Lecture 67: Minimum Time to BURN the Entire Binary Tree || C++ Placement Series

[@govindsuryavanshi6653]

First I have completed all DSA series and now conclude this is Greatest ever DSA series to exist on youtube or paid courses. Your contribution will be remembered. You're God of DSA for us🙇‍♂ Thanks you.

[@its_nawed4611]

If anyone wondering without adding value to map for all the nodes how can we check whether a node is visited or not here is the explanation : If you try to access a key value using indexing operator [], then 2 things can happen : 1->The map contains this key. So it will return the corresponding key value 2->The map doesn't contain the key. In this case it will automatically add a key to the map with key value null.

[@invinishku1542]

i was coding this question by myself and just not stored q.size() before entering into for loop ....int the for loop new elements are getting pushed and whole function was working differently.....but thanks for teaching how to debug code by printing imp things...it took me 40 mins ...but the job was done......keep working hard...cheers..

[@harshpandey4190]

You should be Appreciated for explaining the toughest sum with such a ease

[@verma_jay]

Your explanation was this good that I coded the solution myself by just seeing the algorithm🔥

[@pkyadav6230]

Really??

[@Ash_9202]

​@@pkyadav6230either he is better than us or he could be lying. Don't stress yourself if you weren't able to come up with a solution yourself.

[@yashchauhan7221]

A Single thank you comment is not enough for the efforts bhaiya don't worry about views as people are left behind becoz of your consistent nature and Big Thumbs Up 🔥🔥

[@kumarpawansahh]

Sabd kamm parr jaa rhe hai aapki dedication dekh kr ❤️

[@shashanksingh4708]

I was able to write the code myself after watching the first half of the video. Thanks for the awesome content!

[@anshumansharma1069]

To be honest bhtt aasan laga jaise apne sikhaya bhaiyaa💖💖

[@akshithabharde05]

Finally I am finishing my series of Binary trees thanks a lot bhaiyaa ji i have understood this concept finally now i am moving towards BST

[@janvikundnani9203]

Absolutely love the way you make these hard problems seem so simple!! Thank you so much for such efforts!!🔥

[@RohitKumar-oo8lo]

Agree

[@RohitKumar-oo8lo]

Dhamaal machaa rha hai Course bhaiya!!!

[@AA-wy9vw]

Loving the series bhaiya! Aapki consistency hume inspire karti hai aur mehnat marne ke liye

[@mr.himanshu4299]

thoda piche reh gya hun pr dheere dheere cover kr rha hun end takk complete krunga course ko keep uploading babbar bhai

[@aqusatabassumsyedsadiqueal5920]

You're continuing despite less views in later vedios , 👏👏👏

[@Shyara2308]

sir aaj meko 1 and half month ho gya consist hu course kafi accha bana hai i like it . samaz nahi aa rha to me 3-4 bar piche ja kr dekh rha hu but koi boriat nahi ho rahi , love babber ke course ko love dher sara

[@ramanjaneyuluv479]

babbar bhayya is love 💝

[@prashantshukla6870]

I can say today😜 Maine toh aag lga di binary tree ko 😂😂😂😂

[@lazyemperor5182]

🔥🔥🔥🔥

[@keshavgambhir9894]

Present bhaiya Consistency op 🔥🔥🔥🔥 Commenting for a better reach

[@utkarshsingh1457]

Hard level question bhi easy laga... aur pehli baar me he samajh aagya.. thanks to your explanation and to your dedication to give us such a good content!!!

[@pranayavnish8028]

bro easy is liye lga because he taught the brute force soln. optimized m dfs lagana h without using any extra space. Aise hi h gfg pe HARD ka tag nhi lga hua.

[@Ash_9202]

​@@pranayavnish8028tbhi itna lamba solution bana hai Par kar bhi kya skte hai jb tk graphs pe na poch jaye hum

[@pranayavnish8028]

@@Ash_9202 vai graphs kaha se aagya bich m?? Ye video wala soln to 1st year ke students v khud se bna lenge, This soln(as the optimal) will surely get u rejected in an interview , check leetcode discussions.

[@Ash_9202]

@@pranayavnish8028 agr koi 1st year student kr skta hai to pkka intelligent hoga bhai. Mujhe to aise solution ki Intuition hi nahi aaegi. I guess I'm not that intelligent.

[@JungleKiKahaniyan8]

Maja aa raha hai Aapki consistency ++

[@dipakkumarsingh.7151]

Aaj to aag lagadi bhai. dhuwa dhuwa kardiya binary tree ka

[@Nishant-n8e]

i did with pair function

[@MohitKumar-ys8jd]

Your explanation is the best!! Thanks for this amazing playlist!!

