that "Papa ji banega" was hilarious 😂😂! btw very informative video ❤❤

The way you explain CONCEPTS is very nice !!

The structure with which you create every video is commendable. System design resources are all over the places on the internet, however, with old and rushed explanations especially LLD. This is a humble request if you can try putting in more system design-related content whenever you get some time, please. :!

Thank you so much Bhaiya ji 🙏🙏🙏

Graph playlist 💪💪💪💪

I spent a lot of time figuring out how to solve this without calling BFS every time. I guess this was the most optimal approach.

public int[]shortestDistanceAfterQueries(int n,int[][]queries){ int[]next=new int[n]; Arrays.setAll(next,i->i+1); int m=queries.length,dist=n-1; for(int i=0;i<m;i++){ int s=queries[i][0],e=queries[i][1]; while(next[s]<e){ int t=next[s]; next[s]=e; s=t; dist--; } ans[i]=dist; } return ans; } Tc=n+q;🎉❤

You can use l<=r as we do in normal BS; that way you can do high=mid-1 because eating mid bananas/hour will also be considered then. alternatively you can do l<r but while adjusting high, use high=mid so we are considering at 'mid' also

Hi MIK, Can you share which yesterday's problem you are mentioning here?

😂😂

Bhaeya there is a dp approach also for this problem can you expain that.

😂

Assalam malkum mia bhai

🎉❤

Great explanation, MIK. One thing is, I don't understand why pq is not used. We are almost doing Djikstra here?

with every edge addition the path is getting shorter. it can never increase. if we have received the shortest path value of 1 in any loop, do we need to run the loop again? because after this, every returned value will be 1. we can break out of the loop. (we must assign an initial value of 1 in every res[i]). Please correct me if i am wrong.

Hello Bhaiya, I’ve received another offer of 9 LPA from TCS Prime! Thank you so much for your video explanations they’ve been incredibly helpful throughout my journey. Truly grateful for your guidance!

I used the BFS technique and solved it. Your Graph concept playlist is helpful.

bhaiya trees ka playlist haiii?

Knew about your channel last week. Became a huge fan of yours! Especially your motivation at start encourages me to keep trying. Thanks Man! ❤

sir lps pe appki ik video hai please make video on longest proper prefix and suffix .... which means that we common part ka no count kare.... aaa is me ans hoga 1 not 2....lps me iska ans 2 hota hai kyu ki wo middle wala a bhi count karta hai lekin proper me sirf left wala a count in prefix and suffix me sirf right wala .. please solution

😂

Solved on own. All credit to MIK's Graph Playlist. Turning Story into code. MY CODE-> class Solution { public: vector<int> shortestDistanceAfterQueries(int n, vector<vector<int>>& queries) { unordered_map<int, vector<int>> graph; vector<int> result; //Step1 := Form a initial graph (0->1->2->......n-1) for (int i = 0; i <= n - 2; ++i) { graph[i].push_back(i + 1); } //Step 2 := Execute one query at a time and find distance for (vector<int>& query : queries) { int u = query[0], v = query[1]; graph[u].push_back(v); // unidirectional int ans = bfs(0, n - 1, graph); result.push_back(ans); } return result; } int bfs(int src, int dest, unordered_map<int, vector<int>>& graph) { queue<pair<int, int>> que; // Pair of (node, distance from src) unordered_set<int> visited; que.push({src, 0}); visited.insert(src); while (!que.empty()) { int u = que.front().first; int dist = que.front().second; que.pop(); if (u == dest) { return dist; } for (int v : graph[u]) { if (visited.find(v) == visited.end()) { visited.insert(v); que.push({v, dist + 1}); } } } return -1; } };

😂😂😂

😂😂😂

😂😂

how to know when to make adj list and when not to, e.g. yesterday there wasn't any need of adj list...

Your motivation makes my day

did it by myself today! thanks to you and your graph series!

Already did this using BFS, Thank you bhaiya ❤

Graph Concept playlist is for beginners? With 0 knowledge of graphs?

Kya bhai dil pr war krte ho😂😂❤

😅😂🥲

bhaiya raadhe raadhe❤

bhaiya aap bahut hi acche se samzate ho i am so blessing full. bhaiya ek request or h ki aap ye app ka naam bta do jisse ap ye padhate ho tablet me (konsa notes ka app kafi acche feature h)

Finally, I got placed 🥺😇! I've been following you since you had 10k subscribers. Thank you for explaining the real intuition behind the problems. I never thought I would get placed this year. I used to read comments on your channel from other students sharing their placement stories, and I often doubted myself. But I didn’t give up and got placed in a decent company with a 10 LPA package. I know it’s not much, but I’ll work even harder from now on.

Choosing Between the Two: Use Dijkstra’s Algorithm when dealing with weighted graphs where edge distances vary. Use BFS when the graph is unweighted or all edges have equal weight, as it is simpler and faster in these cases. Thanks & Regards, Tumhara MIK ❤

Your Graph concepts playlist is a game changer 👌🏻

your explanation is really wellll

I coded myself but my solution ended up just like yours. The same vector and variable names. The same if statement 😂

Thanks man! I am definitely checking out your Graph Playlist.

Sir i left Daily problems from around 4 months and i was self doubting and i don't know what happened aaj wapis se krne ka man kra then i opened your video and the motivation i needed was the one which you gave in the video , Now will not stop till i get placed.

bhai eak batao mai love babber ki c++ wale playist kaar rha hu aur mai graphs start karne wala hu to wha se padhu ya apki playist se??

nice explanation loved it!❤❤

Do i need to complete graph topic before or I can directly watch your playlist

Graph playlist is the best graph playlist on youtube, helped me the most and I recommend it to everyone. Keep up the great work Mazhar sir.

🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥 Today's Quote won my heart

keep it up bhai

awesome explaination till now on youtube

"Thanks for this detailed explanation! The step-by-step approach and code breakdown made the concept so much easier to understand. Great job on simplifying complex topics! Looking forward to more such videos!"