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python✅

class Solution {   public:     int finalDestination(int N, int M, int K, vector> &Roads) {         // code here          vector>> adjList(N);         for (int i = 0; i < N; i++) {         adjList[i] = vector>();     }         for (vector& road : Roads) {         int cityA = road[0];         int cityB = road[1];         int weight = road[2];         adjList[cityA].push_back(make_pair(cityB, weight));     }         vector> dp(N, vector(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& 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 GFG✅

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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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