GeeksForGeeks - POTD | GFG POTD Answer

11th December : C++ Solution ☝🏼 ———————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution{ public: long maximumSumSubarray(int K, vector &Arr , int N){ long long ans=0; for(int i=0; i

10th December : C++ Solution ☝🏼 ———————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution{ public: bool subArrayExists(int arr[], int n) { unordered_map mp; int i, sum=0; mp[0]=1; for(i=0;i

9th December : C++ Solution ☝🏼 ———————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution { public: int smithNum(int n) { int sum = 0; int num =n; vector primes; for(int i =2;i< n;i++){ while(num%i == 0){ primes.push_back(i); num/=i; } } num =0; while(n){ int digit = n%10; num+=digit; n/=10; } for(int i =0;i< primes.size();i++){ while(primes[i]){ int digit =primes[i]%10; sum+=digit; primes[i]/=10; } } return num == sum; } };

8th December : C++ Solution ☝🏼 ———————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution { public: bool primecheck(int n){ for(int i=2;i

class Solution { public: bool primecheck(int n){ for(int i=2;i

7th December : C++ Solution ☝🏼 ———————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution{ public: long countSubarrays(int a[], int n, int L, int R){ long ans=0,u=0,j=-1,l=0; for(int i=0;i=a[i] && a[i]>=L){ u++; j=i; } if(u>0 && R>=a[i]){ ans+=(j-l); if(i>j)ans++; } if(a[i]>R){ l=i+1; ans+=u; u=0; } } return ans+u; } };

6th December : C++ Solution ☝🏼 ———————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution { public: int countX(int L, int R, int X) { int count =0; int num=0; for(int i =L+1;i

5th December : C++ Solution ☝🏼 ———————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution { public: int getMinDiff(int arr[], int n, int k) { sort(arr,arr+n); int maxi=arr[n-1]; int mini=arr[0]; int ans=maxi-mini; for(int i=1;i

4th December : C++ Solution ☝🏼 ———————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution{ public: string sum(string s1,string s2) { int i=s1.length()-1,j=s2.length()-1; vectorans; int x,y,res,rem,carry; char ch; while(i>=0 and j>=0) { x=s1[i]-'0'; y=s2[j]-'0'; res=x+y+carry; rem=res%10; carry=res/10; ans.push_back(rem+'0'); i--; j--; } while(i>=0) { x=s1[i]-'0'; res=x+carry; rem=res%10; carry=res/10; ans.push_back(rem+'0'); i--; } while(j>=0) { y=s2[j]-'0'; res=y+carry; rem=res%10; carry=res/10; ans.push_back(rem+'0'); j--; } while(carry) { int rem=carry%10; ans.push_back(rem+'0'); carry/=10; } string t; int p; for( p=ans.size()-1;p>=0;p--) { if(ans[i]!='0') break; } while(p>=0) { t+=ans[p]; p--;; } return t; } bool solve(string s,int pos,int l1,int l2) { string s1=s.substr(pos,l1); string s2=s.substr(pos+l1,l2); string s3=sum(s1,s2); int n=s3.length(); if(n>s.length()-pos-l1-l2) return 0; if(s.substr(pos+l1+l2,n)==s3) { if(pos+l1+l2+n==s.length()) return 1; return solve(s,pos+l1,l2,n); } return 0; } int isSumString(string s) { int n=s.length(); for(int i=1;i

3rd December : C++ Solution ☝🏼 ————————————————————— 🙋🏻‍♂️Discussion ⁉️ Join ✅ @GFG_Answer

class Solution { public: void inorder(Node *root,vector&v){ if(!root)return; inorder(root->left,v); v.push_back(root->data); inorder(root->right,v); } int countPairs(Node* root1, Node* root2, int x) { vectorv1,v2; inorder(root1,v1); inorder(root2,v2); unordered_set s(v1.begin(),v1.end()); int ct = 0; for(int i = 0; i < v2.size(); i++){ if(s.find(x - v2[i]) != s.end()) ct++; } return ct; } };