LeetCode, GeeksForGeeks Problem of the day solution

class Solution { public: /*You are required to complete below method */ void inorder(Node *root,int &sum) { if(root==NULL) return ; if(root->left==NULL && root->right==NULL) { sum=sum+root->data; return ;} inorder(root->left,sum); inorder(root->right,sum); } public: int sumOfLeafNodes(Node *root ){ int sum=0; inorder(root,sum); return sum; } };

class Solution { public: int furthestBuilding(vector<int>& heights, int bricks, int ladders) { priority_queue<int, vector<int>, greater<int>> pq; for(int i = 1; i < heights.size(); i++) { int diff = heights[i] - heights[i-1]; if(diff > 0) pq.push(diff); if(pq.size() > ladders) { if(pq.top() <= bricks) bricks -= pq.top(), pq.pop(); else return i-1; } } return heights.size()-1; } };

class Solution{ public: bool isMaxHeap(int arr[], int n) { for(int i=0;i<=(n-2)/2;i++) { if(arr[2*i+1] > arr[i]) { return false; } else if(2*i+2<n && arr[2*i+2] > arr[i]) { return false; } } return true; } };

class Solution { public: int findLeastNumOfUniqueInts(vector& arr, int k) { unordered_mapmp; for(auto it:arr) mp[it]++; vector>dp; for(auto it:mp){ dp.push_back(it); } sort(begin(dp),end(dp),[](auto a,auto b){ return a.second=it.second){ k-=it.second; count++; }else{ break; } } return size(dp)-count; } };

class Solution { public: void helper(Node *root,Node * &newNode,Node * &r){ if(!root) return ; helper(root->left,newNode,r); if(!newNode) newNode = root; if(r){ r->right = root; r->left = NULL; } r = root; helper(root->right,newNode,r); } public: Node *flattenBST(Node *root) { Node * newNode = NULL,*r = NULL; helper(root,newNode,r); r->left = NULL; return newNode; } };

class Solution { public: long long largestPerimeter(vector& nums) { int n = nums.size(); sort(nums.begin(), nums.end()); long long ans = 0, cur = 0; for(int i=0; i nums[i] && i >= 2){ ans = max(cur, ans); ans += nums[i]; } cur += nums[i]; } if(ans == 0) return -1; return ans ; } };

class Solution { public: int isPossible(vector>paths){ for(int i=0;i

class Solution { public: long long largestPerimeter(vector& nums) { int n = nums.size(); sort(nums.begin(), nums.end()); long long ans = 0, cur = 0; for(int i=0; i nums[i] && i >= 2){ ans = max(cur, ans); ans += nums[i]; } cur += nums[i]; } if(ans == 0) return -1; return ans ; } };

class Solution { public: vector rearrangeArray(vector& nums) { vector positive; vector negative; vector ans; int n = nums.size(); for(int i = 0; i < n; i++){ if(nums[i] < 0){ negative.push_back(nums[i]); } else{ positive.push_back(nums[i]); } } for(int i = 0; i < n/2; i++){ ans.push_back(positive[i]); ans.push_back(negative[i]); } return ans; } };

class Solution { public: void dfs(int node , int parent , vector&low ,vector&tin , vector&vis , vector adj[] , int &timer, vector> & bridges ){ vis[node] = 1; low[node] = timer; tin[node] = timer; timer++; for(auto nbr : adj[node]){ if(nbr == parent) continue; else if(!vis[nbr]){ dfs(nbr,node , low , tin , vis , adj , timer , bridges); low[node] = min(low[node] , low[nbr]); if(low[nbr] > tin[node]){ if(nbr> node) bridges.push_back({node,nbr}); else bridges.push_back({nbr,node}); } } else{ low[node] = min(low[node] , low[nbr]); } } } vector>criticalConnections(int v, vector adj[]){ // Code here vector> bridges; vector low(v,0); vector tin(v,0); vector vis(v,0); int parent = -1; int timer = 0; dfs(0 , parent , low , tin , vis , adj , timer , bridges); sort(bridges.begin() , bridges.end()); return bridges; } };