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@lavanyam32243 ай бұрын
Striver doesn't teach us to just solve a problem. He teaches us how to think so that we will be able to solve the future problems on our own. He is an example of this proverb - "Give a man a fish and you feed him for a day. Teach a man to fish and you feed him for a lifetime." He teaches us to fish!🎣
@kshittizbhardwaj4172 жыл бұрын
You explained the previous question so beautifully that i didn't even need to read this question. Thanks striver!!!!
@tps847027 күн бұрын
The most important part of his teaching is to beautifully explain the approach and to build the intuition behind the solving. Thanks!!!!!!!!!!!!!!!!!!!!!1
@sarangtamrakar87232 жыл бұрын
just under stand last question & now able to right the code for this one also by my own.. Excellent teaching skills.. TUF will grow more..
@sayakbasak587 Жыл бұрын
You explained the previous question so beautifully that I solved this one by myself
@anuragojha38712 жыл бұрын
Best teacher !!..providing quality content for free :)
@charlesbabbage67863 ай бұрын
Grateful for these amazing videos!!
@iamnottech891829 күн бұрын
Aaj smjh aaya why this works u explained so beuatifully ki feel aagi.
@de_ansh2 жыл бұрын
Thank you sir for the series. You are doing really a great job.
@dpsmartguy7 ай бұрын
Bht Achche se smjhaye aapne... Thank you...🙂
@aanchalmittal98972 жыл бұрын
I had a doubt.... Like you explain us this approaches in great detail so do we need to do this same thing in interviews too or only briefly?
@user-yg2bo9bb7o17 күн бұрын
Thanks Striver for such amazing explanation!!!
@pramodreddy12142 жыл бұрын
Thanks man helped me a lot!!
@dheerajsaraswat2272 жыл бұрын
please also cover a "construct a binary tree from levelorder and inorder Traversal".
just solved this question by own bcz of prev lecture . thanks striver ❤❤
@stith_pragya8 ай бұрын
Thank You So Much for this wonderful video..........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
@nuraynasirzade6 ай бұрын
thank you VERY MUCH brilliant explanation👏
@ShariqueAkhtar9 ай бұрын
best explanation to this problem
@ritikshandilya70753 ай бұрын
great explanation
@poorpanda903310 ай бұрын
Thank you !
@kambozprav3472 Жыл бұрын
great explanation !!
@NMCSMROHANHEGDE2 жыл бұрын
last TC in LC will work ,if you pass map by reference
@rahulkumarbarik75842 жыл бұрын
thanks for help bro, worked for me
@knowhere60732 жыл бұрын
but what was the problem bro?? can u plz explain
@rahulsrivastava10402 жыл бұрын
@@knowhere6073 Actually it was new map again again that's why
@abhishekdhok52452 жыл бұрын
@@knowhere6073 When we pass anything without reference everytime new copy of that thing is created. But when we pass anything by reference only one copy is created and everytime we refer to the same copy.. That's why it is fast.
@rounakmukherjee95402 жыл бұрын
Thats call quality content
@rydmerlin Жыл бұрын
Before you optimize for space as you have done with the pointers it would help to have just taken a slice of the array and pass that around as a sub array.
@nagavedareddy58912 жыл бұрын
Huge respect....❤👏
@giraffe43752 жыл бұрын
Very beautifully explained bhaiya
@amanbhadani88402 жыл бұрын
Nice explanation.
@mohdnomaankhan24359 ай бұрын
what if two values in the array is same, then the map won't work right? what should we do in that case?
@gyanprakash302 Жыл бұрын
Can someone explain why "ps, ps+ri-is-1" is used in line 17 of github repo? (Java)
@suryakiran29702 жыл бұрын
Great Explanation
@afaqueahmed3882 Жыл бұрын
can anyone tell me that why this question gives me TLE in Leetcode Code:- /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: TreeNode* buildTree(vector& inorder, vector& postorder) { if(inorder.size() != postorder.size()) return NULL; map mpp; for(int i = 0;ipe || is>ie) return NULL; TreeNode* root = new TreeNode(postorder[pe]); int inroot = mpp[postorder[pe]]; int numleft = inroot-is; root->left = build(inorder,is,inroot-1,postorder,ps,ps+numleft-1,mpp); root->right = build(inorder,inroot+1,ie,postorder,ps+numleft,pe-1,mpp); return root; } };
@ITACHIUCHIHA-dr8sz Жыл бұрын
In build function make inorder, postorder, and map as reference that is make it like this, vector&inorder,.....
