.. inherently it is stack only.. is that okay?

[@user-sh9to8xr3g]

what!! you guys able to think soln in recursion?

[@random-xd7cb]

bro send the code

[@rev_krakken70]

class Solution { int index = 0; public String processString(String s, int length, boolean shouldReverse) { StringBuilder sb = new StringBuilder(); while (index < length && s.charAt(index) != ')') { char currentChar = s.charAt(index); if (currentChar == '(') { index++; sb.append(processString(s, length, true)); } else { sb.append(currentChar); } index++; } return shouldReverse ? sb.reverse().toString() : sb.toString(); } public String reverseParentheses(String s) { StringBuilder sb = new StringBuilder(); int length = s.length(); return processString(s, length, false); } }

[@yashparashar6781]

Giving runtime error

[@akashkataria3218]

quality of explanation is now decreasing, i found you from leetcode solutions section and started watching you continuously for your best explanation, but explanation is not as good as your previous videos

[@mohit7717]

37/38 test cases passed , what I am doing wrong ? Can Anyone please help me? class Solution { public: string reverseParentheses(string s) { stackst; queueq; st.push(s[0]); int i = 1; while( i

[@codingkart245]

my code is working but damn messed up LOL :) -----------------------------------------------------------------------------------------------------------------------------string reverseParentheses(string s) { //store the corresponding opposite indices of parentheses map oppoInd; stack st; for(int i = 0;i= 0){ //if we found a parentheses, go to it's opposite direction //and change the direction of extracting answer too if(oppoInd.find(iter) != oppoInd.end()){ iter = oppoInd[iter]; toRight = !toRight; //removing pointing to the parenthesis when changing iter index if(toRight) iter++; else iter--; } //extract the answers from the corresponding directions //until it reaches end or finds another parenthesis if(toRight){ while((iter < s.size()) && (oppoInd.find(iter) == oppoInd.end())) ans += s[iter++]; }else{ while((iter >= 0) && (oppoInd.find(iter) == oppoInd.end())) ans += s[iter--]; } } return ans; }