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Instead of using O(2N). I have solved it in O(N), here is my approach class Solution { public: int pairsum(vector<int> &arr) { // code here int greater1=arr[0]; int greater2=arr[1]; int n=arr.size(); for(int i=2;i<n;i++){ int smaller=(greater1<greater2)?greater1:greater2; int other=greater1+greater2-smaller; if(arr[i]>smaller){ smaller=arr[i]; } if(greater1<greater2){ greater1=smaller; } else if(greater1>greater2){ greater2=smaller; } } return greater1+greater2; } }; BTW: thanks for making consistent videos, it really helps me keep going bro 😀

Let's Connect 🤝 : LinkedIn : www.linkedin.com/in/rochak-vyas17/ Code :- class Solution { public: int pairsum(vector<int> &arr) { int n=arr.size(); int maxi=INT_MIN; for(int i=0;i<n;i++){ if(arr[i]>maxi){ maxi=arr[i]; } } int secmaxi=INT_MIN; for(int i=0;i<n;i++){ if(arr[i]>secmaxi&&arr[i]<maxi){ secmaxi=arr[i]; } } return maxi+secmaxi; } };

Let's Connect 🤝 : LinkedIn : www.linkedin.com/in/rochak-vyas17/ Code :- struct Node *insert(struct Node *head, int data) { Node* newnode=new Node(data); newnode->npx=XOR(NULL,head); if(head!=NULL){ Node* temp=XOR(head->npx,NULL); head->npx=XOR(temp,newnode); } head=newnode; return head; } vector<int> getList(struct Node *head) { vector<int>ans; Node* prev=NULL; while(head!=NULL){ ans.push_back(head->data); Node* temp=head; head=XOR(prev,head->npx); prev=temp; } return ans; }

Let's Connect 🤝 : LinkedIn : www.linkedin.com/in/rochak-vyas17/ Code :- class Solution { public: void dfs(vector<vector<char>>&grid,vector<vector<int>>&vis,int i,int j,int n, int m){ if(i>=n||j>=m||i<0||j<0||grid[i][j]=='0'||vis[i][j]==1){ return; } vis[i][j]=1; dfs(grid,vis,i+1,j,n,m); dfs(grid,vis,i-1,j,n,m); dfs(grid,vis,i,j+1,n,m); dfs(grid,vis,i,j-1,n,m); dfs(grid,vis,i+1,j+1,n,m); dfs(grid,vis,i-1,j-1,n,m); dfs(grid,vis,i+1,j-1,n,m); dfs(grid,vis,i-1,j+1,n,m); } int numIslands(vector<vector<char>>& grid) { int n=grid.size(); int m=grid[0].size(); int ans=0; vector<vector<int>>vis(n,vector<int>(m,0)); for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ if(grid[i][j]=='1'&&!vis[i][j]){ dfs(grid,vis,i,j,n,m); ans++; } } } return ans; } };

but how 1,3 is a subset. the concept of subset is that , all the elements need to be consecutive

Let's Connect 🤝 : LinkedIn : www.linkedin.com/in/rochak-vyas17/ Code :- class Solution { public: long long findSmallest(vector<int> &arr) { long long ans=1; int n=arr.size(); for(int i=0;i<n;i++){ if(arr[i]>ans){ return ans; } else{ ans=ans+arr[i]; } } return ans; } };

Let's Connect 🤝 : LinkedIn : www.linkedin.com/in/rochak-vyas17/ Code : class Solution { public: // Function to reverse a circular linked list Node* reverse(Node* head) { if(head==NULL||head->next==head){ return head; } Node* prev=NULL; Node* curr=head; Node* next; do{ next=curr->next; curr->next=prev; prev=curr; curr=next; }while(curr!=head); head->next=prev; head=prev; return head; } // Function to delete a node from the circular linked list Node* deleteNode(Node* head, int key) { if(head==NULL){ return NULL; } if(head->data==key){ if(head->next==head){ delete head; return NULL; } Node* last=head; while(last->next!=head){ last=last->next; } last->next=head->next; Node* temp=head; head=head->next; delete temp; return head; } Node* prev; Node* curr=head; while(curr->next!=head){ if(curr->data==key){ prev->next=curr->next; delete curr; return head; } prev=curr; curr=curr->next; } return head; } };

