placement?

According to editorial of leet code this can be optimised to O(n*log(sumof elements)) No one on entire youtube is discussin this approach sab easy easy approach likh rahe hai

Time complexity calculation is wrong. you skipped the cost of sorting a (n*(n+1))/2 sized array. which would be O(n^2 log(n^2)). So total time complexity will be O(n^2log(n^2)). I think time/space complexity-wise your approach 2 is no better than approach 1.

line: 18 is wrong, zero[i] = c1 shoud be outside the if block.

why do we need this mod over here?

Comment section has more factful information than video😂

muh utha kr video daldi ....... phle solution cla kr to dekhlia kro

//ALL TC IS PASSED class Solution { public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) { Queue<Integer> q = new LinkedList<>(); HashSet<Integer> hs = new LinkedHashSet<>(); int[] indegree = new int[n]; for (int i = 0; i < n; i++) { if (leftChild[i] != -1) indegree[leftChild[i]]++; if (rightChild[i] != -1) indegree[rightChild[i]]++; } int rootCount = 0; int root = -1; for (int i = 0; i < n; i++) { if (indegree[i] == 0) { rootCount++; root = i; } if (rootCount > 1) return false; } if (root == -1) return false; q.add(root); while (!q.isEmpty()) { int curr = q.remove(); if (hs.contains(curr)) return false; hs.add(curr); if (leftChild[curr] != -1) q.add(leftChild[curr]); if (rightChild[curr] != -1) q.add(rightChild[curr]); } return hs.size() == n; } }

worst way to waste time

wrong solution

n = 4 leftChild = [3,-1,1,-1] rightChild = [-1,-1,0,-1] Output false Expected true explain me about this

wrong solution Input n = 4 leftChild = [3,-1,1,-1] rightChild = [-1,-1,0,-1] Use Testcase Output false Expected true

Here is Correct code with impovement of this testcase 4 [3,-1,1,-1] [3,-1,1,-1] ``` class Solution { public: int findroot(int n, vector<int>& leftChild, vector<int>& rightChild){ vector<bool> visited(n, false); // visit the both array and tick true for(int i=0; i<n; i++){ if(leftChild[i] != -1) visited[leftChild[i]] = true; if(rightChild[i] != -1) visited[rightChild[i]] = true; } int root = -1; // find the index where it's false, it will be the root which has no parent, for(int i=0; i<n; i++){ if(root == -1 and visited[i] == false){ root = i; // if found more than 1 false index it means more than 1 parent exist, then simlpy return false }else if(visited[i] == false){ return -1; } } return root; } bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) { queue<int> q; set<int> hs; int root = findroot(n, leftChild, rightChild); // had more then 1 parent or cycle so return false if(root == -1) return false; q.push(root); while(!q.empty()){ int node = q.front(); q.pop(); if(hs.find(node) != hs.end()){ return false; } hs.insert(node); if(leftChild[node] != -1){ q.push(leftChild[node]); } if(rightChild[node] != -1){ q.push(rightChild[node]); } } return hs.size() == n; } }; ```

Good explanation, here is one with greedy approach: public int movesToMakeZigzag(int[] nums) { int evenMoves=0,oddMoves=0; for(int i=0;i<nums.length;i++){ int left = i-1 < 0? Integer.MAX_VALUE:nums[i-1]; int curr = nums[i]; int right= i+1>=nums.length?Integer.MAX_VALUE:nums[i+1]; int min=Math.min(left,right); if(i%2==0) evenMoves+= curr-min>=0? curr-min+1:0; else oddMoves+= curr-min>=0? curr-min+1:0; } return Math.min(evenMoves,oddMoves); }

c++ solution int maxNumberOfFamilies(int n, vector<vector<int>>& reservedSeats) { unordered_map<int, vector<int>> mp; int totalCount = n*2; for(int i=0; i<reservedSeats.size(); i++){ mp[reservedSeats[i][0]].push_back(reservedSeats[i][1]); } for(auto &m: mp){ int dec=0; vector<int> p = m.second; if(find(p.begin(), p.end(), 2) != p.end() ||find(p.begin(), p.end(), 3) != p.end() || find(p.begin(), p.end(),4) != p.end() || find(p.begin(), p.end(), 5) != p.end() ){ dec++; totalCount--; } if(find(p.begin(), p.end(), 6) != p.end() ||find(p.begin(), p.end(), 7) != p.end() || find(p.begin(), p.end(),8) != p.end() || find(p.begin(), p.end(), 9) != p.end() ){ dec++; totalCount--; } if(find(p.begin(), p.end(), 4) == p.end() && find(p.begin(), p.end(), 5) == p.end() && find(p.begin(), p.end(),6) == p.end() && find(p.begin(), p.end(), 7) == p.end() ){ if(dec == 2) totalCount++; } } return totalCount; }

Thank you, your solution is very helpful.

why we are doing arr[i-k] to delete first element??