@laveshgoyal9974 Жыл бұрын
@@ITACHIUCHIHA-dr8sz wow, never noticed passing by ref will make such big difference
@guptashashwat Жыл бұрын
@@laveshgoyal9974 Yes, it avoids making copies
@karunasharma9416 Жыл бұрын
@@ITACHIUCHIHA-dr8sz thanks yaar , it really helped.
@amarjeetkumarsingh733 Жыл бұрын
@@ITACHIUCHIHA-dr8sz Thanks Bro
@CharitraAgarwal26 күн бұрын
A simpler recursive solution without map. Based on simple idea that the right segment of inorder range will constitute the right subtree and left one will constitute the left subtree. Idea is to choose wisely which recursion will be called first (in postorder, we call for right subtree first). Here's my solution with comments ;) TreeNode* getNode(vector& inor, vector& post, int s, int e, int &i) { // the inorder range is incorrect if(s > e) return NULL; // the current postorder is the root at this subtree int root = post[i--]; TreeNode *node = new TreeNode(root); // find the root element in the inorder range int ind = s; while(inor[ind] != root) ind++; // elements to the right of root in inorder seq are to the right subtree node->right = getNode(inor, post, ind+1, e, i); // elements to the left of root in inorder seq are to the left subtree node->left = getNode(inor, post, s, ind-1, i); return node; } TreeNode* buildTree(vector& inorder, vector& postorder) { int i = postorder.size()-1; // postorder index. root is at right return getNode(inorder, postorder, 0, inorder.size()-1, i); } PS: I tried this question myself and this is what I came up with!! Then after that I watched Striver's solution.
@omtayade97582 жыл бұрын
In c++, Instead of map, we can use unordered_map which will do read operation in O(1). map is used when we want keys to be sorted on insert
@ekanshsanger83562 жыл бұрын
In worst case unordered map will take O(N) :p
@googlepay4295 Жыл бұрын
@@ekanshsanger8356 yes but that wont matter on LC ig coz its not CF
@Ayush372623 ай бұрын
@@ekanshsanger8356But still its always preferred to use unordered_map because that worst case happens 1 in a million times
If 2 or more elements are same in the each of the postorder or inorder traversal, then how to map. Such situation is resulting in segmentation faults. How to cover this edge case ?
@mdshahriarhossain63332 жыл бұрын
How many videos you are planning to put in this series?
@ManojKumar-jb4sc2 жыл бұрын
One video on topic construct binary tree from preorder and postorder please (leetcode 889)
@shreyasnagabhushan4918 Жыл бұрын
thanks sir
@himanshidafouty34721 күн бұрын
Understood
@mriduljain1981 Жыл бұрын
completed lecture 35 of free ka tree series.
@jayyy3112 жыл бұрын
💚
@bhavya8608 Жыл бұрын
understoodo!!!!
@krishnarajs80122 жыл бұрын
Can anyone tell me the actual time complexity . I cant understand that
@rishikchakravarty79866 ай бұрын
class Solution { public: int index=0; TreeNode* createTree(vectorpreorder, vectorinorder, int start, int end) { if(start>end) return NULL; TreeNode *node = new TreeNode(preorder[index++]); int pos; for(int i =0;ival) { pos=i; break; } } node->left = createTree(preorder, inorder, start, pos-1); node->right=createTree(preorder, inorder, pos+1,end); return node; } TreeNode* buildTree(vector& preorder, vector& inorder) { return createTree(preorder,inorder,0,inorder.size()-1); } }; how about this, this sems a tad simpler?
@prateekjoshi64252 жыл бұрын
the C++ code is not working for the last TC(202th) on LC...please have a look
@takeUforward2 жыл бұрын
Copy my code, submit, its working, all codes are tested.
@prateekjoshi64252 жыл бұрын
@@takeUforward thanks...this worked...but for preorder its not working
@kwanikar72 жыл бұрын
@@prateekjoshi6425 try passing the map by reference
@rishabhgupta9846 Жыл бұрын
able to solve by myself
@jitinroy2246 Жыл бұрын
genius
@satvrii Жыл бұрын
❤❤
@supratimbhattacharjee53242 жыл бұрын
class Solution { public: TreeNode* create(vector& inorder, vector& postorder, unordered_map& hash, int is, int ie, int ps, int pe) { if(is>ie || ps>pe) return nullptr; int rootIndxInInorder=hash[postorder[pe]]; int lps=ps; int lpe=lps+rootIndxInInorder-is-1; int lis=is; int lie=rootIndxInInorder-1; int rps=lpe+1; int rpe=pe-1; int ris=rootIndxInInorder+1; int rie=ie; TreeNode* root=new TreeNode(postorder[pe]); root->left=create(inorder,postorder,hash,lis,lie,lps,lpe); root->right=create(inorder,postorder,hash,ris,rie,rps,rpe); return root; } TreeNode* buildTree(vector& inorder, vector& postorder) { int n=inorder.size(); unordered_map hash; for(int i=0;i
while traversing in right it should be poe-1 and not pos -1. Dry run and recheck.
