We are given an array of integers where all elements are between 0-9. have to start from the 1st position and reach end in minimum no of moves such that we can from an index i move 1 position back and forward i.e i-1 and i+1 and jump to any index having the same value as index i. Time Limit : 1 second Max input size : 100000 I have tried to solve this problem use a single source shortest path approach using Breadth First Search and though BFS itself is O(V+E) and runs in time the adjacency list creation takes O(n2) time and therefore overall complexity becomes O(n2). is there any way i can decrease the time complexity of adjacency list creation? or is there a better and more efficient way of solving the problem? int main(){ vector<int> v; string str; vector<int> sets[10]; cin>>str; int in; for(int i=0;i<str.length();i++){ in=str[i]-'0'; v.push_back(in); sets[in].push_back(i); } int n=v.size(); if(n==1){ cout<<"0\n"; return 0; } if(v[0]==v[n-1]){ cout<<"1\n"; return 0; } vector<int> adj[100001]; for(int i=0;i<10;i++){ for(int j=0;j<sets[i].size();j++){ if(sets[i][j]>0) adj[sets[i][j]].push_back(sets[i][j]-1); if(sets[i][j]<n-1) adj[sets[i][j]].push_back(sets[i][j]+1); for(int k=j+1;k<sets[i].size();k++){ if(abs(sets[i][j]-sets[i][k])!=1){ adj[sets[i][j]].push_back(sets[i][k]); adj[sets[i][k]].push_back(sets[i][j]); } } } } queue<int> q; q.push(0); int dist[100001]; bool visited[100001]={false}; dist[0]=0; visited[0]=true; int c=0; while(!q.empty()){ int dq=q.front(); q.pop(); c++; for(int i=0;i<adj[dq].size();i++){ if(visited[adj[dq][i]]==false){ dist[adj[dq][i]]=dist[dq]+1; visited[adj[dq][i]]=true; q.push(adj[dq][i]); } } } cout<<dist[n-1]<<"\n"; return 0; }