GeeksForGeeks - POTD | GFG POTD Answer

24th November : C++ Solution ☝🏼 ————————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution { public: vector nthRowOfPascalTriangle(int n) { long long mod = 1000000007; vector> v(n + 1); v[0].push_back(1); for (int i = 1; i <= n; i++) { v[i].push_back(1); for (int j = 1; j < i; j++) { long long val = (v[i - 1][j - 1] + v[i - 1][j]) % mod; v[i].push_back(val); } v[i].push_back(1); } return v[n-1]; } };

23rd November : C++ Solution ☝🏼 ————————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution{ public: int height(Node *node) { if (node == NULL) { return 0; } return (node->height); } int getBalance(Node *node) { if (node == NULL) { return 0; } return (height(node->left) - height(node->right)); } Node *leftRotate(Node *node) { Node *x = node->right; Node *y = x->left; x->left = node; node->right = y; node->height = max(height(node->left), height(node->right)) + 1; x->height = max(height(x->left), height(x->right)) + 1; return x; } Node *rightRotate(Node *node) { Node *x = node->left; Node *y = x->right; x->right = node; node->left = y; node->height = max(height(node->left), height(node->right)) + 1; x->height = max(height(x->left), height(x->right)) + 1; return x; } /*You are required to complete this method */ Node* insertToAVL(Node* node, int data) { if (node == NULL) { return (new Node(data)); } if (data < node->data) { node->left = insertToAVL(node->left, data); } else if (data > node->data) { node->right = insertToAVL(node->right, data); } else { return node; } node->height = 1 + max(height(node->left), height(node->right)); int balance = getBalance(node); if ((balance > 1) && (data < node->left->data)) { return rightRotate(node); } if ((balance < -1) && (data > node->right->data)) { return leftRotate(node); } if ((balance > 1) && (data > node->left->data)) { node->left = leftRotate(node->left); return rightRotate(node); } if ((balance < -1) && (data < node->right->data)) { node->right = rightRotate(node->right); return leftRotate(node); } return node; } };

21st November : C++ Solution ☝🏼 ————————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution{ public: bool solve(struct Node* r1, struct Node* r2){ if((r1==NULL and r2!=NULL ) or (r1!=NULL and r2==NULL)) return false; if(r1==NULL and r2==NULL) return true; if(r1->data!=r2->data) return false; bool s1=solve(r1->left, r2->right); if(s1==false) return false; bool s2=solve(r1->right, r2->left); return s2; } // return true/false denoting whether the tree is Symmetric or not bool isSymmetric(struct Node* root) { if(root==NULL) return true; return solve(root->left, root->right); } };

21st November : C++ Solution ☝🏼 ————————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer ⚡Support Guys YouTube.com/@afzalmiryt

class Solution { public: //Function to check if two trees are identical. bool isIdentical(Node *r1, Node *r2) { if(r1 == NULL && r2 == NULL) return true; else if((r1 == NULL && r2 != NULL) || (r1 != NULL && r2 == NULL)) return false; if(r1->data != r2->data) return false; bool ans = isIdentical(r1->left, r2->left); ans &= isIdentical(r1->right, r2->right); return ans; } };

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20th November : C++ Solution ☝🏼 ————————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer ⚡Support Guys YouTube.com/@afzalmiryt

class Solution{ public: int solve(Node *root, int k, map &s, int sum) { if(root == NULL) return 0; int c = 0; sum += root -> data; if(s.find(sum-k) != s.end()) c += s[sum - k]; s[sum]++; int left = solve(root -> left, k, s, sum); int right = solve(root -> right, k, s, sum); s[sum]--; return c + left + right; } int sumK(Node *root,int k) { map s; s[0]++; return solve(root, k, s, 0); } };

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18th November : C++ Solution ☝🏼 ————————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution { public: Node* findIntersection(Node* head1, Node* head2) { Node *start=NULL, *q=NULL; int d1,d2; while(head1!=NULL && head2!=NULL){ d1=head1->data; d2=head2->data; if(d1==d2){ Node* node = new Node(d1); if(start==NULL){ start=node; q=node; } else{ q->next=node; q=node; } head1=head1->next; head2=head2->next; } else if(d1next; } else if(d2next; } } return start; } };

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class Solution { public: Node* reverseDLL(Node * head) { if(head==NULL || head->next==NULL ){ return head; } Node* previ=NULL; Node* curr=head; Node* forw=NULL; while(curr!=NULL){ forw=curr->next; curr->next=previ; curr->prev=forw; previ=curr; curr=forw; } return previ; } };

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17th November : C++ Solution ☝🏼 ————————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer ⚡ My laptop Review ⚡ https://youtu.be/D-EzAzrBRls

class Solution { pair solve(Node* root) { if(root -> left == NULL and root -> right == NULL) return {root, root}; if(root -> left == NULL) { pair right = solve(root -> right); root -> right = right.first; right.first -> left = root; return {root, right.second}; } if(root -> right == NULL) { pair left = solve(root -> left); left.second -> right = root; root -> left = left.second; return {left.first, root}; } pair left = solve(root -> left); pair right = solve(root -> right); left.second -> right = root; root -> left = left.second; root -> right = right.first; right.first -> left = root; return {left.first, right.second}; } public: //Function to convert binary tree into circular doubly linked list. Node *bTreeToCList(Node *root) { if(root == NULL) return NULL; pair res = solve(root); res.first -> left = res.second; res.second -> right = res.first; return res.first; } };