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Leetcode D 🥳🥳🥳🥳 Enjoy guys

class Solution { public:     int count;     int dfs(int i,vector> &adj,vector &a,int k,int par){         int val=0;         for(int j=0;j>& edges, vector& values, int k) {         vector> adj(n);         for(int i=0;i

In case if it is not working then try this class Solution { public:     int maxSubarrays(vector &ll) {         int r = INT_MAX;                 for (auto i : ll)             r &= i;         if (r != 0) return 1;                         int c = INT_MAX, seg = 0;                 for (auto i : ll) {             c &= i;             if (c == r) {                 seg++;                 c = INT_MAX;             }         }                 return seg;     } };

Leetcode C 🥳🥳🥳🥳🥳

#include <vector> #include <climits> class Solution { public: int maxSubarrays(vector<int> &ll) { int r = INT_MAX; for (auto i : ll) r &= i; if (r != 0) return 1; int c = INT_MAX, seg = 0; for (auto i : ll) { c &= i; if (c == r) { seg++; c = INT_MAX; } } return seg; } };

class Solution { public: int minOperations(vector<int>& v) { map<int,int>m; for(auto it:v){ m[it]++; } int ans=0; for(auto it:m){ if(it.second<2)return -1; ans+=((it.second+2)/3); } return ans; } }; Leetcode B

class Solution { public:     int minOperations(vector& nums, int k) {         reverse(nums.begin(),nums.end());                  int count = 0;         set s;         int f = 0;         for(auto val:nums){             if(val<=k){                 s.insert(val);             }             count++;             if(s.size()==k){                 f = 1;                 break;             }         }                  if(f==1){             return count;         }                  return -1;     } }; Leetcode A

Leetcode A class Solution { public: int minOperations(vector<int>& v) { map<int,int>m; for(auto it:v){ m[it]++; } int ans=0; for(auto it:m){ if(it.second<2)return -1; ans+=((it.second+2)/3); } return ans; } }; Leetcode B

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