Thanks, the volume is toooo low

good detail explanation. please keep it up same way

It was really helpful

In java return Integer.parseInt(String.valueOf(num).replaceFirst("6","9"));

for this problem this is better video than others

this code is not valid if in a one row only seat no. 8 is reservd

Awesome bro!

suffle is not creating new array and modifying the existing

Bsdk apni original awaj me samja na angrej kyu ban raha he

not going to work if you are using set, cause it is possible for two trans to be the same, and we need them both in final answer.

thnks

this code is not running ,it giving an error that- expected ) in foreach(var kvp in map). I copied the code from the github link you provided in the comments.

thank you

Can anyone pls find out the error (logical) in my smaallll code?: class Solution { public: int tupleSameProduct(vector<int>& nums) { int n = nums.size(); vector<int> hash(1e8, 0); int ans = 0; for(int i = 0; i < n - 1; i++) { for(int j = i + 1; j < n; j++) { int ind = nums[i] * nums[j]; hash[ind]++; if(hash[ind] >= 2) ans += 8; } } return ans; } };

horrible

C++ Code : int minFlips(int a, int b, int c) { int count=0; while(c || a || b){ if(c&1){ if((a&1)==0 && (b&1)==0) count++; } else{ if(a&1) count++; if(b&1) count++; } //moving through the bits right to left a >>= 1; b >>= 1; c >>= 1; } return count; }

class Solution { public: bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) { int root=-1; int count=0; vector<int> adj[n]; vector<int> indeg(n,0); vector<bool> vis(n,false); queue<int> q; for(int i=0;i<n;i++) { if(leftChild[i]!=-1) { adj[i].push_back(leftChild[i]); indeg[leftChild[i]]++; } if(rightChild[i]!=-1) { adj[i].push_back(rightChild[i]); indeg[rightChild[i]]++; } } for(int i=0;i<n;i++) { if(!indeg[i]) { count++; root=i; } } if(count>1 || root==-1) return false; q.push(root); vis[root]=true; while(!q.empty()) { int curNode=q.front(); q.pop(); for(auto x: adj[curNode]) { if(vis[x]) return false; q.push(x); vis[x]=true; } } for(int i=0;i<n;i++) if(!vis[i]) return false; return true; } };

Nice Explanation

Same solution after finding the root node. public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) { //kzbin.info/www/bejne/qoKaqpaOrbNpepo //Declare set to find unique node in Tree Set<Integer> uniqueNode = new HashSet(); for(int left : leftChild){ if(left!=-1) uniqueNode.add(left); } for(int right : rightChild){ if(right!=-1) uniqueNode.add(right); } //If any node is not child of any other node means it's root. int rootNode = -1; for(int i=0; i<n; i++){ if(uniqueNode.contains(i)) continue; rootNode = i; break; } if(rootNode==-1){ return false; } Queue<Integer> queue = new LinkedList(); Set<Integer> visitedSet = new HashSet(); //Adding root in queue queue.add(rootNode); while(!queue.isEmpty()){ int size = queue.size(); while(size>0){ int node = queue.poll(); if(visitedSet.contains(node)) return false; visitedSet.add(node); if(leftChild[node]!=-1){ queue.add(leftChild[node]); } if(rightChild[node]!=-1){ queue.add(rightChild[node]); } size--; } } return visitedSet.size()==n; }

The explaination was awesome....💯

Should this be time: O(n + m)? Because you looped twice on each?

Thanku

Hi

cool question and great walk through, thank you so much for sharing your knowledge!

woow. really nice solution and explained well too 😃

Why we do in opposite direction????

what if q queries are given like a range of l to r how will you solve this?

great walktrhough. thanks for sharing.

thank you

Feels like you're reading the solution off the book or something. You could explain better with pen and paper.

Really simple code and explanation . Great work 👍

Bhai tu ye fake accent mat kiya kar. Isliye dislikes zyada h tere video me

This is not an expected solution in an interview we have to do a lazy increment ie increment only during pop if that element is within bottom k