LeetCode, GeeksForGeeks Problem of the day solution

class Solution { public: int maxSumAfterPartitioning(vector& arr, int k) { int n = arr.size(); vector dp(n+1, 0); for(int i=n-1;i>=0;i--){ int maxParti = INT_MIN; int ans = INT_MIN; int len = 0; for(int j=i;j

class Solution { public: // Should return decimal equivalent modulo 1000000007 of binary linked list long long unsigned int decimalValue(Node *head) { // Your Code Here long long unsigned int ans=0; Node *temp=head; while(temp!=NULL) { ans=ans<<1; ans=ans|temp->data; ans=ans%1000000007; temp=temp->next; } return ans; } };

class Solution { public: void update_next_num(int &num, int &base_num, int &digit_mask) { num += digit_mask; if (num % 10 == 0) { digit_mask = digit_mask * 10 + 1; base_num = base_num * 10 + base_num % 10 + 1; num = base_num; } } vector sequentialDigits(int low, int high) { vector result; int len = 1; while (low / len > 0) { len *= 10; } // Get smallest base int base_num = 1; int digit_mask = 1; while (base_num <= len / 10) { base_num = base_num * 10 + base_num % 10 + 1; digit_mask = digit_mask * 10 + 1; } // Get smallest number greater than low int num = base_num; while (num < low) { update_next_num(num, base_num, digit_mask); } while (num <= high) { result.push_back(num); update_next_num(num, base_num, digit_mask); } return result; } };

class Solution{ public: /*You are required to complete this method */ int atoi(string s) { bool neg=0; int n=s.length(); int ans=0; for(int i=0; i='0' && s[i]<='9'){ ans=(ans*10)+(s[i]-'0'); }else return -1; } return neg?(-1)*ans:ans; } };

class Solution { public: vector> divideArray(vector& nums, int k) { sort(nums.begin(),nums.end()); vector> ans; for(int i=0;i

class Solution { public: //Function to check if a string is Pangram or not. bool checkPangram (string s) { // your code here vector alphabet(26,false); for(int i = 0; i < s.size(); i++) { if(s[i] >= 'A' && s[i] <= 'Z') { char lowercase = (s[i]^(1 << 5)); int index = lowercase - 'a'; alphabet[index] = true; } else if(s[i] >= 'a' && s[i] <= 'z') { int index = s[i]-'a'; alphabet[index] = true; } } for(int i = 0; i < 26; i++) { if(!alphabet[i]) return false; } return true; } };

class Solution { public: vector<int> dailyTemperatures(vector<int>& temp) { int n=temp.size(); stack<int>st; vector<int>arr(n,0); for(int i=0;i<n;i++){ while(!st.empty() and temp[st.top()]<temp[i] ){; int j=st.top(); st.pop(); arr[j]=i-j; } st.push(i); } // reverse(arr.begin(),arr.end()); return arr; } };

class Solution { public: //Function to insert string into TRIE. void insertWord(struct TrieNode *root, string key){ if(key.size() == 0){ root->isLeaf = true; return; } int index = key[0] - 'a'; struct TrieNode * child; if(root->children[index] != NULL){ child = root->children[index]; } else{ child = getNode(); root->children[index] = child; } insertWord(child, key.substr(1)); } void insert(struct TrieNode *root, string key) { insertWord(root, key); } //Function to use TRIE data structure and search the given string. bool searchKey(struct TrieNode *root, string key){ if(key.size()==0){ return root->isLeaf; } int index = key[0] - 'a'; struct TrieNode * child; if(root->children[index] !=NULL){ child = root->children[index]; } else{ return false; } return searchKey(child, key.substr(1)); } bool search(struct TrieNode *root, string key) { return searchKey(root, key); } };

class Solution { public: int evalRPN(vector<string>& tokens) { stack<long long>s; for(string a: tokens){ if(a=="+"){ long val1,val2; val1=s.top(); s.pop(); val2=s.top(); s.pop(); s.push(val1+val2); }else if(a=="-"){ int val1,val2; val1=s.top(); s.pop(); val2=s.top(); s.pop(); s.push(val2-val1); }else if(a=="*"){ long val1=s.top(); s.pop(); long val2=s.top(); s.pop(); s.push(val1*val2); }else if(a=="/"){ long val1,val2; val1=s.top(); s.pop(); val2=s.top(); s.pop(); s.push(val2/val1); }else{ s.push(stoi(a)); } } return (int)s.top(); } };

class Solution { public: int LCSof3 (string A, string B, string C, int n1, int n2, int n3) { int dp[n1+1][n2+1][n3+1]; memset(dp, 0, sizeof(dp)); for(int i=1; i<=n1; i++) { for(int j=1; j<=n2; j++) { for(int k=1; k<=n3; k++) { if (A[i-1] == B[j-1] && B[j-1] == C[k-1]) { dp[i][j][k]=dp[i-1][j-1][k-1]+1; } else { dp[i][j][k]=max(max(dp[i-1][j][k], dp[i][j-1][k]),dp[i][j][k-1]); } } } } return dp[n1][n2][n3]; } };