Let's Connect 🤝 : LinkedIn : www.linkedin.com/in/rochak-vyas17/ Code : class Solution { public: // Function to find the majority elements in the array vector<int> findMajority(vector<int>& nums) { int candidate1=nums[0]; int candidate2=nums[1]; int vote1=0; int vote2=0; int n=nums.size(); vector<int>ans; for(int i=0;i<n;i++){ if(nums[i]==candidate1){ vote1++; } else if(nums[i]==candidate2){ vote2++; } else if(vote1==0){ candidate1=nums[i]; } else if(vote2==0){ candidate2=nums[i]; } else{ vote1--; vote2--; } } vote1=0; vote2=0; int flag=0; for(int i=0;i<n;i++){ if(nums[i]==candidate1){ vote1++; } else if(nums[i]==candidate2){ vote2++; } } if(vote1>n/3){ ans.push_back(candidate1); flag=1; } if(vote2>n/3){ ans.push_back(candidate2); flag=1; } if(!flag){ return {-1}; } return ans; } };

Let's Connect 🤝 : LinkedIn : www.linkedin.com/in/rochak-vyas17/ Code : class Solution { public: long long maxSubarraySum(vector<int> &arr) { int n=arr.size(); long long sum=0; long long maxi=INT_MIN; for(int i=0;i<n;i++){ sum+=arr[i]; maxi=max(maxi,sum); if(sum<0){ sum=0; } } return maxi; } };

return int(n/2) +1

Let's Connect 🤝 : LinkedIn : www.linkedin.com/in/rochak-vyas17/ Code : class Solution{ public: int nthStair(int n){ int dp[n+1]; dp[1]=1; dp[2]=2; for(int i=3;i<=n;i++){ dp[i]=dp[i-2]+1; } return dp[n]; } };

Bhai solution link bhi dal diya kr

Easy Peasy ☺☺

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Bhai solution github pe dalke uska link share kr diya kr,baaki sb thik hai. sahi ja raha hai tu👍

TY :)

Great Explanation Bhaiya , Love You 🧡🧡🧡🧡 Bhaiya i think there will be no need of visited vector as we are using priority_queue ...

why not DP?

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HWLLO BHAI

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thank.. u bhai..

class Solution { public: int t=0; void gettime(map<Node*,Node*>&mp,Node* start){ queue<Node*>q; q.push(start); map<Node*,int>vis; vis[start]=1; while(!q.empty()){ int s=q.size(); int flag=0; for(int i=0;i<s;i++){ Node* node=q.front(); q.pop(); if(node->left&&!vis[node->left]){ flag=1; vis[node->left]=1; q.push(node->left); } if(node->right&&!vis[node->right]){ flag=1; vis[node->right]=1; q.push(node->right); } if(mp[node]&&!vis[mp[node]]){ flag=1; vis[mp[node]]=1; q.push(mp[node]); } } if(flag){ t++; } } } Node* bfs(Node* root, int target,map<Node*,Node*>&mp){ queue<Node*>q; Node* start; q.push(root); while(!q.empty()){ Node* node=q.front(); q.pop(); if(node->data==target){ start=node; } if(node->left){ mp[node->left]=node; q.push(node->left); } if(node->right){ mp[node->right]=node; q.push(node->right); } } return start; } int minTime(Node* root, int target) { map<Node*,Node*>mp; Node* start=bfs(root,target,mp); gettime(mp,start); return t; } };

Bro solution link dedo

Code is giving tle now

amazing explanation

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Very clear explanation, Thankyou.

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bhai link toh de de previous video kaa description me, pura maja kharab kar diya

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thanks for the video bro..this is the most simple yet the best approach in youtube 💛

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👌

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your sliding windows approach is very helpful brother🙏🙏💥

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