@prakharmangal11522 жыл бұрын
Python Solution class Solution: def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]: hm = dict() for i in range(len(inorder)): hm[inorder[i]] = i return self.buildTreePostIn(0, len(inorder)-1, inorder, 0, len(postorder)-1, postorder, hm) def buildTreePostIn(self, ist, ie, inorder, pst, pe, postorder, hm): if ist > ie or pst > pe: return None root = TreeNode(postorder[pe]) inroot = hm[postorder[pe]] nums_left = inroot - ist root.left = self.buildTreePostIn(ist, inroot - 1, inorder, pst, pst + nums_left - 1,postorder, hm) root.right = self.buildTreePostIn(inroot+1, ie, inorder, pst + nums_left, pe-1,postorder, hm) return root
@girikgarg8 Жыл бұрын
Done!
@rks35222 жыл бұрын
13:00
@pratikdas1780 Жыл бұрын
both this and the previous question will mess with your brain.
@anuj8855Ай бұрын
Exactly 💯
@Aryan-fi2qf2 жыл бұрын
I am getting TLE if I don't pass the map by reference can anyone explain why?
@takeUforward2 жыл бұрын
Creates a copy, so more time. Reference means usung the same!!
@Aryan-fi2qf2 жыл бұрын
@@takeUforward Thanks for the reply.
@guptashashwat Жыл бұрын
Efforts ++
@I_Keshav_Prajapati Жыл бұрын
tle on testCase 201(leetcode) for both pre and postorder can anyone help?
@shivamkumar5857 Жыл бұрын
pass every thing by reference
@heyprashant2 жыл бұрын
great content man. I have one query, what if there are duplicate nodes?
@em_ashutosh2 жыл бұрын
While hashing use whole node.
@maneeshguptanalluru78072 жыл бұрын
@@em_ashutosh could you please elaborate your approach?
@rishabhgupta9846 Жыл бұрын
when searching for root->val search from instart to inend in inorder array
@muthupandideivamsanmugam1774 Жыл бұрын
@Aryan Bharat but the input is only a integer type vector 😀
@lavudyabharath87832 жыл бұрын
can we just reverse the post order array and apply the method that we used in preorder??
@krishnarajs80122 жыл бұрын
Not possible if you reverse pre order you will not get post order
@your_name962 жыл бұрын
I did it after reversing the post order as well as it was easier for me to visualise, but ofcourse the pointer increment decrement will be different. TreeNode * hlp(vector&postorder,int postStart,int postEnd, vector&inorder,int inStart, int inEnd,map&inPos){ // base cases if(postStart > postEnd or inStart > inEnd)return NULL; // first create the root of the tree TreeNode* root = new TreeNode(postorder[postStart]); int inRoot = inPos[root->val]; int numsInRight = inEnd - inRoot; root->left = hlp(postorder,postStart+numsInRight+1,postEnd,inorder,inStart,inRoot-1,inPos); root->right =hlp(postorder,postStart+1,postStart + numsInRight,inorder,inRoot+1,inEnd,inPos); return root; } TreeNode* buildTree(vector& inorder, vector& postorder) { mapinPos; reverse(postorder.begin(),postorder.end()); for(int i=0; i < inorder.size(); i++)inPos[inorder[i]] = i; TreeNode *root = hlp(postorder,0,postorder.size()-1,inorder,0,inorder.size()-1,inPos); return root; }
@avicr47272 жыл бұрын
yes you can do it but keep in after reversing you will get root right left so first half will denote right ans second half will denote left
@HarivanshKripaVlogs2 жыл бұрын
In video u have used(giving TLE):- ps+numleft-1 But in github solution it is(CORRECT):- ps+numleft-1-is I am not getting the logic of the github code. Can you please explain the reason?
@vishalgupta9572 жыл бұрын
did u get the answer??
@piyushacharya7696 Жыл бұрын
reach++
@cenacr0079 ай бұрын
us
@JohnWick-kh7ow2 жыл бұрын
Why solution with unordered_map is taking more runtime than map?
@neelpatel1222 жыл бұрын
Ideally unordered_map solution is faster. But, leetcode test cases are poor, and when you submit your code again and again you'll see change in runtime.
@amanbhadani88402 жыл бұрын
In best case unordered map takes O(1) time but in worst case it takes O(n) time which is more than the average time complexity of ordered map i.e log(n).
