GeeksForGeeks - POTD | GFG POTD Answer

class Solution{ public: vector generateNextPalindrome(int num[], int n) { bool ch=false; for(int i=0,j=n-1;i=0; i++,j--) { if(num[i]num[j]) { num[j]=num[i]; ch=true; } } if(ch==false) { int s,e; if(n%2==0) { s=n/2-1; e=n/2; } else { s=n/2; e=n/2; } while( s>=0 && e=0 && e ans; ans.push_back(1); for(int i=1;i ans; for(int i=0;i

16th August : C++ Solution ☝🏼

class Solution { public: //Function to find the nth catalan number. int mod=1e9+7; int rec(int i) { if(i==0 || i==1) return 1; long long val=0; for(int j=0;j&dp) { if(i==0 || i==1) return dp[i]=1; if(dp[i]!=-1) return dp[i]; long long val=0; for(int j=0;jdp(n+1,-1); // return rec1(n,dp); // dp tabulation vectordp(n+1,-1); dp[1]=dp[0]=1; for(int i=2;i<=n;i++) { long long val=0; for(int j=0;j

15th August : C++ Solution ☝🏼

class Solution{ public: int maxOnes(int a[], int n) { int mx=0; int one=0, z=0; for(int i=0; i

14th August : C++ Solution ☝🏼

class Solution { public: vector singleNumber(vector nums) { int numsXor = 0; for(int i=0;i 0) { mask = (1<

13th August : C++ Solution ☝🏼

class Solution { public: int nthFibonacci(int n){ int a = 0, b = 1, c; if (n == 0) return a; for (int i = 2; i <= n; i++) { c = (a + b)%1000000007; a = b; b = c; } return b; } };

12th August : C++ Solution ☝🏼

class Solution { public: //Function to find length of longest increasing subsequence. int longestSubsequence(int n, int a[]) { vectorlis; lis.push_back(a[0]); for(int i=1; ilis.back()){ lis.push_back(a[i]); } else{ // int index = lower_bound(lis.begin(),lis.end(),a[i]); auto it = lower_bound(lis.begin(), lis.end(), a[i]); int index = it - lis.begin(); // Convert iterator to index lis[index] = a[i]; } } return lis.size(); } };

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11th August : C++ Solution ☝🏼

class Solution { public: long long find(int coins[], int N, int sum,int idx,vector>&dp){ if(idx==0){ if(sum%coins[0]==0)return 1; return 0; } if(sum<0)return 0; if(sum==0)return 1; if(dp[idx][sum]!=-1)return dp[idx][sum]; long long take=0,notTake=find(coins,N,sum,idx-1,dp); if(sum>=coins[idx])take=find(coins,N,sum-coins[idx],idx,dp); return dp[idx][sum]=take+notTake; } long long int count(int coins[], int N, int sum) { vector>dp(N,vector(sum+1,-1)); return find(coins,N,sum,N-1,dp); } };

10th August : C++ Solution ☝🏼

class Solution { public: //Function to find the length of longest common subsequence in two strings. int lcs(int n, int m, string s1, string s2) { vector> dp(n+1,vector(m+1,-1)); for(int i=0;i<=n;i++) dp[i][0]=0; for(int j=0;j<=m;j++) dp[0][j]=0; for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ if(s1[i-1]==s2[j-1]) dp[i][j]=1+dp[i-1][j-1]; else dp[i][j]=max(dp[i-1][j],dp[i][j-1]); } } return dp[n][m]; } };

9th August : C++ Solution ☝🏼

class Solution{ public: long long int largestPrimeFactor(int N){ for(int i=2;i<=sqrt(N);i++) { if((N%i) == 0)return largestPrimeFactor(N/i); } return N; } };

8th August : C++ Solution ☝🏼

class Solution { public: int countFractions(int n, int numerator[], int denominator[]) { map<pair<int,int>,int>m; int ans=0; for(int i=0;i<n;i++) { int gcd=__gcd(numerator[i],denominator[i]); numerator[i]=numerator[i]/gcd; denominator[i]=denominator[i]/gcd; int x=numerator[i]; int y=denominator[i]; int z=y-x; if(m[{z,y}]) { ans+=m[{z,y}]; } m[{x,y}]++; } return ans; } };