I tried to solve the question using topological sort using BFS .
Here’s the code

class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
        int n = graph.length;
        int m = graph[0].length;
        int V [][]= new int[n][m];

        List<List<Integer>> adj=new ArrayList<>();
        List<List<Integer>> adjRev = new ArrayList<>();
        for(int i=0;i<n;i++){
            for(int j=0;j<m;i++){
                adjRev.add(new ArrayList<>());
            }       
        }

        int indeg[] = new int[n];

        for(int i=0;i<n;i++){
            for(int j=0;j<m;i++){
                for(int it : adj.get(i)){
                adjRev.get(it).add(i);
                indeg[i]++;
            }
            }
        }

        Queue<Integer> q = new LinkedList<>();
        List<Integer> safeNodes = new ArrayList<>();
        for(int i=0;i<n ;i++){
            for(int j=0;j<m;i++){
                if(indeg[i]==0){
                q.add(i);
                }
            }
           
        }

        while(!q.isEmpty()){
            int node = q.peek();
            q.remove();
            safeNodes.add(node);
            for(int it : adjRev.get(node)){
                indeg[it]-- ;
                if(indeg[it] == 0) q.add(it);
            }
        }
        Collections.sort(safeNodes);
        return safeNodes;
    }
}

Earlier the code was facing index out of bound error .So I tried to fox it by taking a two dimensional Array named ‘V’ with ‘n’ number of rows and ‘m’ number of columns which had the count of adjacency matrix of the given graph.