@suvanshmahajan59022 жыл бұрын
"us"
@pragatiagrawal201 Жыл бұрын
I am getting TLE...Out of 202 cases only 201 got passed :(
@vamsimadugula8524 Жыл бұрын
me too did you find the mistake?
@pragatiagrawal201 Жыл бұрын
@@vamsimadugula8524 Ab to main sab kuch bhool sa gyi hoon haan pr ye na maine check kiya Leetcode pr abhi to dekha ki mera accept ho gya tha. Solution shayad yhi tha pr phir bhi yahan paste kre deti hoon.. mera mood nhin h compare krne ka..shayad maine hi galti ki thi jitna mujhe yad hai. Code jo accept ho gya tha wo ye hai ::: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: TreeNode* buildTree(vector& inorder, vector& postorder) { int postStart = 0, postEnd = postorder.size() - 1; int inStart = 0, inEnd = inorder.size() - 1; map < int, int > mp; for (int i = inStart; i & postorder, int postStart, int postEnd, vector < int > & inorder, int inStart, int inEnd, map < int, int > & mp) { if (postStart > postEnd || inStart > inEnd) return NULL; TreeNode * root = new TreeNode(postorder[postEnd]); int inRoot = mp[root -> val]; int numsLeft = inRoot - inStart; root -> left = constructTree(postorder, postStart, postStart + numsLeft - 1, inorder, inStart, inRoot - 1, mp); root -> right = constructTree(postorder, postStart + numsLeft, postEnd-1, inorder, inRoot + 1, inEnd, mp); return root; } };
@pranjal-barnwal Жыл бұрын
@@pragatiagrawal201 The only difference is that Map is passed as reference in 2nd function, so value doesn't get copied again and again. Instead same map is used in multiple calls.
@pragatiagrawal3599 Жыл бұрын
@@pranjal-barnwal Okayyy.. Thanks a lot:)
@vivekshrivastav36742 жыл бұрын
Similar code but bit simpler way to represent : // for finding index of an element int find(int ele, int in[], int l, int r) { for(int i=l; i r or idxright = build(in, post, pos+1, r , --idx); if(node->right == NULL) idx++; node->left = build(in, post,l, pos -1, --idx ); if(node->left == NULL) idx++; return node; } //Function to return a tree created from postorder and inoreder traversals. Node *buildTree(int in[], int post[], int n) { // Your code here int idx = n-1, l=0, r = n-1; int pos = find(post[idx], in, l, r); Node* node = new Node(post[idx]); node->right = build(in, post, pos+1, r, --idx); if(node->right == NULL) idx++; node->left = build(in, post, l, pos -1, --idx ); return node; }
@codewithom11 Жыл бұрын
It is giving TLE😭
@_PRANAYMATE5 ай бұрын
With the knowledge Of Previous video i done this question on my own here is the code public static TreeNode buildTree2(int[] postorder, int[] inorder) { //Store the inorder in the map Map map=new HashMap(); for(int i=0;i postEnd || inStart > inEnd) { return null; } TreeNode root = new TreeNode(postOrder[postEnd]); //index of the root node int inRoot = map.get(root.val); //nodes in left int numsRight = inEnd - inRoot; root.right = constructBinaryTree2(postOrder, postEnd - numsRight, postEnd - 1, inOrder, inRoot + 1, inEnd, map); root.left = constructBinaryTree2(postOrder, postStart, postEnd - numsRight - 1, inOrder, inStart, inRoot - 1, map); return root; }
@nileshsinha78692 жыл бұрын
why this code is giving TLE in leetcode??
@neelpatel1222 жыл бұрын
If your logic is correct and still receiving TLE then try passing the vectors and map in the helper function by "reference".
@amitkoushik55042 жыл бұрын
@@neelpatel122 Thanks buddy, it actually works.
@amitchaurasia5922 жыл бұрын
Bhai hindi me bol sakte ho kya english samaj ni aati
@amanbhadani88402 жыл бұрын
English seekh lene ka re baba,simple.
@amitchaurasia5922 жыл бұрын
@@amanbhadani8840 bhai tm English medium me padhe hoge hm hindi medium me padhe hai
@amanbhadani88402 жыл бұрын
@@amitchaurasia592 Bhai ye bhi 1 essential skill ban chuka hai aajkal and company me phir conversation kaise kroge.Thats why it's better to improve yourself rather than finding excuse.
@amitchaurasia5922 жыл бұрын
@@amanbhadani8840 aacha bhai phir English seekh leta hu
@amitchaurasia5922 жыл бұрын
Par philhal ye hindi bole thoda samjhu phir to English ka dekh linga
@radiokepler Жыл бұрын
overactor bro
@adityaprasad66932 жыл бұрын
It's a humble suggestion please refine your explanations it sometimes seems like you are just talking to yourself.