[@shubhamrao5952]

This question was asked by me 2 years back in Paytm SDE 1 interview.

[@KPCoder]

Thanks you sir, excellent content, no one can explain like you from basic 🙏 god bless you

[@crazyduniya128]

bohot badiya question maja aa gya! wah babbar ji wah

[@code_sleep_live]

I think a better solution is solving using recursion after creating mapping with parent and finding target node as done in the video. What we can do is pass cur node and prev node in recursion and recursively check for left right and parent unless any of them is equal to prev node and then take the max time from the generated branches.The code will look like class Solution { int minBurnTime(Node* cur,Node* prev,map &m) { if(cur==NULL) return 0; int time=0; if(m[cur]!=prev) time=minBurnTime(m[cur],cur,m); if(cur->left!=prev) time=max(time,minBurnTime(cur->left,cur,m)); if(cur->right!=prev) time=max(time,minBurnTime(cur->right,cur,m)); return time+1; } public: int minTime(Node* root, int target) { if(root==NULL) return 0; map m; queue q; Node* tar; q.push(root); m[root]=NULL; while(!q.empty()) { Node* t=q.front(); q.pop(); if(t->data==target) tar=t; if(t->left) { m[t->left]=t; q.push(t->left); } if(t->right) { m[t->right]=t; q.push(t->right); } } return minBurnTime(tar,NULL,m)-1; } };

[@ashurajput3526]

best explanation for burning tree on utube

[@mriduljain6809]

Loving the series bhaiya... Sab samajh aa rha h

[@sachinrathod8319]

Sahi questions tha bhai maja aa gaya solve karne ka.

[@vivek25467]

save trees !! save nature !! save lives 🙃

[@RohitKumar-lt2bo]

TC: O(n) //when unordered_map is used insted of map SC: O(n)

[@Phantomex984]

Here is the code with some minor updates : class Solution { private: Node* createParentMapping(Node* root, int target, map& nodeToParent) { Node* res = nullptr; queue q; q.push(root); nodeToParent[root] = nullptr; while (!q.empty()) { Node* front = q.front(); q.pop(); if (front->data == target) { res = front; } if (front->left) { nodeToParent[front->left] = front; q.push(front->left); } if (front->right) { nodeToParent[front->right] = front; q.push(front->right); } } return res; } int burnTree(Node* root, map& nodeToParent) { if (!root) { return 0; } unordered_map visited; queue q; q.push(root); visited[root] = true; int ans = 0; while (!q.empty()) { int size = q.size(); bool flag = false; for (int i = 0; i < size; i++) { Node* front = q.front(); q.pop(); if (front->left && !visited[front->left]) { flag = true; q.push(front->left); visited[front->left] = true; } if (front->right && !visited[front->right]) { flag = true; q.push(front->right); visited[front->right] = true; } if (nodeToParent[front] && !visited[nodeToParent[front]]) { flag = true; q.push(nodeToParent[front]); visited[nodeToParent[front]] = true; } } if (flag) { ans++; } } return ans; } public: int minTime(Node* root, int target) { map nodeToParent; Node* targetNode = createParentMapping(root, target, nodeToParent); int ans = burnTree(targetNode, nodeToParent); return ans; } };

[@RahulKumar-wt9qc]

Jabardast bhaiya 👍👍

[@amishapandey6875]

Thankyou so much Bhaiya. Can't be more thankful !

[@rachit_joshi]

Thank You So Much BHRATA SHREE !!!!!!!

[@SauravKumar-yp1fy]

class Solution { public: void getAdjacents(Node*root,Node*parent,unordered_map&mp) { if(!root) { return; } vectortemp; if(parent!=NULL) { temp.push_back(parent->data); } if(root->left) { temp.push_back(root->left->data); } if(root->right) { temp.push_back(root->right->data); } mp[root->data]=temp; getAdjacents(root->left,root,mp); getAdjacents(root->right,root,mp); } int minTime(Node* root, int start) { // Your code goes here unordered_mapmp; getAdjacents(root,NULL,mp); unordered_mapvis; queueq; int count=0; q.push(start); while(!q.empty()) { int size=q.size(); for(int i=0;i

[@madetolaugh3476]

Consistency....Prathishta...Anushaashan....

[@shashikantverma9931]

Bhai Bhai Bhai consistency oppppoo

[@suyashjain3223]

You made the Hard problem easy, Thanks a lot

[@monubhargav346]

space complexity is becoming n but allowed space complexity is O(height of tree)

[@parikshitgupta7711]

Instead of using vis as a map, you can use a set and check ki agar set mein voh node hai ki nahi, it will save space

[@ishansingh6241]

I am thinking mai ise adjancy list bana kr bsf chala dun usse bhi ho jayega sayad.

