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GeeksForGeeks - POTD | GFG POTD Answer

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class Solution { public: int minimumSum(string s) { int n=s.size(); for(int i=0; i=0; i--){ if (s[i]!='?'){ if (ind>n-1-i){ ind = min(ind, n-1-i); foundatlast = true; } break; } } if (foundatlast) ch = s[n-1-ind]; else ch = s[ind]; if (ind==n) ch='a'; for(int i=0; i

10th May : C++ Solution ☝🏼

class Solution{ public: #define insert(n) push_back(n) int totalCuts(int n,int k,vector<int> &v){ vector<int> s1; multiset<int> s2(v.begin(), v.end()); int sum=0; int mn, mx=INT_MIN; for(int i=0; i<n-1; i++){ s1.insert(v[i]); s2.erase(s2.find(v[i])); mx=max(mx, v[i]); mn=*s2.begin(); if (mn + mx >= k) sum++; } return sum; } };

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9th May : C++ Solution ☝🏼

class Solution { public: const long long M = 1000000007; // modulo map F; long long f(long long n) { if(F.count(n)) return F[n]; long long k=n/2; if (n%2==0) { // n=2*k return F[n] = (f(k)*f(k) + f(k-1)*f(k-1)) % M; } else { // n=2*k+1 return F[n] = (f(k)*f(k+1) + f(k-1)*f(k)) % M; } } int countStrings(long long int N) { if(N==0) return 0; F[0]=F[1]=1; return f(N+1); } };

8th May : C++ Solution ☝🏼

class Solution{ public: int modulo(string s,int m) { int ans=0; int t=1; for(int i=s.size()-1;i>=0;i--) { if(s[i]=='1') { ans=(ans+t)%m; } t*=2; t=t%m; } ans=ans%m; return ans; } };

7th May : C++ Solution ☝🏼

class Solution{ public: string stringMirror(string str){ int n=str.size(); if(n==1) return str+str; string pre=""; pre+=str[0]; for(int i=1;istr[i]){ pre+=str[i]; } else if(pre.length()>1 && str[i-1]==str[i]){ pre+=str[i]; } else break; } string rev=pre; reverse(rev.begin(),rev.end()); return pre+rev; } };

6th May : C++ Solution ☝🏼

class Solution { public: bool makeChanges(int N, int K, int target, vector &coins) { int is_p[1010][110]={}; function can = [&](int t,int k)->int { if (is_p[t][k]!=0) return is_p[t][k]; if (k==0) return t?-1:1; for (int i=0;i=coins[i]&&can(t-coins[i],k-1)==1) return is_p[t][k]=1; return is_p[t][k]=-1; }; return can(target,K)==1?true:false; } };

5th May : C++ Solution ☝🏼

class Solution{ public: int ans=0; int dfs(Node* node, int &k){ if(node==nullptr) return 0; int leftSet=dfs(node->left,k); int rightSet=dfs(node->right,k); int set=(leftSet|rightSet); set |= (1<data); int copy=set; int mems=0; while(copy){ copy=(copy&(copy-1)); mems++; } if(mems<=k) ans++; return set; } int goodSubtrees(Node *root,int k){ dfs(root, k); return ans; } };

4th May : C++ Solution ☝🏼

class Solution{ public: int maxCoins(int n,vector> &ranges){ sort(ranges.begin(), ranges.end(), [&](vector &a, vector &b){ if(a[0] == b[0]) return a[1] < b[1]; return a[0] < b[0]; }); vector post(n); post[n - 1] = ranges[n - 1][2]; for(int i = n - 2; i > -1; i--) post[i] = max(post[i + 1], ranges[i][2]); int answer = 0; for(int i = 0; i < n; i++){ int low= i; int high = n; // Returns the maximum coins while(low < high - 1){ int mid = low + (high - low) / 2; if(ranges[mid][0] >= ranges[i][1]) high = mid; else low = mid; } int cur = 0; if(high < n) cur = post[high]; answer = max(answer, ranges[i][2] + cur); } return answer; } };

3rd May : C++ Solution ☝🏼

class Solution{ public: bool makePalindrome(int n,vector &arr){ unordered_set st(arr.begin(), arr.end()); int pairs = 0; for(string tmp : st) { reverse(begin(tmp), end(tmp)); if(st.find(tmp) != st.end()) ++pairs; } return pairs == n; } };

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2nd May : C++ Solution ☝🏼