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class Solution{
    public:
    long long refueling(int X)
    {
        long long a, b;
        for(int i=0; i<X; i++){
            a = pow(2,i);
            if(a == X)return a;
            else if(a>X){
                b = pow(2,i-1);
                if(abs(X-a) == abs(X-b))return a;
                if(abs(X-a) < abs(X-b))return a;
                else return b;
            }
        }
        return -1;
    }
};

C++
Refueling

class Solution{
    public:
    string kPeriodic(string s, int K){
        // code here
        map<char,int> hm;
        for(int i=0;i<s.size();i++){
          hm[s[i]]++;
        }
        int n= s.size();
        int res= n/__gcd(n,K);
       
    for(auto t:hm){
        if(t.second % res)
            return "-1";
        }
       
        string p="";
       
        for(auto t:hm){
              int pre=t.second/res;
              while(pre--)
               p+=t.first;
          }
         
        string ans="";
        while(res--)
            ans+=p;
        
       return ans;

    }
};

C++
K-Periodic Circular String
Second
GFG Weekly Coding Contest 103

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class Solution {
  public:
    int finalDestination(int N, int M, int K, vector<vector<int>> &Roads) {
        // code here
         vector<vector<pair<int, int>>> adjList(N);
   
    for (int i = 0; i < N; i++) {
        adjList[i] = vector<pair<int, int>>();
    }
   
    for (vector<int>& road : Roads) {
        int cityA = road[0];
        int cityB = road[1];
        int weight = road[2];
        adjList[cityA].push_back(make_pair(cityB, weight));
    }
   
    vector<vector<int>> dp(N, vector<int>(K + 1, INT_MAX));
   
    dp[0][0] = 0;
   
    for (int k = 0; k <= K; k++) {
        for (int i = 0; i < N; i++) {
            if (dp[i][k] != INT_MAX) {
                for (pair<int, int>& neighbor : adjList[i]) {
                    int nextCity = neighbor.first;
                    int weight = neighbor.second;
                   
                    if (k + 1 <= K) {
                        dp[nextCity][k + 1] = min(dp[nextCity][k + 1], dp[i][k] + weight / 2);
                    }
                   
                    dp[nextCity][k] = min(dp[nextCity][k], dp[i][k] + weight);
                }
            }
        }
    }
   
    int minFuel = INT_MAX;
   
    for (int k = 0; k <= K; k++) {
        minFuel = min(minFuel, dp[N - 1][k]);
    }
   
    return minFuel == INT_MAX ? -1 : minFuel;
    }
};

Final destination Code C++
GFG
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