[@suranjandhara7535]

Hello Bhaiya ❤️😍😘🥰😘🥰.... Aa gya hu..... Ab to college khul chuka hai... Abhi college me hu Thanks bhaiya ❤️❤️😍😍😍😍😍😍

[@tusharvaish8096]

bhaiya apke is video ne toh dil jeet liya thankyou for this awesome series ;

[@daipayannit-s415]

Commenting for better reach!!

[@unboxtheuniverse5336]

Congratulations for 125K family 🤩🥳🥳🥳 Mehnat he aapki 🤍 🚨🚨🚨🚨 Bhaiya plz make a Roadmap For InternShip 🥺 with DSA busted Course as resource 🤩

[@prikshit_sharma4071]

Bhaiya ham apko nam roshan krege ❤️

[@shivkumar-og4ow]

Thank you bhaiya.. Consistency ++

[@azaankhan5208]

Superbbbbb !!! Thank you so much for all the efforts.... I have one query, how many more videos will this series going to have.. ?

[@trojanhorse8278]

I think we should simply create a graph from the given tree and thereby the bfs will be bit more easy. That's the way I solved this question.

[@SirAkPandey]

Best solution for this problem 👍🏻♥️🙏

[@tejashmaurya2149]

Salute You Bhaiya💯💯💯

[@naveensingh_0735]

Words aren't enough bade bhaiya 🙏

[@prashantbirajdar9271]

very nice bhaiyaa everything is understood...... bit tricky but after practicing 2 or three times it would be easy to impliment and understand ...thanks bhaiyaa... keep it up!! your dedication is inspiration to all of us...!!!

[@naveensingh_0735]

We will make you proud

[@dwivedi6037]

Bhaiya aap great ho I have no words ♥️♥️ next level 😘😘😘😘😘😘😘😘😘😘😘😘😘😘😘😘

[@Bhagwati_Jewellers.]

Amazing bhiaya 🙏

[@jenithpatel5594]

Here, you solved using O(N) space complexity, but in question, we require O(H) space complexity.

[@letsstartwithrahul8841]

in this code no significance of bool flag because by help of flag we only save last iteration and last iteration will be save by only q.size() if no element will push in queue then this is the case when queue is empty

[@Vishal-cm4dl]

Yeah, simply we can increment time every time when !q.empy()

[@anshikasingh4158]

boht mazza aaya pehle toh lga nhi ho paega pher yaad aaya -Just ask 1 question “ Why I started ?

[@ihateracistandblackpeople4272]

I also try to take that line to motivate me but that just doesn't work for me.Can you please tell what's that keeps you motivated or I should say why you started.

[@RohitSingh-hc8yi]

thanks a lot bhaiya

[@anuragpatel7787]

hnji this is anurag patel Lec 67 completed successfully ✅

[@ravivishvas3676]

best content ever...

[@sherebanoburhani8939]

Present bhaiya❤

[@sumitchakrabortystudy651]

labrador traversal🤣🤣🤣🤣. btw noicely explained

[@sukhpreetsingh5200]

done bhaiya

[@abhishekshimpi214]

Maza hi agya bhaiyya🤩🤩🔥🔥

[@tanishkayadav7351]

learning from the best!

[@vivekrathore6030]

full code :) class Solution { public: //step-1 create mapping and return the target node Node* createParentMapping(Node* root,int target,map &nodeToParent){ Node* res = NULL; //to store the result; queue q; q.push(root); nodeToParent[root] = NULL; while(!q.empty()){ Node* front = q.front(); q.pop(); if(front->data==target){ res = front; } if(front->left){ nodeToParent[front->left] = front; q.push(front->left); } if(front->right){ nodeToParent[front->right] = front; q.push(front->right); } } return res; } int burnTree(Node* root,map &nodeToParent){ map visited; queue q; q.push(root); visited[root] = true; int ans = 0; while(!q.empty()){ bool flag = 0; int size = q.size(); for(int i=0;ileft && !visited[front->left]){ flag = 1; q.push(front->left); visited[front->left] = 1; } if(front->right && !visited[front->right]){ flag = 1; q.push(front->right); visited[front->right] = 1; } if(nodeToParent[front] && !visited[nodeToParent[front]]){ flag = 1; q.push(nodeToParent[front]); visited[nodeToParent[front]] = 1; } } if(flag==1){ ans++; } } return ans; } int minTime(Node* root, int target) { map nodeToParent; Node* targetNode = createParentMapping(root,target,nodeToParent); int ans = burnTree(targetNode,nodeToParent); return ans; } };

[@afeefuddin8883]

Learnt coding, now need to learn the intiution myself P.S: Coded on my own.

