Best explanation for monotonic stack

Thnk u bhaiya :heart

Furthur space optimization is possible TC-O(n) SC-O(1) class Solution { public: vector<int> finalPrices(vector<int>& prices) { int n = prices.size(); vector<int> ans (n); ans[n-1]=n; for( int i = n-2 ; i>=0 ; i--){ int j=i+1; while(j<n&&prices[j]>prices[i]){ j=ans[j]; } ans[i]=j; } for(int i=0;i<n;i++){ if(ans[i]>=n){ ans[i]=prices[i]; }else{ ans[i]=prices[i]-prices[ans[i]]; } } return ans; } };

space optimization , Mind blown

no DSU :((

can we use dp here?

Amazing explanation! was able to implement it on my own, the breakdown of logic into steps really helps a lot.

Great explanation

Thanks man .I was stuck on this but your one statement explained me the whole problem solution .

nice video

Please make a video on last LC weekly contests solutions. B was really interesting and code heavy

Thanks Aryan for this tutorial <3 I wouldn't know that we can use map greater<char>. Thankyou again for this amazing tutorial.

Thank you for posting videos on a daily basis! Even though I often struggle with the daily challenges, watching your approach to problem-solving has helped me overcome my fear and start thinking about the best possible solutions. I have one request: could you please solve it in Java or at least provide an overview? Thanks, and keep coding 😄!

thanks bro for this amazing content please keep it we really need your kind of peoples and your explaining style is also very well i totally get the idea

when i=s.length(); then It will give ArrayOutOfBoundIndex

loved the way of explanation..... crystal clear ....

Great Intuition

great video with the dry run example. the deque diagram was a little confusing but good explanation overall

Tell me one Thing ...... The hook is at (6,8) , Bishop is at (6,6) and the queen is at (6,3) Now Would you Please write just the moves here ....

for the last solution, we don't need two sets instead, we can have one set and run loop over the array. Once the element is found in the set we can push it to our answer array and delete that element from the set. So that, next time same elements comes which we have already inserted in our array, we will not push because it is not in the set.

I recently started watching your video and I love how to explain things. Keep going BOSS 👌👌

@AryamMital Can you comeup with some c++ cour se

Love your Content bro. 👊 👊 😎 😎 Keep doing this,

I am using Python 🤡

In the priority queue loop, why you are calculating the gain with adding 1 to pass and total?? You are adding it in the calculategain method right??

Who else has seen Aryan's id in today's leetcode solution section.😅

THE GOAT IS BACK TO CPP LETSGOOOOOOOOOOOOOOOOOO

please solve locked problems as well .

has he started writing code in Cpp now?

great explanation bro thanks for the video :)

bhaiya plz do leetcode 1595, I need ur explanation

amazing and detailed explanation 👏👏

I want to know what is/was your highest POTD streak on leetcode???

Very good explanation!!!

awesome, let me ping the entire code here struct FT{ vector<int> bit; int n; FT(int sz){ n = sz; bit.resize(sz + 1, 0); } void add(int indx, int val){ while(indx <= n){ bit[indx] = max(bit[indx], val); indx += indx & -indx; } } int get(int indx){ int mx_val = 0; while(indx > 0){ mx_val = max(mx_val, bit[indx]); indx -= indx & -indx; } return mx_val; } }; class Solution { public: vector<bool> getResults(vector<vector<int>>& queries) { ios_base::sync_with_stdio(false); int n = min(3 * (int)queries.size(), (int)5e4) + 1; set<int> obstacles; FT ft(n); obstacles.insert(0), obstacles.insert(n); for(const auto &q: queries){ if(q[0] == 1){ obstacles.insert(q[1]); } } for(auto it = obstacles.begin(); it != obstacles.end(); ++it){ if(it == obstacles.begin()) continue; ft.add(*it, *it - *prev(it)); } vector<bool> res; for(int i = queries.size() - 1; i >= 0; --i){ vector<int> &q = queries[i]; if(q[0] == 1){ int x = q[1]; auto it = obstacles.find(x); auto nxt = next(it); auto prv = prev(it); ft.add(*nxt, *nxt - *prv); obstacles.erase(it); } else{ int x = q[1], sz = q[2]; auto it = prev(obstacles.upper_bound(x)); int mx_sz = max(ft.get(*it), x - *it); res.emplace_back(mx_sz >= sz); } } reverse(res.begin(), res.end()); return res; } };

love the two pointer explanation 💝

Nice to see you Back

in your approach 5 i thought we could do better by storing just the last index of curmin and curmax..that'd speed it up a bit i guess but idk

I think in deque approach shrinking the window by hop is same as shrinking it one by one. Please help if anyone have code of shrinking window one by one.

vielen dank

when i struggle , your videos r wat to go

which mic are you using ? thanks in adanvace