GeeksForGeeks - POTD | GFG POTD Answer

13th April : C++ Solution ☝🏼

class Solution { public: typedef pair pll; pll minmax(long long pre[], int l, int r) { int lo = l, hi = r-1; long long minSum = 0, maxSum = pre[r] - pre[l-1]; while(lo <= hi) { int mid = lo + (hi - lo) / 2; long long lsum = pre[mid] - pre[l-1], rsum = pre[r] - pre[mid]; if(abs(rsum - lsum) < maxSum - minSum) { minSum = min(lsum, rsum); maxSum = max(lsum, rsum); } if(lsum < rsum) lo = mid + 1; else hi = mid - 1; } return {minSum, maxSum}; } long long minDifference(int N, vector &A) { // code here long long presum[N+1]; presum[0] = 0; for(int i = 1; i <= N; i++) presum[i] = presum[i-1] + A[i-1]; long long ans = presum[N]; for(int i = 1; i < N-2; i++) { pll mml = minmax(presum, 1, i+1), mmr = minmax(presum, i+2, N); ans = min(ans, max(mml.second, mmr.second) - min(mml.first, mmr.first)); } return ans; } };

12th April : C++ Solution ☝🏼

class Solution{ public: int dominantPairs(int n,vector &arr){ sort(arr.begin(), arr.begin()+n/2); sort(arr.begin()+n/2, arr.end()); int ans=0; for(int i=n/2;i

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11th April : C++ Shortest Solution ☝🏼

class Solution { public: int solve(int a, int b, int c) { int maxi = max(a,max(b,c)); int rem = a+b+c - maxi; // b+c; if(2*rem+2

10th April : C++ Solution ☝🏼

class Solution { public: int maxIntersections(vector> lines, int N) { map mp; for(auto it:lines) { // (s,e) freq[s]++, freq[e+1]-- int s=it[0],e=it[1]; mp[s]++; mp[e+1]--; } int maxi=1; int presum=0; // it shows the frequency of every element for(auto it:mp) { presum+=it.second; maxi=max(maxi,presum); } return maxi; } };

9th April : C++ Solution ☝🏼

int done=0; vectorf, inv; class Solution { const long long M=1e9+7; public: int bestNumbers(int N, int A, int B, int C, int D) { if(!done) pre(); if(A==B) { long long sum=(long long)N*A; bool flag=false; while(sum>0) { if((sum%10)==C or (sum%10)==D) { flag=true; break; } sum/=10; } if(flag) return 1; return 0; } long long ans=0; for(long long i=0;i<=N;i++) { long long sum = i*A+(N-i)*B; bool flag=false; while(sum>0) { if((sum%10)==C or (sum%10)==D) { flag=true; break; } sum/=10; } if(flag) ans=(ans+nCr(N, i))%M; } return ans; } void pre() { done=1; f.resize(1e5+1, 1); inv.resize(1e5+1); for(int i=2;i<=1e5;i++) f[i]=(f[i-1]*i)%M; inv[1e5]=fastpow(f[1e5], M-2); for(int i=1e5-1;i>=0;i--) inv[i]=(inv[i+1]*(i+1))%M; } long long fastpow(long long a, long long n) { long long res=1; while(n>0) { if(n&1) res=(res*a)%M; a=(a*a)%M; n>>=1; } return res; } long long nCr(int n, int r) { return (((f[n]*inv[n-r])%M)*inv[r])%M; } };