[@ayushp4471]

Thank you bhaiya for your efforts and also your explanation is the BEST❤

[@deepak44222]

happy married life to you bhaiya. 😁😁😁😁😂😂😂😂

[@AbhishekGupta-tn8sp]

class Solution { private: unordered_map node2Parent; unordered_map visited; queueq; Node* targetNode = NULL; int time = 0; //step 1: create node 2 parent mapping via level order traversal //and also try to find target node void getNodeToParent(Node* root, int target){ queueaux; aux.push(root); while(!aux.empty()){ Node* curr = aux.front(); aux.pop(); //checking if current node is target node if(curr->data == target){ targetNode = curr; } //set parent of root to null if(curr==root){ node2Parent[root]=NULL; } if(curr->left){ node2Parent[curr->left] = curr; aux.push(curr->left); } if(curr->right){ node2Parent[curr->right] = curr; aux.push(curr->right); } } } void burning(){ while(!q.empty()){ Node* curr = q.front(); q.pop(); if(curr==NULL){ time++; if(!q.empty()){ q.push(NULL); } } else{ visited[curr]=true; // Check for left node if(curr->left && !visited[curr->left]){ q.push(curr->left); } // Check for right node if(curr->right && !visited[curr->right]){ q.push(curr->right); } // Check for parent node Node* parent = node2Parent[curr]; if(parent && !visited[parent]){ q.push(parent); } } } } public: int minTime(Node* root, int target) { getNodeToParent(root,target); q.push(targetNode); q.push(NULL); burning(); return time-1; } };

[@anuptewary3016]

Present bhaiya with lot's of fire 🔥

[@yashchampaneri1972]

26:27 😁😁😂😂 kyu tum log ne vahi suna jo mene sunna ?

[@zishan53]

Great 😃👍👍👍 fantastic

[@bheemsingh-fh1zu]

Present bhaiya Reach++😊

[@muhammadsuleman8936]

Thanks brother for such amazing contenet

[@DivyanshuYadav-t7l]

better than striver's explanation

[@kalpeshmali1476]

Shandar bhaiya

[@vinayaksinghal]

class Solution { private: TreeNode*f(TreeNode* root, int start,map&mpp){ mpp[root]=NULL; TreeNode*target=NULL; queueq; q.push(root); while(!q.empty()){ TreeNode*front=q.front(); q.pop(); if(front->val==start){ target=front; } if(front->right){ mpp[front->right]=front; q.push(front->right); } if(front->left){ mpp[front->left]=front; q.push(front->left); } } return target; } int g(TreeNode*target,map&mpp){ mapvis; vis[target]=1; queueq; int ans=0; q.push(target); while(!q.empty()){ int size=q.size(); bool flag=0; for(int i=0;iright && !vis[front->right]){ flag=1; vis[front->right]=1; q.push(front->right); } if(front->left && !vis[front->left]){ flag=1; vis[front->left]=1; q.push(front->left); } if(mpp[front] && !vis[mpp[front]]){ flag=1; vis[mpp[front]]=1; q.push(mpp[front]); } } if(flag==1){ ans+=1; } } return ans; } public: int amountOfTime(TreeNode* root, int start) { mapmpp; TreeNode*target=f(root,start,mpp); int ans=g(target,mpp); return ans; } };

[@divyamvijay9071]

You are really doing great for us bhaiya ❤❤❤

[@aryankr]

bohot bohot shukriya

[@jericodcosta2049]

Attandance marked thankyou

[@codingaid2753]

Thanks bhaiya for easy and nice explanation👏

[@RohitRana-tz2lr]

Thanks bhaiya for all of your efforts

[@aditichourasia2990]

Hats off!

[@RohitSingh-hc8yi]

Bhaiya in beginning itself you said it is a hard problem let's see if I'll be understand it or not

[@mritunjay7065]

Awesome explanation !👏

[@srajittanwar636]

Present Bhaiya

[@torus151]

present sirji

[@vinaysiwach9734]

i code this problem after watching just your algo

[@RohitSingh-hc8yi]

Attendance marked

[@sunny-gb6xh]

thank you bhaiya for your grt content :)

[@KratosProton]

Great Explaination!

[@devangmishra4506]

maza la diya bhaiya ne

[@APOORVSAHU-ro6vc]

I watch full video Thanks 🙏

[@sainikkhaddar6064]

GOD LEVEL SIR

[@BhaiRayyanKiDuniya]

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