this problem is minimal task Link can be solved by dp, i am using bfs or djikstra u can say , so i made a priority_queue and when first time when my string will be having the size of 2*n-1 means i had reached to end so this will be the answer lexicographically

see the code i am getting TLE no wrong answer i think worst complexity will be O(N^2) ,Please help

struct cmp {
    bool operator()(const pair<pair<int,int>,string> &x, const pair<pair<int,int>,string> &y) const {
        return x.second > y.second ;
    }
};
int check(int i,int j,int n){
    return i>=0 && j>=0 && i<n && j<n;
}
string ans;
vector<pair<int,int>>dir={{1,0},{0,1}};
string  bfs(int i,int j,vector<vector<char>>&vec,vector<vector<int>>&vis,int n){
    priority_queue<pair<pair<int,int>,string>,vector<pair<pair<int,int>,string>>,cmp>pq;
    string s;
    s+=vec[i][j];
    pq.push({{i,j},s});
    while(!pq.empty()){
        auto it=pq.top();pq.pop();
        string temp=it.second;
        if(it.second.size()==(n*2 - 1)){
            return it.second;
        }
        for(auto &i:dir){
            int new_x=it.first.first+i.first;
            int new_y=it.first.second+i.second;
            if(check(new_x,new_y,n) && vis[new_x][new_y]==-1){
                vis[new_x][new_y]=1;
                pq.push({{new_x,new_y},temp+vec[new_x][new_y]});
            }
        }
    }
}
void solve() {
    // executing code from here 
    int t=1;
    while (t--) {
        int n;
        cin >> n;
        vector<vector<char>> vec(n,vector<char>(n,'.'));
        vector<vector<int>> vis(n,vector<int>(n,-1));
        for (int i = 0; i < n; i++){
            for(int j=0;j<n;j++){
                cin>>vec[i][j];
            }
        }
        cout<<bfs(0,0,vec,vis,n)<<endl;
    }
}
int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    // sieve();
    solve();
    return 0;
}